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638. Shopping Offers
In LeetCode Store, there are n items to sell. Each item has a price. However, there are some special
offers, and a special offer consists of one or more different kinds of items with a sale price.
You are given an integer array price where price[i] is the price of the ith item, and an integer
array needs where needs[i] is the number of pieces of the ith item you want to buy.
You are also given an array special where special[i] is of size n + 1 where special[i][j] is the number of
pieces of the jth item in the ith offer and special[i][n] (i.e., the last integer in the array) is the price of
the ith offer.
Return the lowest price you have to pay for exactly certain items as given, where you could make
optimal use of the special offers. You are not allowed to buy more items than you want, even if that
would lower the overall price. You could use any of the special offers as many times as you want.
Example 1:
Input: price = [2,5], special = [[3,0,5],[1,2,10]], needs = [3,2]
Output: 14
Explanation: There are two kinds of items, A and B. Their prices are $2 and $5 respectively.
In special offer 1, you can pay $5 for 3A and 0B
In special offer 2, you can pay $10 for 1A and 2B.
You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
Example 2:
Input: price = [2,3,4], special = [[1,1,0,4],[2,2,1,9]], needs = [1,2,1]
Output: 11
Explanation: The price of A is $2, and $3 for B, $4 for C.
You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C.
You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4
for 1C.
You cannot add more items, though only $9 for 2A ,2B and 1C.
Constraints:
n == price.length == needs.length
1 <= n <= 6
0 <= price[i], needs[i] <= 10
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1 <= special.length <= 100
special[i].length == n + 1
0 <= special[i][j] <= 50
The input is generated that at least one of special[i][j] is non-zero for 0 <= j <= n - 1.
CODE:
class Solution {
int res;
public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
res = Integer.MAX_VALUE;
backtrack(price, special, needs, 0, 0);
return res;
private void backtrack(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int min,
int pos) {
int totalPrice = 0;
for (int i = 0; i < needs.size(); i++) {
totalPrice += needs.get(i) * price.get(i);
res = Math.min(res, min + totalPrice);
for (int i = pos; i < special.size(); i++) {
boolean valid = true;
List<Integer> offer = special.get(i);
for (int j = 0; j < needs.size(); j++) {
if (needs.get(j) < offer.get(j)) {
valid = false;
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break;
if (valid) {
List<Integer> dummy = new ArrayList<>(needs);
for (int j = 0; j < dummy.size(); j++) {
dummy.set(j, dummy.get(j) - offer.get(j));
backtrack(price, special, dummy, min + offer.get(offer.size()-1), i);
99. Recover Binary Search Tree
You are given the root of a binary search tree (BST), where the values of exactly two nodes of the
tree were swapped by mistake. Recover the tree without changing its structure.
Example 1:
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Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
Example 2:
Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST
valid.
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Constraints:
The number of nodes in the tree is in the range [2, 1000].
-231 <= Node.val <= 231 – 1
CODE:
class Solution {
TreeNode prev=null,first=null,second=null;
void inorder(TreeNode root){
if(root==null)
return ;
inorder(root.left);
if(prev!=null&&root.val<prev.val){
if(first==null)
first=prev;
second=root;
prev=root;
inorder(root.right);
public void recoverTree(TreeNode root) {
if(root==null)
return ;
inorder(root);
int temp=first.val;
first.val=second.val;
second.val=temp;
}
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95. Unique Binary Search Trees II
Given an integer n, return all the structurally unique BST's (binary search trees), which has
exactly n nodes of unique values from 1 to n. Return the answer in any order.
Example 1:
Input: n = 3
Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]
Example 2:
Input: n = 1
Output: [[1]]
Constraints:
1 <= n <= 8
CODE:
class Solution {
public List<TreeNode> generateTrees(int n) {
if (n == 0) {
return new ArrayList<>();
Map<String, List<TreeNode>> memo = new HashMap<>();
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return generateTreesHelper(1, n, memo);
private List<TreeNode> generateTreesHelper(int start, int end, Map<String, List<TreeNode>>
memo) {
String key = start + "-" + end;
if (memo.containsKey(key)) {
return memo.get(key);
List<TreeNode> trees = new ArrayList<>();
if (start > end) {
trees.add(null);
return trees;
for (int rootVal = start; rootVal <= end; rootVal++) {
List<TreeNode> leftTrees = generateTreesHelper(start, rootVal - 1, memo);
List<TreeNode> rightTrees = generateTreesHelper(rootVal + 1, end, memo);
for (TreeNode leftTree : leftTrees) {
for (TreeNode rightTree : rightTrees) {
TreeNode root = new TreeNode(rootVal);
root.left = leftTree;
root.right = rightTree;
trees.add(root);
}
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memo.put(key, trees);
return trees;
