Memoization (1D, 2D and 3D)

Last Updated : 10 Aug, 2022

Most of the Dynamic Programming problems are solved in two ways:

Tabulation: Bottom Up
Memoization: Top Down

One of the easier approaches to solve most of the problems in DP is to write the recursive code at first and then write the Bottom-up Tabulation Method or Top-down Memoization of the recursive function. The steps to write the DP solution of Top-down approach to any problem is to:

Write the recursive code
Memoize the return value and use it to reduce recursive calls.

1-D Memoization
The first step will be to write the recursive code. In the program below, a program related to recursion where only one parameter changes its value has been shown. Since only one parameter is non-constant, this method is known as 1-D memoization. E.g., the Fibonacci series problem to find the N-th term in the Fibonacci series. The recursive approach has been discussed here.
Given below is the recursive code to find the N-th term:

C++

// C++ program to find the Nth term
// of Fibonacci series
#include <bits/stdc++.h>
using namespace std;
 
// Fibonacci Series using Recursion
int fib(int n)
{
 
    // Base case
    if (n <= 1)
        return n;
 
    // recursive calls
    return fib(n - 1) + fib(n - 2);
}
 
// Driver Code
int main()
{
    int n = 6;
    printf("%d", fib(n));
    return 0;
}

Java

// Java program to find the 
// Nth term of Fibonacci series
import java.io.*;
 
class GFG
{
     
// Fibonacci Series 
// using Recursion
static int fib(int n)
{
 
    // Base case
    if (n <= 1)
        return n;
 
    // recursive calls
    return fib(n - 1) +
           fib(n - 2);
}
 
// Driver Code
public static void main (String[] args) 
{
    int n = 6;
    System.out.println(fib(n));
}
}
 
// This code is contributed
// by ajit

Python3

# Python3 program to find the Nth term
# of Fibonacci series
 
# Fibonacci Series using Recursion
def fib(n):
 
 
    # Base case
    if (n <= 1):
        return n
 
    # recursive calls
    return fib(n - 1) + fib(n - 2)
 
# Driver Code
if __name__=='__main__':
    n = 6
    print (fib(n))
  
# This code is contributed by 
# Shivi_Aggarwal

C#

// C# program to find 
// the Nth term of 
// Fibonacci series
using System;
 
class GFG
{
     
// Fibonacci Series 
// using Recursion
static int fib(int n)
{
    // Base case
    if (n <= 1)
        return n;
 
    // recursive calls
    return fib(n - 1) +
           fib(n - 2);
}
// Driver Code
static public void Main ()
{
    int n = 6;
    Console.WriteLine(fib(n));
}
}
 
// This code is contributed
// by akt_mit

Javascript

<script>
// Javascript program to find the Nth term
// of Fibonacci series
 
// Fibonacci Series using Recursion
function fib(n)
{
 
    // Base case
    if (n <= 1)
        return n;
 
    // recursive calls
    return fib(n - 1) + fib(n - 2);
}
 
// Driver Code
let n = 6;
document.write(fib(n));
 
// This code is contributed by subhammahato348.
</script>

PHP

<?php
// PHP program to find 
// the Nth term of 
// Fibonacci series
// using Recursion
 
function fib($n)
{
 
    // Base case
    if ($n <= 1)
        return $n;
 
    // recursive calls
    return fib($n - 1) +
           fib($n - 2);
}
 
// Driver Code
$n = 6;
echo fib($n);
 
// This code is contributed 
// by ajit
?>

Output:

Time Complexity: O(2^n)

As at every stage we need to take 2 function calls and the height of the tree will be of the order of n.

Auxiliary Space: O(n)

The extra space is used due to recursion call stack.

A common observation is that this implementation does a lot of repeated work (see the following recursion tree). So this will consume a lot of time for finding the N-th Fibonacci number if done.

                            fib(5)   
                     /                 \        
               fib(4)                  fib(3)   
             /      \                /       \
         fib(3)      fib(2)         fib(2)    fib(1)
        /   \          /    \       /      \ 
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /    \ 
fib(1) fib(0) 

In the above tree fib(3), fib(2), fib(1), fib(0) all are called more than once.

The following problem has been solved using the Tabulation method.
In the program below, the steps to write a Top-Down approach program have been explained. Some modifications in the recursive program will reduce the complexity of the program and give the desired result. If fib(x) has not occurred previously, then we store the value of fib(x) in an array term at index x and return term[x]. By memoizing the return value of fib(x) at index x of an array, reduce the number of recursive calls at the next step when fib(x) has already been called. So without doing further recursive calls to compute the value of fib(x), return term[x] when fib(x) has already been computed previously to avoid a lot of repeated work as shown in the tree.
Given below is the memoized recursive code to find the N-th term.

