Minimum Size Subarray Sum LeetCode

 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


Optimal Solution:


Optimized Sliding Window Approach (O(N))

Instead of checking all subarrays, we use a sliding window:

  1. Expand the window (right pointer) until sum ≥ target.

  2. Shrink the window (left pointer) while sum ≥ target.

  3. Keep track of the minimum valid window size.

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int left = 0, sum = 0, minLength = Integer.MAX_VALUE;

        for (int right = 0; right < nums.length; right++) {
            sum += nums[right];  // Expand the window

            while (sum >= target) { // Shrink from left if condition met
                minLength = Math.min(minLength, right - left + 1); // Update min length
                sum -= nums[left];  // Remove the leftmost element
                left++;  // Shrink the window
            }
        }
        return (minLength == Integer.MAX_VALUE) ? 0 : minLength;
    }
}


Brute Force :

Issue -Time limit exceeded

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        int minLength = Integer.MAX_VALUE;

        for (int i = 0; i < n; i++) {
            int sum = 0; // Reset sum for new subarray

            for (int j = i; j < n; j++) {
                sum += nums[j];

                if (sum >= target) {
                    minLength = Math.min(minLength, j - i + 1); // Fix length calculation
                    break; // No need to continue checking larger subarrays
                }
            }
        }
        return (minLength == Integer.MAX_VALUE) ? 0 : minLength; // Handle case when no subarray is found
    }
}

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