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 <= 1091 <= nums.length <= 1051 <= 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
}
}