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2.3 Daa

The document describes an algorithm to solve the 0-1 knapsack problem using dynamic programming. It involves creating a 2D array dp to store the maximum values for each item and capacity. Two nested loops iterate through items and capacities, considering subproblems. The base cases and recursive formulas to fill the dp array are provided.

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42 views2 pages

2.3 Daa

The document describes an algorithm to solve the 0-1 knapsack problem using dynamic programming. It involves creating a 2D array dp to store the maximum values for each item and capacity. Two nested loops iterate through items and capacities, considering subproblems. The base cases and recursive formulas to fill the dp array are provided.

Uploaded by

Shubhankar Kr
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Course Name: DAA Lab Course Code: 21ITH-311/21CSH-311

Experiment 2.3

Aim: Develop a program and analyze complexity to implement 0-1 Knapsack using dynamic
programming.

Objectives: to implement 0-1 Knapsack using Dynamic Programming

Input/Apparatus Used: Online GDB compiler.

Algorithm:

1. The method knapsack takes three arguments: an array of values, an array of weights, and the capacity

of the knapsack. It returns the maximum value that can be obtained.

2. n represents the number of items, which is determined by the length of the values array.

3. An array dp is created to store the maximum value that can be obtained at each capacity for different

numbers of items.

4. The nested loops iterate through each item and each possible capacity, considering all subproblems from

0 to the current number of items and capacity.

5. The base case is handled first. When there are no items (i = 0) or the knapsack capacity is 0 (w = 0), the

maximum value that can be obtained is 0.

6. If the current item's weight is less than or equal to the current capacity, the maximum value is either the

maximum of the value excluding the current item or the value including the current item and the

maximum value obtained from the remaining capacity.

7. If the current item's weight is greater than the current capacity, the maximum value is the same as the

maximum value obtained without considering the current item.

Name: Harsh Gupta UID: 21BCS10721


Course Name: DAA Lab Course Code: 21ITH-311/21CSH-311
Sample Code:
public class knapsack {
public static int knapsack(int[] values, int[] weights, int capacity) {
int n = values.length;
int[][] dp = new int[n + 1][capacity + 1];

for (int i = 0; i <= n; i++) {


for (int w = 0; w <= capacity; w++) {
if (i == 0 || w == 0) {
dp[i][w] = 0;
} else if (weights[i - 1] <= w) {
dp[i][w] = Math.max(dp[i - 1][w], values[i - 1] + dp[i - 1][w - weights[i - 1]]);
} else {
dp[i][w] = dp[i - 1][w];
}
}
}

return dp[n][capacity];
}

public static void main(String[] args) {


int[] values = {60, 100, 120};
int[] weights = {10, 20, 30};
int capacity = 50;

int maxValue = knapsack(values, weights, capacity);


System.out.println("Maximum value that can be obtained: " + maxValue);
}
}
Output:

Time Complexity:
O(n*capacity)

Name: Harsh Gupta UID: 21BCS10721

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