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++){ cu
代码(Python3) class Solution: def checkSubarraySum(self, nums: List[int], k: int) -> bool: # remain_to_first_index 维护每一个前缀和 % k 第一次出现的下标 remain_to_first_index: Dict[int, int] = {} # 初始前缀和 0 % k 的下标为 -1 ,方便处理整个前缀和满足题意的情况 remain_to...
Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K. Example 1: Note: 给定一个集合,求这个集合中子集的个数,其中对子集的要求是子集中元素的和能被k整除。 记数组pre[i+1]表示前 i 位元素的和mo... ...
题目地址: https://leetcode.com/problems/minimum-size-subarray-sum/description/题目描述:Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead....
题目地址:https://leetcode.com/problems/binary-subarrays-with-sum/description/题目描述In an array A of 0s and 1s, how many non-empty subarrays have sum S?Example 1:Input: A = [1,0,1,0,1], S = 2 Output: 4 Explanation: The 4 subarrays are bolded below: [1,0,1,0,1] [1,...
Version 2,采用动态规划求解,首先定义状态,dp[i]是以nums[i]为结尾的连续子数组的最大和,状态转移方程为:如果dp[i-1]>0,无论nums[i]是什么数,以nums[i]为结尾的连续子数组的最大和都为dp[i-1]+nums[i],如果dp[i-1]<0,无论nums[i]是什么数,以nums[i]为结尾的连续子数组的最大和都为nums[i...
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: [4, 5, 0, -2, -3, 1]...
链接:https://leetcode-cn.com/problems/minimum-size-subarray-sum python # 长度最小的子数组 # 找出该数组中满足其和 ≥ target 的长度最小的 连续子数组[numsl, numsl+1, ..., numsr-1, numsr] ,并返回其长度。如果不存在符合条件的子数组,返回 0 。
Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead. Example: Input: s = 7, nums = [2,3,1,2,4,3] ...
9. So each of the 9 arrays will have 3 possible sums with equal probability. For an instance, [1 2 3], we can get 1+2, 2+3 and 1+2+3. And there are 27 outcome in total for this input, the expected value can be calculated by finding sum of all 27S and dividing it by 27...