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1st approachclass Solution {public: vector> subsets(vector& nums) { int n = nums.size(); int tablesize = pow(2,n); vector> res; vector arr(tablesize); for (int i = 0; i  solclass Solution {public: vector> subsets(vector& nums) { int n = nums.size(); int tablesize = pow(2, n); // 2^n possible subsets vector> res(tablesize); // Initial..

class Solution {public: int findKthLargest(vector& nums, int k) { priority_queue, greater> minheap; for(int i =0; iminheap.top()){ minheap.pop(); minheap.push(nums[i]); } } return minheap.top(); }};  1.create a min heap using priority queue. first parameter int is identifying the type of the variables in the heap...

You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed.Return the number of consistent strings in the array words. Example 1:Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"] Output: 2 Explanation: Strings "aaab" and "baa" are consistent since they only conta..

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You must write an algorithm with O(log n) runtime complexity.Example 1:Input: nums = [1,3,5,6], target = 5 Output: 2Example 2:Input: nums = [1,3,5,6], target = 2 Output: 1Example 3:Input: nums = [1,3,5,6], target = 7 Output..

You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti].For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ).Return an array answer where answer[i] is the answer to the ith query.Example 1:Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]Output..