In this article, we'll explore multiple approaches to solving the problem of finding a subarray with a given sum in Java. Problem Statement Given an array of integers and a target sum, find a continuous subarray
A in the input array, n is the length of the array & s in the given sum. Initialize vector b. (for storing indexes of subarray) Initialize a variable cur_sum=0 for i=0:n-1 set cur_sum=A[i] for j=i+1:n-1 add A[j] to cur_sum if(cur_sum==s) i is the starting index...
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... ...
Java Here, we are going to learn how to find the maximum subarray sum using Kadane's Algorithm in Java?Problem StatementGiven an array with size N, write a Java program to find the maximum subarray sum using Kadane's Algorithm.Example...
209. Minimum Size Subarray Sum java solutions Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array[2,3,1,2,4,3]ands = 7,...
代码运行次数:0 运行 AI代码解释 classSolution:defmaxSubArray(self,nums:List[int])->int:n=len(nums)dp=[0]*n dp[0]=nums[0]maximum=dp[0]foriinrange(1,n):dp[i]=max(dp[i-1]+nums[i],nums[i])maximum=max(maximum,dp[i])returnmaximum...
For example, given the array[2,3,1,2,4,3]ands = 7, the subarray[4,3]has the minimal length under the problem constraint. 解题思路: 用一个指针维护左边界即可,JAVA实现如下: 1 2 3 4 5 6 7 8 9 10 11 12 13 publicintminSubArrayLen(ints,int[] nums) { ...
Java class Solution { public int numSubarraysWithSum(int[] A, int S) { if (A.length == 0) return 0; int result = 0; int[] sumOne = new int[A.length]; int st = 0; int ed = 0; sumOne[0] = A[0] == 1 ? 1 : 0; ...
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]...
遍历map,如果有value>1的,则存在,返回true,否则返回false; Runtime:1 ms, faster than96.91%of Java online submissions for Find Subarrays With Equal Sum. Memory Usage:40.1 MB, less than86.76%of Java online submissions for Find Subarrays With Equal Sum....