Minimum Size Subarray Sum

The Minimum Size Subarray Sum problem is a great example of using the sliding window technique to solve a real-world scenario - finding the smallest set of contiguous actions or values that meet a threshold. It helps build your skills with optimizing time and space over brute-force solutions and comes up frequently in interviews at all levels.

Problem Statement

Given an array of positive integers and an integer target, return the minimal length of a contiguous subarray of which the sum is greater than or equal to the 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

1. Clarify Requirements Before Jumping Into Code

Let's make sure we understand the requirements:

• We need to return the length of the subarray.

• Subarrays must be contiguous.

• If no such subarray exists, we should return 0

• Only positive integers are in the array (important for sliding window guarantees).

2. Identify the Category of the Question

This is a classic sliding window problem, and a very common one for learning how to minimize the size of a valid window that meets some condition.

3. Consider a Brute‑Force Approach to Understand the Problem

The brute-force solution here, as it so often is, would be to try all possible subarrays and track the minimum length of those that meet the condition.

For each starting index i, compute all subarrays ending at j >= i and sum them.

If sum >= target, update the minimum length.

Time Complexity: O(n²)

4. Brainstorm More Solutions

Like in the Longest Substring question, we are looking for a window - only unlike that question, this time we're looking for the smallest possible window. But the technique we relied on in that question, a sliding window, works here in much the same way.

• We'll start the window at elements 0 and 1

• We'll move the right pointer to expand our window

• Once the window sum becomes ≥ target, we'll try shrinking the window from the left to find the smallest valid subarray.

• Track the minimum window size during this process.

Time Complexity: O(n)

Each element is visited at most twice: once when added, once when removed.

But we should note, this works because the input array contains only positive integers. Once a sum exceeds the target, we can reliably predict that the sum will shrink by moving the left pointer. There will never be a situation where the sum stays the same or increases by removing an element.

5. Discuss Trade‑Offs Between Your Solutions

Approach	Time Complexity	Space Complexity	Pros	Cons
Brute-force	O(n²)	O(1)	Easy to write, intuitive	Too slow for large arrays
Sliding Window	O(n)	O(1)	Fast, optimal, clean	Only works with positive numbers

6. Write Pseudocode to Structure Your Thoughts

initialize left = 0, sum = 0, minLen = infinity

for right in 0 to n:
    add nums[right] to sum
    while sum >= target:
        update minLen = min(minLen, right - left + 1)
        subtract nums[left] from sum
        increment left

if minLen == infinity:
    return 0
else:
    return minLen

7. Consider Edge Cases

• No subarray meets the condition → return 0.

• The entire array is needed → return nums.length.

• Only one element is enough → return 1.

• Large array with all 1s and high target → test performance.

8. Write Full Code Syntax (Java)

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

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

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

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

9. Test Your Code

System.out.println(minSubArrayLen(7, new int[]{2,3,1,2,4,3})); // 2
System.out.println(minSubArrayLen(4, new int[]{1,4,4}));       // 1
System.out.println(minSubArrayLen(11, new int[]{1,1,1,1,1,1})); // 0
System.out.println(minSubArrayLen(15, new int[]{1,2,3,4,5}));   // 5
System.out.println(minSubArrayLen(11, new int[]{1,2,3,4,5}));   // 3

10. Key Lessons to Remember for Future Questions

• When all elements are positive, the sliding window technique is a perfect fit for subarray problems.

• For problems involving "minimum" or "shortest" subarrays with constraints, look for a dynamic shrinking window pattern.

• Brute-force is useful for building intuition, but sliding windows unlock linear-time solutions in the right scenarios.

• For an extra challenge, think about how you would adapt your solution if the input did contain negative numbers.

  • Hint: maybe you could adapt Kadane's algorithm

Good Luck and Happy Coding!