C++

// CPP program to find the Nth term
// of Fibonacci series
#include <bits/stdc++.h>
using namespace std;
int term[1000];
// Fibonacci Series using memoized Recursion
int fib(int n)
{
    // base case
    if (n <= 1)
        return n;
 
    // if fib(n) has already been computed
    // we do not do further recursive calls
    // and hence reduce the number of repeated
    // work
    if (term[n] != 0)
        return term[n];
 
    else {
 
        // store the computed value of fib(n)
        // in an array term at index n to
        // so that it does not needs to be
        // precomputed again
        term[n] = fib(n - 1) + fib(n - 2);
 
        return term[n];
    }
}
 
// Driver Code
int main()
{
    int n = 6;
    printf("%d", fib(n));
    return 0;
}

Java

// Java program to find 
// the Nth term of 
// Fibonacci series
import java.io.*;
 
class GFG 
{
 
static  int []term = new int [1000]; 
// Fibonacci Series using
// memoized Recursion
static int fib(int n)
{
    // base case
    if (n <= 1)
        return n;
 
    // if fib(n) has already 
    // been computed we do not
    // do further recursive 
    // calls and hence reduce 
    // the number of repeated
    // work
    if (term[n] != 0)
        return term[n];
 
    else
    {
 
        // store the computed value
        // of fib(n) in an array 
        // term at index n to so that 
        // it does not needs to be 
        // precomputed again
        term[n] = fib(n - 1) + 
                  fib(n - 2);
 
        return term[n];
    }
}
 
// Driver Code
public static void main (String[] args) 
{
    int n = 6;
    System.out.println(fib(n));
}
}
 
// This code is contributed by ajit

Python3

# Python program to find the Nth term
# of Fibonacci series
term = [0 for i in range(1000)]
 
# Fibonacci Series using memoized Recursion
def fib(n):
   
  # base case
  if n <= 1:
    return n
 
  # if fib(n) has already been computed
  # we do not do further recursive calls
  # and hence reduce the number of repeated
  # work
  if term[n] != 0:
    return term[n]
 
  else:
     
    # store the computed value of fib(n)
    # in an array term at index n to
    # so that it does not needs to be
    # precomputed again
    term[n] = fib(n - 1) + fib(n - 2)
    return term[n]
 
# Driver Code
n = 6
print(fib(n))
 
# This code is contributed by rohitsingh07052

C#

// C# program to find 
// the Nth term of 
// Fibonacci series
 
using System; 
class GFG 
{
      
// Fibonacci Series using
// memoized Recursion
static int fib(int n)
{
    int[] term = new int [1000];
      
    // base case
    if (n <= 1)
        return n;
  
    // if fib(n) has already 
    // been computed we do not
    // do further recursive 
    // calls and hence reduce 
    // the number of repeated
    // work
    if (term[n] != 0)
        return term[n];
  
    else
    {
  
        // store the computed value
        // of fib(n) in an array 
        // term at index n to so that 
        // it does not needs to be 
        // precomputed again
        term[n] = fib(n - 1) + 
                  fib(n - 2);
  
        return term[n];
    }
}
  
// Driver Code
public static void Main () 
{
    int n = 6;
    Console.Write(fib(n));
}
}

Javascript

<script>
    // Javascript program to find
    // the Nth term of
    // Fibonacci series
     
    // Fibonacci Series using
    // memoized Recursion
    function fib(n)
    {
        let term = new Array(1000);
        term.fill(0);
 
        // base case
        if (n <= 1)
            return n;
 
        // if fib(n) has already
        // been computed we do not
        // do further recursive
        // calls and hence reduce
        // the number of repeated
        // work
        if (term[n] != 0)
            return term[n];
 
        else
        {
 
            // store the computed value
            // of fib(n) in an array
            // term at index n to so that
            // it does not needs to be
            // precomputed again
            term[n] = fib(n - 1) +
                      fib(n - 2);
 
            return term[n];
        }
    }
 
    let n = 6;
    document.write(fib(n));
     
    // This code is contributed by mukesh07.
</script>

Output:

Time Complexity: O(n)

As we have to calculate values for all function calls just once and there will be n values of arguments so the complexity will reduce to O(n).

Auxiliary Space: O(n)

The extra space is used due to recursion call stack.

If the recursive code has been written once, then memoization is just modifying the recursive program and storing the return values to avoid repetitive calls of functions that have been computed previously.

2-D Memoization
In the above program, the recursive function had only one argument whose value was not constant after every function call. Below, an implementation where the recursive program has two non-constant arguments has been shown.
For e.g., Program to solve the standard Dynamic Problem LCS problem when two strings are given. The general recursive solution of the problem is to generate all subsequences of both given sequences and find the longest matching subsequence. The total possible combinations will be 2ⁿ. Hence, the recursive solution will take O(2ⁿ). The approach to writing the recursive solution has been discussed here.
Given below is the recursive solution to the LCS problem:

C++

// A Naive recursive implementation of LCS problem
#include <bits/stdc++.h>
 
int max(int a, int b);
 
