int getLongestSubarry(vector<int> &vec, int k){ int sum=0, res=0; map<int, int> sum_map; for(int idx=0; idx<vec.size(); ++idx){ sum += vec[idx]; if(sum<=k) res = idx+1; //此处修改 else { //前n项和大于sum-k并且需要最早的,此时子数组长度最长 map<int,int>::iterat...
2.Smallest Subarray with a given sum(easy) 2.1 问题描述 2.2 解决方法 2.3 代码 3.课后回顾 4.参考链接 sliding window pattern 1.原理描述 滑动窗口模式(sliding window pattern)是用于在给定数组或链表的特定窗口大小上执行所需的操作,比如寻找包含所有 1 的最长子数组。从第一个元素开始滑动窗口并逐个元素...
Finding a subarray with a given sum is a common problem that often appears in coding interviews and competitive programming. This problem can be solved using various techniques, each with its own trade-offs regarding time complexity and space complexity. In this article, we'll explore multiple ...
C++ implementation to find subarray with given sum#include <bits/stdc++.h> using namespace std; vector<int> find(vector<int> a,int n,int s){ vector<int> b; //output vector int cur_sum=0; for(int i=0;i<n;i++){ cur_sum=a[i];//cur_suming element for(int j=i+1;j<n;j...
There may be more than one subarrays with sum as the given sum, return the first such subarray. Solution 1. O(n^2) runtime: enumerate all possible subarrays and check if there is at least subarray that sums to the given sum.
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.Example 1:Input: A = [4,5,0,-2,-3,1], K = 5Output: 7Explanation: There are 7 subarrays with a sum divisible by K = 5:[4, 5, 0, -2, -3, 1], [5]...
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: ...
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Input: A=[4,5,0,-2,-3,1], K=5 Output:7 Explanation: There are7subarrays with a sum divisible by K=5: ...
sum[i..j] = sum[0..j] - sum[0..i-1]. We use a hashmap to check a previous matching index with a given number. classSolution {public: vector<int> subarraySum(vector<int>nums){ size_t len=nums.size(); unordered_map<long, vector<int>>hm; ...
Given a binary array arr[] and an integer target, return the number of non-empty subarrays with a sum equal to the target. Note: A subarray is the contiguous part of the array. Examples: Input: arr[] = [1, 0, 1, 0, 1], target = 2 ...