Minimum Size Subarray Sum

Medium Arrays & Hashing Sliding Window · Variable Original on LeetCode

Minimum Size Subarray Sum is a medium arrays & hashing problem solved with the sliding window · variable pattern. The best approach, optimal (variable sliding window), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from brute force up.

Problem

Given an array of positive integers nums and a positive integer target, return the length of the shortest contiguous subarray whose sum is at least target. If none exists, return 0.

Examples

Example 1

Input: target = 8, nums = [3, 1, 2, 5, 2, 4]
Output: 3
Why: [1, 2, 5] sums to 8 with length 3. No pair of neighbours reaches 8 (the largest is 5 + 2 = 7).

Example 2

Input: target = 20, nums = [2, 3, 4]
Output: 0
Why: Even the whole array sums to only 9.

Constraints

1 <= target <= 10^9
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4 (all positive)

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Brute force	`O(n²)`	`O(1)`
Optimal (variable sliding window)	`O(n)`	`O(1)`

1Brute force

TimeO(n²)

SpaceO(1)

For each start, extend until the sum reaches target and record the length.

For each i, add nums[j] for j >= i until sum >= target.
Track the smallest j - i + 1.

Java

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int best = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            int sum = 0;
            for (int j = i; j < nums.length; j++) {
                sum += nums[j];
                if (sum >= target) { best = Math.min(best, j - i + 1); break; }
            }
        }
        return best == Integer.MAX_VALUE ? 0 : best;
    }
}

2Optimal (variable sliding window)

TimeO(n)l and r each move at most n times.

SpaceO(1)

Grow the window on the right. While its sum is at least target, record the length and shrink from the left. Because values are positive, each element enters and leaves the window once.

For each r: sum += nums[r].
While sum >= target: best = min(best, r - l + 1); sum -= nums[l++].
Return best, or 0 if it never changed.

Java

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int l = 0, sum = 0, best = Integer.MAX_VALUE;
        for (int r = 0; r < nums.length; r++) {
            sum += nums[r];
            while (sum >= target) {
                best = Math.min(best, r - l + 1);
                sum -= nums[l++];
            }
        }
        return best == Integer.MAX_VALUE ? 0 : best;
    }
}

Edge cases to test

No subarray reaches the target (return 0)
A single element already reaches the target

Hints

Hint 1

All numbers are positive, so growing the window only increases the sum and shrinking only decreases it.

FAQ

What is the best time complexity for Minimum Size Subarray Sum?

Optimal (variable sliding window) runs in O(n) time and O(1) extra space. l and r each move at most n times.

Which pattern does Minimum Size Subarray Sum use?

It is a arrays & hashing problem that uses the sliding window · variable pattern. Other problems with the same pattern: Count of Subarrays with Product Less Than K.

Is there a brute force solution for Minimum Size Subarray Sum?

Yes. Brute force takes O(n²) time and O(1) space. For each start, extend until the sum reaches target and record the length.

Which edge cases should I test for Minimum Size Subarray Sum?

No subarray reaches the target (return 0); A single element already reaches the target.