// Returns length of LCS for X[0..m-1], Y[0..n-1]
int lcs(char* X, char* Y, int m, int n)
{
    if (m == 0 || n == 0)
        return 0;
    if (X[m - 1] == Y[n - 1])
        return 1 + lcs(X, Y, m - 1, n - 1);
    else
        return max(lcs(X, Y, m, n - 1),
                  lcs(X, Y, m - 1, n));
}
 
// Utility function to get max of 2 integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
// Driver Code
int main()
{
    char X[] = "AGGTAB";
    char Y[] = "GXTXAYB";
 
    int m = strlen(X);
    int n = strlen(Y);
 
    printf("Length of LCS is %dn", lcs(X, Y, m, n));
 
    return 0;
}

Java

// A Naive recursive implementation of LCS problem
import java.io.*;
class GFG
{
 
    // Utility function to get max of 2 integers
    static int max(int a, int b) { return (a > b) ? a : b; }
 
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    static int lcs(String X, String Y, int m, int n)
    {
        if (m == 0 || n == 0)
            return 0;
        if (X.charAt(m - 1) == Y.charAt(n - 1))
            return 1 + lcs(X, Y, m - 1, n - 1);
        else
            return max(lcs(X, Y, m, n - 1),
                       lcs(X, Y, m - 1, n));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String X = "AGGTAB";
        String Y = "GXTXAYB";
 
        int m = X.length();
        int n = Y.length();
 
        System.out.print("Length of LCS is "
                         + lcs(X, Y, m, n));
    }
}
 
// This code is contributed by subhammahato348

Python3

# A Naive recursive implementation of LCS problem
 
# Returns length of LCS for X[0..m-1], Y[0..n-1]
def lcs(X, Y, m, n):
    if (m == 0 or n == 0):
        return 0;
    if (X[m - 1] == Y[n - 1]):
        return 1 + lcs(X, Y, m - 1, n - 1);
    else:
        return max(lcs(X, Y, m, n - 1),
                  lcs(X, Y, m - 1, n));
 
# Driver Code
if __name__=='__main__':
    X = "AGGTAB";
    Y = "GXTXAYB";
    m = len(X);
    n = len(Y);
    print("Length of LCS is {}n".format(lcs(X, Y, m, n)))
 
# This code is contributed by rutvik_56.

C#

// A Naive recursive implementation of LCS problem
using System;
class GFG
{
 
  // Utility function to get max of 2 integers
  static int max(int a, int b) { return (a > b) ? a : b; }
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1]
  static int lcs(string X, string Y, int m, int n)
  {
    if (m == 0 || n == 0)
      return 0;
    if (X[m - 1] == Y[n - 1])
      return 1 + lcs(X, Y, m - 1, n - 1);
    else
      return max(lcs(X, Y, m, n - 1),
                 lcs(X, Y, m - 1, n));
  }
 
  // Driver Code
  public static void Main()
  {
    string X = "AGGTAB";
    string Y = "GXTXAYB";
 
    int m = X.Length;
    int n = Y.Length;
 
    Console.Write("Length of LCS is "
                  + lcs(X, Y, m, n));
  }
}
 
// This code is contributed by subhammahato348

Javascript

<script>
// A Naive recursive implementation of LCS problem
 
    // Utility function to get max of 2 integers
    function max(a,b)
    {
         return (a > b) ? a : b;
    }
     
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    function lcs(X,Y,m,n)
    {
        if (m == 0 || n == 0)
            return 0;
        if (X[m-1] == Y[n-1])
            return 1 + lcs(X, Y, m - 1, n - 1);
        else
            return max(lcs(X, Y, m, n - 1),
                       lcs(X, Y, m - 1, n));
    }
     
    // Driver Code
    let X = "AGGTAB";
    let Y = "GXTXAYB";
    let m = X.length;
    let n = Y.length;
    document.write("Length of LCS is "
                         + lcs(X, Y, m, n));
     
     
 
// This code is contributed by avanitrachhadiya2155
</script>

Output:

Length of LCS is 4

Considering the above implementation, the following is a partial recursion tree for input strings “AXYT” and “AYZX”

                  lcs("AXYT", "AYZX")
                       /                 \
         lcs("AXY", "AYZX")            lcs("AXYT", "AYZ")
         /           \                   /               \
lcs("AX", "AYZX") lcs("AXY", "AYZ")   lcs("AXY", "AYZ") lcs("AXYT", "AY")

In the above partial recursion tree, lcs(“AXY”, “AYZ”) is being solved twice. On drawing the complete recursion tree, it has been observed that there are many subproblems that are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. The tabulation method has been discussed here.
A common point of observation to use memoization in the recursive code will be the two non-constant arguments M and N in every function call. The function has 4 arguments, but 2 arguments are constant which does not affect the Memoization. The repetitive calls occur for N and M which have been called previously. So use a 2-D array to store the computed lcs(m, n) value at arr[m-1][n-1] as the string index starts from 0. Whenever the function with the same argument m and n are called again, we do not perform any further recursive call and return arr[m-1][n-1] as the previous computation of the lcs(m, n) has already been stored in arr[m-1][n-1], hence reducing the recursive calls that happen more than once.
Below is the implementation of the Memoization approach of the recursive code.

