# Minimum Size Subarray Sum LeetCode Solution

Here, We see Minimum Size Subarray Sum LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given an array of positive integers `nums` and a positive integer `target`, return the minimal length of a

subarray whose sum is greater than or equal to`target`. If there is no such subarray, return `0` instead.

Example 1:
Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:
Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:
Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

## Minimum Size Subarray Sum LeetCode SolutionC++

``````class Solution {
public:
int minSubArrayLen(int target, vector<int>& nums) {
int l = 0, r = 0, n = nums.size(), sum = 0, len = INT_MAX;
while (r < n) {
sum += nums[r++];
while (sum >= target) {
len = min(len, r - l);
sum -= nums[l++];
}
}
return len == INT_MAX ? 0 : len;
}
};```Code language: PHP (php)```

## Minimum Size Subarray Sum LeetCode SolutionJava

``````class Solution {
public int minSubArrayLen(int target, int[] nums) {
if (nums == null || nums.length == 0)
return 0;
int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
while (j < nums.length) {
sum += nums[j++];
while (sum >= target) {
min = Math.min(min, j - i);
sum -= nums[i++];
}
}
return min == Integer.MAX_VALUE ? 0 : min;
}
}```Code language: JavaScript (javascript)```

## Minimum Size Subarray Sum SolutionJavaScript

``````var minSubArrayLen = function(target, nums) {
let left = 0;
let right = 0;
let sum = 0;
let result = 0;
while(right < nums.length) {
sum += nums[right];
while(sum >= target) {
let len = right - left +1;
if(result === 0 || len < result) result = len;
sum -= nums[left];
left++
}
right++;
}
return result;
};```Code language: JavaScript (javascript)```

## Minimum Size Subarray Sum SolutionPython

``````class Solution(object):
def minSubArrayLen(self, target, nums):
total = left = right = 0
res = len(nums) + 1
while right < len(nums):
total += nums[right]
while total >= target:
res = min(res, right-left+1)
total -= nums[left]
left += 1
right += 1
return res if res <= len(nums) else 0```Code language: HTML, XML (xml)```
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