How do you solve Minimum Size Subarray Sum on LeetCode?

Minimum Size Subarray Sum (LeetCode #209) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window can significantly reduce time complexity.

What is the brute force solution for Minimum Size Subarray Sum?

The brute force approach for Minimum Size Subarray Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if its sum meets or exceeds the target. This is straightforward but inefficient, especially for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Size Subarray Sum?

Minimum Size Subarray Sum uses the following concepts: Array, Binary Search, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Minimum Size Subarray Sum?

Explain your thought process clearly while coding. Consider edge cases, such as when no valid subarray exists. Practice explaining the sliding window technique beforehand.

What are common mistakes when solving Minimum Size Subarray Sum?

Not resetting the current sum correctly when moving the left pointer. Failing to check all subarrays in the brute force approach.

#209

Minimum Size Subarray Sum

Medium

ArrayBinary SearchSliding WindowPrefix SumSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses the sliding window technique to efficiently find the minimum length of a subarray. This approach expands and contracts the window based on the current sum, ensuring we only traverse the array once.

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Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) and a variable to track the current sum.
2Step 2: Expand the right pointer to increase the current sum until it meets or exceeds the target.
3Step 3: Once the target is met, try to contract the left pointer to find the minimal length while still meeting the target.
4Step 4: Update the minimum length if a shorter valid subarray is found.
5Step 5: Repeat until the right pointer reaches the end of the array.

solution.py11 lines

1def minSubArrayLen(target, nums):
2    left = 0
3    current_sum = 0
4    min_length = float('inf')
5    for right in range(len(nums)):
6        current_sum += nums[right]
7        while current_sum >= target:
8            min_length = min(min_length, right - left + 1)
9            current_sum -= nums[left]
10            left += 1
11    return min_length if min_length != float('inf') else 0

ℹ

Complexity note: The time complexity is O(n) because each element is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) since we only use a few variables.

1Using a sliding window can significantly reduce time complexity.
2Understanding when to expand and contract the window is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.