LeetCode in Kotlin

209. Minimum Size Subarray Sum

Medium

Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to 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

Constraints:

• 1 <= target <= 109
• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

Solution

class Solution {
    fun minSubArrayLen(s: Int, nums: IntArray): Int {
        var sum = 0
        var start = 0
        var minLength = Integer.MAX_VALUE
        var end = 0
        if (nums.size < 1) {
            return 0
        }
        while (end < nums.size) {
            sum += nums[end]
            while (sum >= s) {
                minLength = Math.min(minLength, end - start + 1)
                sum -= nums[start++]
            }
            end++
        }
        return if (start > 0) minLength else 0
    }
}

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