Why does a sliding window work here?

All numbers are positive, so adding a number always increases the sum and removing one always decreases it. Once a window reaches the target, any longer window starting at the same left end is not useful, so shrinking from the left is the only way to find something shorter.

Why is it O(n) with a while loop inside a for loop?

Each index is added once when right reaches it and removed at most once when left passes it. Across the whole run the inner loop does at most n removals in total.

What if the array could contain negative numbers?

Then the window's sum is no longer monotonic and shrinking can make it bigger, so the sliding window breaks. That version needs prefix sums with a monotonic deque.

Minimum Size Subarray Sum

Find the shortest subarray whose sum reaches a target in O(n) with a sliding window that grows on the right and shrinks from the left.

MediumSolve on LeetCode

Problem Understanding

Minimum Size Subarray Sum: given an array of positive integers and a target, return the length of the shortest contiguous subarray whose sum is at least target, or 0 if there is none.

Example: target = 7, nums = [2, 3, 1, 2, 4, 3] returns 2: [4, 3].

The Intuition: Stretch, Then Squeeze

Because every number is positive, a window's sum goes up when it grows and down when it shrinks. That makes a variable-size sliding window work:

Move right one step and add nums[right] to total.
While total ≥ target, the window is good enough: record its length, then remove nums[left] and move left on, looking for something shorter.
When the sum drops below the target, go back to growing.

Each number enters and leaves the window at most once, so the whole scan is O(n). Compare Longest Substring Without Repeating Characters, which shrinks for the opposite reason: to make an invalid window valid again.

Interactive Walkthrough

The shaded cells are the current window L..R; the number just added or dropped is filled. The best window so far is underlined in green. Watch the window stretch until the sum is enough, then pull in from the left.

Dry Run Table

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

Add	Window	Total	Shrink?	Best
2, 3, 1	0..2	6	no	—
2	0..3	8	record 4, drop 2	4
4	1..4	10	record 4, drop 3; 7: record 3, drop 1	3
3	3..5	9	record 3, drop 2; 7: record 2, drop 4	2

Return 2.

The Solution Template

Variable-size sliding window: grow right, shrink left while the sum is enough.

Minimum Size Subarray Sum Code

1var minSubArrayLen = function(target, nums) {
2    let best = Infinity, total = 0, left = 0;
3    for (let right = 0; right < nums.length; right++) {
4        total += nums[right];
5        while (total >= target) {
6            best = Math.min(best, right - left + 1);
7            total -= nums[left++];
8        }
9    }
10    return best === Infinity ? 0 : best;
11};

Edge Cases & Common Mistakes

No window reaches the target: return 0, not infinity.
A single number is enough: the answer is 1.
Using if instead of while to shrink: the window may need to drop several numbers before it stops being enough.
Recording the length after shrinking: record it while the sum still reaches the target, before removing nums[left].

Summary

With only positive numbers, a window's sum grows and shrinks predictably. Extend the window on the right; while its sum reaches the target, record its length and drop numbers from the left. Each number enters and leaves once: O(n) time, O(1) space.

More practice: DSA interview questions with answers · patterns cheat sheet

Minimum Window Substring

Max Consecutive Ones III

Minimum Size Subarray Sum

Problem Understanding

The Intuition: Stretch, Then Squeeze

Interactive Walkthrough

Dry Run Table

The Solution Template

Minimum Size Subarray Sum Code

Edge Cases & Common Mistakes

Summary

The approach, step by step

Grow on the right

Shrink while it is enough

Return the best

Frequently asked questions

Why does a sliding window work here?

Why is it O(n) with a while loop inside a for loop?

What if the array could contain negative numbers?

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