C++

// C++ program to memoize
// recursive implementation of LCS problem
#include <bits/stdc++.h>
int arr[1000][1000];
int max(int a, int b);
 
// Returns length of LCS for X[0..m-1], Y[0..n-1] */
// memoization applied in recursive solution
int lcs(char* X, char* Y, int m, int n)
{
    // base case
    if (m == 0 || n == 0)
        return 0;
 
    // if the same state has already been
    // computed
    if (arr[m - 1][n - 1] != -1)
        return arr[m - 1][n - 1];
 
    // if equal, then we store the value of the
    // function call
    if (X[m - 1] == Y[n - 1]) {
 
        // store it in arr to avoid further repetitive 
        // work in future function calls
        arr[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1);
        return arr[m - 1][n - 1];
    }
    else {
        // store it in arr to avoid further repetitive 
        // work in future function calls
        arr[m - 1][n - 1] = max(lcs(X, Y, m, n - 1),
                                lcs(X, Y, m - 1, n));
        return arr[m - 1][n - 1];
    }
}
 
// Utility function to get max of 2 integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
// Driver Code
int main()
{
    memset(arr, -1, sizeof(arr));
    char X[] = "AGGTAB";
    char Y[] = "GXTXAYB";
 
    int m = strlen(X);
    int n = strlen(Y);
 
    printf("Length of LCS is %d", lcs(X, Y, m, n));
 
    return 0;
}

Java

// Java program to memoize
// recursive implementation of LCS problem
import java.io.*;
import java.lang.*;
class GFG 
{
  public static int arr[][] = new int[1000][1000];
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
  public static int lcs(String X, String Y, int m, int n)
  {
 
    // base case
    if (m == 0 || n == 0)
      return 0;
 
    // if the same state has already been
    // computed
    if (arr[m - 1][n - 1] != -1)
      return arr[m - 1][n - 1];
 
    // if equal, then we store the value of the
    // function call
    if ( X.charAt(m - 1) == Y.charAt(n - 1)) 
    {
 
      // store it in arr to avoid further repetitive 
      // work in future function calls
      arr[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1);
      return arr[m - 1][n - 1];
    }
    else
    {
 
      // store it in arr to avoid further repetitive 
      // work in future function calls
      arr[m - 1][n - 1] = max(lcs(X, Y, m, n - 1),
                              lcs(X, Y, m - 1, n));
      return arr[m - 1][n - 1];
    }
  }
 
  // Utility function to get max of 2 integers
  public static int max(int a, int b)
  {
    return (a > b) ? a : b;
  }
 
  // Driver code
  public static void main (String[] args) 
  {
    for(int i = 0; i < 1000; i++)
    {
      for(int j = 0; j < 1000; j++)
      {
        arr[i][j] = -1;
      }
    }
    String X = "AGGTAB";
    String Y = "GXTXAYB";
 
    int m = X.length();
    int n = Y.length();
 
    System.out.println("Length of LCS is " + lcs(X, Y, m, n));
  }
}
 
// This code is contributed by manupathria.

Python3

# Python3 program to memoize
# recursive implementation of LCS problem
 
# Returns length of LCS for X[0..m-1], Y[0..n-1]
# memoization applied in recursive solution
def lcs(X, Y, m, n):
 
    global arr
 
    # base case
    if (m == 0 or n == 0):
        return 0
 
    # if the same state has already been
    # computed
    if (arr[m - 1][n - 1] != -1):
        return arr[m - 1][n - 1]
 
    # if equal, then we store the value of the
    # function call
    if (X[m - 1] == Y[n - 1]):
 
        # store it in arr to avoid further repetitive
        # work in future function calls
        arr[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1)
        return arr[m - 1][n - 1]
 
    else:
 
        # store it in arr to avoid further repetitive
        # work in future function calls
        arr[m - 1][n - 1] = max(lcs(X, Y, m, n - 1),
                                lcs(X, Y, m - 1, n))
        return arr[m - 1][n - 1]
 
 
# Driver code
 
arr = [[0]*1000]*1000
 
for i in range(0, 1000):
    for j in range(0, 1000):
        arr[i][j] = -1
 
X = "AGGTAB"
Y = "GXTXAYB"
 
m = len(X)
n = len(Y)
 
print("Length of LCS is ", lcs(X, Y, m, n))
 
# This code is contributed by Dharanendra L V.

C#

// C# program to memoize
// recursive implementation of LCS problem
using System;
public class GFG
{
 
  public static int[, ] arr = new int[1000, 1000];
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
  public static int lcs(String X, String Y, int m, int n)
  {
 
    // base case
    if (m == 0 || n == 0)
      return 0;
 
    // if the same state has already been
    // computed
    if (arr[m - 1, n - 1] != -1)
      return arr[m - 1, n - 1];
 
