Leetcode Minimum Size Subarray Sum problem solution

YASH PAL, 31 July 202419 January 2026

In this Leetcode Minimum Size Subarray Sum problem solution, we have given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, …, numsr-1, numsr] of which the sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Leetcode Minimum Size Subarray Sum problem solution

Leetcode Minimum Size Subarray Sum problem solution in Python.

class Solution(object):
    def minSubArrayLen(self, s, nums):
        res = float("inf")
        if(nums):
            i=0
            window,currWindowSum = 0,0
            while(i+window<len(nums)):
                currWindowSum+=nums[i+window]
                if(currWindowSum>=s):
                    while(currWindowSum>=s): 
                        res=min(window+1,res)
                        currWindowSum-=nums[i] 
                        i+=1
                        window-=1
                window+=1
        return 0 if(res==float("inf")) else res

Minimum Size Subarray Sum problem solution in Java.

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums.length == 0) return 0;
        int left = 0, right = 0, min = Integer.MAX_VALUE, sum = nums[0];
        while(right < nums.length){
            if(sum < s){
                right++;
                if(right == nums.length) break;
                sum += nums[right];
            }else{
                min = Math.min(min, right-left+1);
                left++;
                sum -= nums[left-1];
            }
        }
        return min == Integer.MAX_VALUE? 0 : min;
    }
}

Problem solution in C++.

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int left = 0, right = 0;
        int res = INT_MAX;
        int cnt = 0;
        while(right < nums.size()){
            cnt += nums[right];
            while(left <= right && cnt >= s){
                res = min(res, right-left+1);
                cnt -= nums[left];
                left++;
            }
            right++;
        }
        return res == INT_MAX ? 0 : res;
    }
};

Problem solution in C.

int minSubArrayLen(int s, int* nums, int numsSize) {
    int *slowRunner, *fastRunner;
    int sumSoFar;
    int shortest;
    int curLength;
    
    slowRunner = fastRunner = nums;
    sumSoFar = 0;
    shortest = numsSize;
    
    while (fastRunner < nums + numsSize) {
        sumSoFar += *fastRunner;
        fastRunner++;   
   
        while (sumSoFar >= s) {
            sumSoFar -= *(slowRunner++);
            curLength = fastRunner - slowRunner + 1;
            shortest = curLength < shortest ? curLength : shortest; 
        }
    }
    return fastRunner - slowRunner == numsSize ? 0 : shortest;
}

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