Sliding Window:

Day 1/75:

Question: Leetcode (Medium)

Minimum Size Subarray Sum: 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 there is no such subarray, return 0 instead.

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.

Intuition: This question is really interesting and help us to solve any Sliding Window Problem with little change. We'll maintain two pointers to track the current subarray's start and end. By adjusting these pointers and updating the sum, we aim to find the smallest subarray satisfying the target condition.

Approach:

Initialization: Initialize start and end pointers, minLength to infinity, and sum to 0.

Here start and end pointer defines the starting and end point of window.

Sliding Window Technique:

• Traverse through the array with the end pointer.
• Add nums[end] to the sum.
• While the sum is greater than or equal to the target:Update minLength as the minimum between current minLength and the window size.Subtract nums[start] from the sum and move the start pointer.Increment the end pointer.

Result: Return minLength if found, else return 0.

GitHub Solution : SlidingWindow/MinimumSIzeSubArraySum.java

Time Complexity: The time complexity of this approach is O(n), as we iterate through the array once with two pointers.

Space Complexity: Space complexity is O(1) since we use a constant amount of extra space for variables.

Related Questions:

• Subarray Product Less Than K(Medium)
• Count Subarrays With Score Less Than K(Hard)
• Binary Subarrays With Sum

Reflection: Today's coding session was productive, reinforcing understanding of the sliding window technique. Excited to explore related questions with the same approach and continue the coding journey in the 75 Days Code Challenge!