    // if equal, then we store the value of the
    // function call
    if ( X[m - 1] == Y[n - 1]) 
    {
 
      // store it in arr to avoid further repetitive 
      // work in future function calls
      arr[m - 1, n - 1] = 1 + lcs(X, Y, m - 1, n - 1);
      return arr[m - 1, n - 1];
    }
    else
    {
 
      // store it in arr to avoid further repetitive 
      // work in future function calls
      arr[m - 1, n - 1] = max(lcs(X, Y, m, n - 1),
                              lcs(X, Y, m - 1, n));
      return arr[m - 1, n - 1];
    }
  }
 
  // Utility function to get max of 2 integers
  public static int max(int a, int b)
  {
    return (a > b) ? a : b;
  }
 
  // Driver code
  static public void Main (){
 
    for(int i = 0; i < 1000; i++)
    {
      for(int j = 0; j < 1000; j++)
      {
        arr[i, j] = -1;
      }
    }
    String X = "AGGTAB";
    String Y = "GXTXAYB";
 
    int m = X.Length;
    int n = Y.Length;
 
    Console.WriteLine("Length of LCS is " + lcs(X, Y, m, n));
  }
}
 
// This code is contributed by Dharanendra L V.

Javascript

<script>
// Javascript program to memoize
// recursive implementation of LCS problem
     
    let arr=new Array(1000);
    for(let i=0;i<1000;i++)
    {
        arr[i]=new Array(1000);
        for(let j=0;j<1000;j++)
        {
            arr[i][j]=-1;
        }
    }
     
    // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
    function lcs(X,Y,m,n)
    {
        // base case
    if (m == 0 || n == 0)
      return 0;
  
    // if the same state has already been
    // computed
    if (arr[m - 1][n - 1] != -1)
        return arr[m - 1][n - 1];
  
    // if equal, then we store the value of the
    // function call
    if ( X[m-1] == Y[n-1])
    {
  
      // store it in arr to avoid further repetitive
      // work in future function calls
      arr[m - 1][n - 1] = 1 + lcs(X, Y, m - 1, n - 1);
      return arr[m - 1][n - 1];
    }
    else
    {
  
      // store it in arr to avoid further repetitive
      // work in future function calls
      arr[m - 1][n - 1] = Math.max(lcs(X, Y, m, n - 1),
                              lcs(X, Y, m - 1, n));
      return arr[m - 1][n - 1];
    }
    }
     
     
     
    // Driver code
    let X = "AGGTAB";
    let Y = "GXTXAYB";
    let m = X.length;
    let n = Y.length;
    document.write("Length of LCS is " + lcs(X, Y, m, n));
 
     
     
// This code is contributed by rag2127
</script>

Output:

Length of LCS is 4

3-D Memoization

In the above program, the recursive function had only two arguments whose values were not constant after every function call. Below, an implementation where the recursive program has three non-constant arguments is done.
For e.g., Program to solve the standard Dynamic Problem LCS problem for three strings. The general recursive solution of the problem is to generate all subsequences of both given sequences and find the longest matching subsequence. The total possible combinations will be 3ⁿ. Hence, a recursive solution will take O(3ⁿ).
Given below is the recursive solution to the LCS problem:

C++

// A recursive implementation of LCS problem
// of three strings
#include <bits/stdc++.h>
int max(int a, int b);
 
// Returns length of LCS for X[0..m-1], Y[0..n-1]
int lcs(char* X, char* Y, char* Z, int m, int n, int o)
{
    // base case
    if (m == 0 || n == 0 || o == 0)
        return 0;
 
    // if equal, then check for next combination
    if (X[m - 1] == Y[n - 1] and Y[n - 1] == Z[o - 1]) {
 
        // recursive call
        return 1 + lcs(X, Y, Z, m - 1, n - 1, o - 1);
    }
    else {
 
        // return the maximum of the three other
        // possible states in recursion
        return max(lcs(X, Y, Z, m, n - 1, o),
                   max(lcs(X, Y, Z, m - 1, n, o),
                       lcs(X, Y, Z, m, n, o - 1)));
    }
}
 
// Utility function to get max of 2 integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
// Driver Code
int main()
{
 
    char X[] = "geeks";
    char Y[] = "geeksfor";
    char Z[] = "geeksforge";
    int m = strlen(X);
    int n = strlen(Y);
    int o = strlen(Z);
    printf("Length of LCS is %d", lcs(X, Y, Z, m, n, o));
 
    return 0;
}

Java

// A recursive implementation of LCS problem
// of three strings 
class GFG
{
  // Utility function to get max of 2 integers
  static int max(int a, int b)
  {
    return (a > b) ? a : b;
  }
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1]
  static int lcs(char[] X, char[] Y, char[] Z, 
                 int m, int n, int o)
  {
 
    // base case
    if (m == 0 || n == 0 || o == 0)
      return 0;
 
