LeetCode Challenge: 209. Minimum Size Subarray Sum - JavaScript Solution 🚀

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The Minimum Size Subarray Sum problem is an excellent example of sliding window and binary search techniques. Let’s tackle this problem and explore both O(n) and O(nlogn) solutions.

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🚀 Problem Description

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 no such subarray exists, return 0.

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💡 Examples

Example 1

Input: target = 7, nums = [2,3,1,2,4,3]  
Output: 2  
Explanation: The subarray `[4,3]` has a sum of 7, which is the minimal length.

Example 2

Input: target = 4, nums = [1,4,4]  
Output: 1  
Explanation: The subarray `[4]` has a sum of 4, which is the minimal length.

Example 3

Input: target = 11, nums = [1,1,1,1,1,1,1,1]  
Output: 0  
Explanation: No subarray meets the condition.

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🏆 JavaScript Solutions

Approach 1: Sliding Window (O(n))

The sliding window technique allows us to dynamically adjust the window size as we traverse the array.

Implementation

var minSubArrayLen = function(target, nums) {
    let left = 0;
    let sum = 0;
    let minLength = Infinity;

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

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

    return minLength === Infinity ? 0 : minLength;
};

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🔍 How It Works

1. Expand the Window:

   • Add elements to the sum by moving the right pointer.

2. Shrink the Window:

   • When the sum becomes greater than or equal to target, update the minLength and shrink the window by moving the left pointer.

3. Edge Case:

   • If no subarray meets the condition, return 0.

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Complexity Analysis

• Time Complexity: O(n), where n is the length of the array.
  • Each element is processed at most twice (once by right and once by left).
• Space Complexity: O(1), as we use constant extra space.

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Approach 2: Binary Search (O(nlogn))
An alternative approach uses binary search to find the minimal subarray for each possible starting point.

Implementation

var minSubArrayLen = function(target, nums) {
    const n = nums.length;
    const prefixSum = Array(n + 1).fill(0);

    for (let i = 1; i <= n; i++) {
        prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
    }

    let minLength = Infinity;

    for (let i = 0; i < n; i++) {
        const desiredSum = prefixSum[i] + target;
        let left = i + 1, right = n;

        while (left <= right) {
            const mid = Math.floor((left + right) / 2);
            if (prefixSum[mid] >= desiredSum) {
                minLength = Math.min(minLength, mid - i);
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
    }

    return minLength === Infinity ? 0 : minLength;
};

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🔍 How It Works

1. Prefix Sum:

   • Compute a prefix sum array where prefixSum[i] is the sum of the first i elements.

2. Binary Search:

   • For each starting index i, find the smallest ending index j such that prefixSum[j] - prefixSum[i] >= target.

3. Update Result:

   • Update minLength with the smallest valid subarray length.

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Complexity Analysis

• Time Complexity: O(nlogn), due to the binary search for each starting index.
• Space Complexity: O(n), for the prefix sum array.

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📋 Dry Run

Input: target = 7, nums = [2,3,1,2,4,3]

Sliding Window Steps:

Output: 2

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✨ Pro Tips for Interviews

1. Explain Sliding Window: Highlight its efficiency in adjusting window size dynamically.

2. Discuss Edge Cases:

   • nums with a single element equal to target.
   • Subarrays with all elements smaller than target.

3. Follow-Up: Mention how binary search is an alternative if asked for optimization.

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📚 Learn More

Check out the previous full explanation and code walkthrough on my Dev.to post:
👉 3Sum - JavaScript Solution

Which approach would you use? Let’s discuss below! 🚀

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Comments

[Rahul Kumar Barnwal]

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