209. Minimum Size Subarray Sum 🌐

Difficulty: Medium - Tags: Array, Two Pointers, Sliding Window

LeetCode Problem Link

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.

Examples 🌟

🔹 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

Explanation: No subarray sums up to or exceeds 11.

Constraints ⚙️

• 1 <= target <= 10^9

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^4

Solution 💡

To solve this problem, we can use the Sliding Window approach:

1. Initialize Pointers: Use two pointers left and right to represent the current window, and initialize sum to keep track of the current sum of the window.

2. Expand and Shrink the Window:

   • Expand the right pointer to include elements in the window.

   • When the sum of the window is greater than or equal to target, try to shrink the window from the left by moving the left pointer to minimize the subarray length.

3. Update the Result:

   • Keep track of the minimal length of such subarrays.

   • Return the minimal length found, or 0 if no valid subarray is found.

Java Solution

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

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

            while (sum >= target) {
                minLength = Math.min(minLength, right - left + 1);
                sum -= nums[left++];
            }
        }

        return minLength == Integer.MAX_VALUE ? 0 : minLength;
    }
}

Time Complexity ⏳

• O(n): Each element is added to the sum and removed from the sum at most once.

Space Complexity 💾

• O(1): No additional space is used except for the variables to store the current sum, pointers, and the result.

You can find the full solution here.