    // if equal, then check for next combination
    if (X[m - 1] == Y[n - 1] && Y[n - 1] == Z[o - 1]) 
    {
 
      // recursive call
      return 1 + lcs(X, Y, Z, m - 1, n - 1, o - 1);
    }
    else
    {
 
      // return the maximum of the three other
      // possible states in recursion
      return Math.max(lcs(X, Y, Z, m, n - 1, o),
                      Math.max(lcs(X, Y, Z, m - 1, n, o),
                               lcs(X, Y, Z, m, n, o - 1)));
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    char[] X = "geeks".toCharArray();
    char[] Y = "geeksfor".toCharArray();
    char[] Z = "geeksforge".toCharArray();
    int m = X.length;
    int n = Y.length;
    int o = Z.length;
    System.out.println("Length of LCS is " + lcs(X, Y, Z, m, n, o));
  }
}
 
// This code is contributed by divyesh072019.

Python3

# A recursive implementation of LCS problem of three strings
 
# Returns length of LCS for X[0..m-1], Y[0..n-1]
def lcs(X, Y, Z, m, n, o):
    # base case
    if m == 0 or n == 0 or o == 0:
      return 5
     
    # if equal, then check for next combination
    if X[m - 1] == Y[n - 1] and Y[n - 1] == Z[o - 1]:
      # recursive call
      return 1 + lcs(X, Y, Z, m - 1, n - 1, o - 1)
    else:
      # return the maximum of the three other
      # possible states in recursion
      return max(lcs(X, Y, Z, m, n - 1, o), max(lcs(X, Y, Z, m - 1, n, o), lcs(X, Y, Z, m, n, o - 1)))
 
X = "geeks".split()
Y = "geeksfor".split()
Z = "geeksforge".split()
m = len(X)
n = len(Y)
o = len(Z)
print("Length of LCS is", lcs(X, Y, Z, m, n, o))
 
# This code is contributed by rameshtravel07.

C#

// A recursive implementation of LCS problem
// of three strings
using System;
class GFG {
     
    // Utility function to get max of 2 integers
    static int max(int a, int b)
    {
        return (a > b) ? a : b;
    }
 
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    static int lcs(char[] X, char[] Y, char[] Z, int m, int n, int o)
    {
       
        // base case
        if (m == 0 || n == 0 || o == 0)
            return 0;
      
        // if equal, then check for next combination
        if (X[m - 1] == Y[n - 1] && Y[n - 1] == Z[o - 1]) 
        {
      
            // recursive call
            return 1 + lcs(X, Y, Z, m - 1, n - 1, o - 1);
        }
        else
        {
      
            // return the maximum of the three other
            // possible states in recursion
            return Math.Max(lcs(X, Y, Z, m, n - 1, o),
                       Math.Max(lcs(X, Y, Z, m - 1, n, o),
                           lcs(X, Y, Z, m, n, o - 1)));
        }
    }
 
  // Driver code
  static void Main()
  {
    char[] X = "geeks".ToCharArray();
    char[] Y = "geeksfor".ToCharArray();
    char[] Z = "geeksforge".ToCharArray();
    int m = X.Length;
    int n = Y.Length;
    int o = Z.Length;
    Console.WriteLine("Length of LCS is " + lcs(X, Y, Z, m, n, o));
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript

<script>
 
// A recursive implementation of LCS problem
// of three strings
     
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    function lcs(X,Y,Z,m,n,o)
    {
        // base case
    if (m == 0 || n == 0 || o == 0)
      return 0;
  
    // if equal, then check for next combination
    if (X[m - 1] == Y[n - 1] && Y[n - 1] == Z[o - 1])
    {
  
      // recursive call
      return 1 + lcs(X, Y, Z, m - 1, n - 1, o - 1);
    }
    else
    {
  
      // return the maximum of the three other
      // possible states in recursion
      return Math.max(lcs(X, Y, Z, m, n - 1, o),
                      Math.max(lcs(X, Y, Z, m - 1, n, o),
                               lcs(X, Y, Z, m, n, o - 1)));
    }
    }
     
    // Driver code
    let X = "geeks".split("");
    let Y = "geeksfor".split("");
    let Z = "geeksforge".split("");
    let m = X.length;
    let n = Y.length;
    let o = Z.length;
    document.write(
    "Length of LCS is " + lcs(X, Y, Z, m, n, o)
    );
     
 
// This code is contributed by unknown2108
 
</script>

Output:

Length of LCS is 5

The tabulation method has been shown here. On drawing the recursion tree completely, it has been noticed that there are many overlapping sub-problems which are been calculated multiple times. Since the function parameter has three non-constant parameters, hence a 3-D array will be used to memorize the value that was returned when lcs(x, y, z, m, n, o) for any value of m, n, and o was called so that if lcs(x, y, z, m, n, o) is again called for the same value of m, n, and o then the function will return the already stored value as it has been computed previously in the recursive call. arr[m][n][o] stores the value returned by the lcs(x, y, z, m, n, o) function call. The only modification that needs to be done in the recursive program is to store the return value of (m, n, o) state of the recursive function. The rest remains the same in the above recursive program.
Below is the implementation of the Memoization approach of the recursive code:

