LeetCode in Kotlin
1493. Longest Subarray of 1’s After Deleting One Element
Medium
Given a binary array nums, you should delete one element from it.
Return the size of the longest non-empty subarray containing only 1’s in the resulting array. Return 0 if there is no such subarray.
Example 1:
Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1’s.
Example 2:
Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1’s is [1,1,1,1,1].
Example 3:
Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.
Constraints:
1 <= nums.length <= 105nums[i]is either0or1.
Solution
class Solution {
fun longestSubarray(nums: IntArray): Int {
var s = 0
var e = 0
var max = Int.MIN_VALUE
var extraZero = false
var allOne = true
while (e < nums.size) {
if (nums[e] == 1) {
e++
} else if (!extraZero) {
allOne = false
extraZero = true
e++
} else {
allOne = false
max = Math.max(max, e - s - 1)
while (nums[s] != 0) {
s++
}
s++
extraZero = false
}
}
if (nums[e - 1] == 1) {
max = Math.max(max, e - s - 1)
}
if (allOne) {
return nums.size - 1
}
return if (max == Int.MIN_VALUE) 0 else max
}
}