C++

// A memoize recursive implementation of LCS problem
#include <bits/stdc++.h>
int arr[100][100][100];
int max(int a, int b);
 
// Returns length of LCS for X[0..m-1], Y[0..n-1] */
// memoization applied in recursive solution
int lcs(char* X, char* Y, char* Z, int m, int n, int o)
{
    // base case
    if (m == 0 || n == 0 || o == 0)
        return 0;
 
    // if the same state has already been
    // computed
    if (arr[m - 1][n - 1][o - 1] != -1)
        return arr[m - 1][n - 1][o - 1];
 
    // if equal, then we store the value of the
    // function call
    if (X[m - 1] == Y[n - 1] and Y[n - 1] == Z[o - 1]) {
 
        // store it in arr to avoid further repetitive work
        // in future function calls
        arr[m - 1][n - 1][o - 1] = 1 + lcs(X, Y, Z, m - 1,
                                            n - 1, o - 1);
        return arr[m - 1][n - 1][o - 1];
    }
    else {
 
        // store it in arr to avoid further repetitive work
        // in future function calls
        arr[m - 1][n - 1][o - 1] =
                               max(lcs(X, Y, Z, m, n - 1, o),
                                 max(lcs(X, Y, Z, m - 1, n, o),
                                    lcs(X, Y, Z, m, n, o - 1)));
        return arr[m - 1][n - 1][o - 1];
    }
}
 
// Utility function to get max of 2 integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}
 
// Driver Code
int main()
{
    memset(arr, -1, sizeof(arr));
    char X[] = "geeks";
    char Y[] = "geeksfor";
    char Z[] = "geeksforgeeks";
    int m = strlen(X);
    int n = strlen(Y);
    int o = strlen(Z);
    printf("Length of LCS is %d", lcs(X, Y, Z, m, n, o));
 
    return 0;
}

Java

// A memoize recursive implementation of LCS problem
import java.io.*;
class GFG
{
   
  public static int[][][] arr = new int[100][100][100];
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
  static int lcs(String X, String Y, String Z,
                 int m, int n, int o)
  {
     
      // base case
      if (m == 0 || n == 0 || o == 0)
          return 0;
 
      // if the same state has already been
      // computed
      if (arr[m - 1][n - 1][o - 1] != -1)
          return arr[m - 1][n - 1][o - 1];
 
      // if equal, then we store the value of the
      // function call
      if (X.charAt(m - 1) == Y.charAt(n - 1) && 
          Y.charAt(n - 1) == Z.charAt(o - 1)) {
 
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1][n - 1][o - 1] = 1 + lcs(X, Y, Z, m - 1,
                                              n - 1, o - 1);
          return arr[m - 1][n - 1][o - 1];
      }
      else
      {
 
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1][n - 1][o - 1] =
                                 max(lcs(X, Y, Z, m, n - 1, o),
                                   max(lcs(X, Y, Z, m - 1, n, o),
                                      lcs(X, Y, Z, m, n, o - 1)));
          return arr[m - 1][n - 1][o - 1];
      }
  }
 
  // Utility function to get max of 2 integers
  static int max(int a, int b)
  {
      return (a > b) ? a : b;
  }
 
  // Driver Code
  public static void main (String[] args)
  {  
    for(int i = 0; i < 100; i++) 
    {
      for(int j = 0; j < 100; j++) 
      {
        for(int k = 0; k < 100; k++)
        {
          arr[i][j][k] = -1;
        }
      }
    }
     
    String X = "geeks";
    String Y = "geeksfor";
    String Z = "geeksforgeeks";
    int m = X.length();
    int n = Y.length();
    int o = Z.length();
    System.out.print("Length of LCS is " + lcs(X, Y, Z, m, n, o));
  }
}
 
// This code is contributed by Dharanendra L V.

Python3

# A memoize recursive implementation of LCS problem
 
# Returns length of LCS for X[0..m-1], Y[0..n-1] */
# memoization applied in recursive solution
def lcs(X, Y, Z, m, n, o):
    global arr
 
    # base case
    if(m == 0 or n == 0 or o == 0):
        return 0
 
    # if the same state has already been
    # computed
    if (arr[m - 1][n - 1][o - 1] != -1):
        return arr[m - 1][n - 1][o - 1]
 
    # if equal, then we store the value of the
    # function call
    if (X[m - 1] == Y[n - 1] and
            Y[n - 1] == Z[o - 1]):
 
        # store it in arr to avoid further repetitive work
        # in future function calls
        arr[m - 1][n - 1][o - 1] = 1 + lcs(X, Y, Z, m - 1,
                                           n - 1, o - 1)
        return arr[m - 1][n - 1][o - 1]
 
    else:
 
        # store it in arr to avoid further repetitive work
        # in future function calls
        arr[m - 1][n - 1][o - 1] = max(lcs(X, Y, Z, m, n - 1, o),
                                       max(lcs(X, Y, Z, m - 1, n, o), lcs(X, Y, Z, m, n, o - 1)))
        return arr[m - 1][n - 1][o - 1]
 
# Driver Code
arr = [[[0 for k in range(100)] for j in range(100)] for i in range(100)]
 
for i in range(100):
    for j in range(100):
        for k in range(100):
            arr[i][j][k] = -1
 
X = "geeks"
Y = "geeksfor"
Z = "geeksforgeeks"
m = len(X)
n = len(Y)
o = len(Z)
print("Length of LCS is ", lcs(X, Y, Z, m, n, o))
 
# This code is contributed by Dharanendra L V.

C#

// A memoize recursive implementation of LCS problem
using System;
public class GFG{
 
  public static int[, , ] arr = new int[100, 100, 100];
 
  // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
  static int lcs(String X, String Y, String Z, int m, int n, int o)
  {
      // base case
      if (m == 0 || n == 0 || o == 0)
          return 0;
 
      // if the same state has already been
      // computed
      if (arr[m - 1, n - 1, o - 1] != -1)
          return arr[m - 1, n - 1, o - 1];
 
      // if equal, then we store the value of the
      // function call
      if (X[m - 1] == Y[n - 1] && 
          Y[n - 1] == Z[o - 1]) {
 
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1, n - 1, o - 1] = 1 + lcs(X, Y, Z, m - 1,
                                              n - 1, o - 1);
          return arr[m - 1, n - 1, o - 1];
      }
      else {
 
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1, n - 1, o - 1] =
                                 max(lcs(X, Y, Z, m, n - 1, o),
                                   max(lcs(X, Y, Z, m - 1, n, o),
                                      lcs(X, Y, Z, m, n, o - 1)));
          return arr[m - 1, n - 1, o - 1];
      }
  }
 
  // Utility function to get max of 2 integers
  static int max(int a, int b)
  {
      return (a > b) ? a : b;
  }
 
  // Driver Code
  static public void Main (){
 
    for(int i = 0; i < 100; i++) {
      for(int j = 0; j < 100; j++) {
        for(int k = 0; k < 100; k++) {
          arr[i, j, k] = -1;
        }
      }
    }
     
    String X = "geeks";
    String Y = "geeksfor";
    String Z = "geeksforgeeks";
    int m = X.Length;
    int n = Y.Length;
    int o = Z.Length;
    Console.WriteLine("Length of LCS is " + lcs(X, Y, Z, m, n, o));
  }
}
 
// This code is contributed by Dharanendra L V.

Javascript

<script>
// A memoize recursive implementation of LCS problem
     
    let arr=new Array(100);
    for(let i=0;i<100;i++)
    {
        arr[i]=new Array(100);
        for(let j=0;j<100;j++)
        {
            arr[i][j]=new Array(100);
            for(let k=0;k<100;k++)
            {
                arr[i][j][k]=-1;
            }
        }
    }
     
    // Returns length of LCS for X[0..m-1], Y[0..n-1] */
  // memoization applied in recursive solution
    function lcs(X,Y,Z,m,n,o)
    {
        // base case
      if (m == 0 || n == 0 || o == 0)
          return 0;
  
      // if the same state has already been
      // computed
      if (arr[m - 1][n - 1][o - 1] != -1)
          return arr[m - 1][n - 1][o - 1];
  
      // if equal, then we store the value of the
      // function call
      if (X[m-1] == Y[n-1] &&
          Y[n-1] == Z[o-1]) {
  
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1][n - 1][o - 1] = 1 + lcs(X, Y, Z, m - 1,
                                              n - 1, o - 1);
          return arr[m - 1][n - 1][o - 1];
      }
      else
      {
  
          // store it in arr to avoid further repetitive work
          // in future function calls
          arr[m - 1][n - 1][o - 1] =
                                 max(lcs(X, Y, Z, m, n - 1, o),
                                   max(lcs(X, Y, Z, m - 1, n, o),
                                      lcs(X, Y, Z, m, n, o - 1)));
          return arr[m - 1][n - 1][o - 1];
      }
    }
     
    // Utility function to get max of 2 integers
    function max(a,b)
    {
         return (a > b) ? a : b;
    }
     
    // Driver Code
    let X = "geeks";
    let Y = "geeksfor";
    let Z = "geeksforgeeks";
    let m = X.length;
    let n = Y.length;
    let o = Z.length;
    document.write("Length of LCS is " + lcs(X, Y, Z, m, n, o));
     
     
 
 
// This code is contributed by patel2127
</script>

Output:

Length of LCS is 5

Note: The array used to Memoize is initialized to some value (say -1) before the function call to mark if the function with the same parameters has been previously called or not.

Striver

Improve

Emulating a 2-d array using 1-d array

Memoization (1D, 2D and 3D)

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