LeetCode in Kotlin
3507. Minimum Pair Removal to Sort Array I
Easy
Given an array nums, you can perform the following operation any number of times:
- Select the adjacent pair with the minimum sum in
nums. If multiple such pairs exist, choose the leftmost one. - Replace the pair with their sum.
Return the minimum number of operations needed to make the array non-decreasing.
An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).
Example 1:
Input: nums = [5,2,3,1]
Output: 2
Explanation:
- The pair
(3,1)has the minimum sum of 4. After replacement,nums = [5,2,4]. - The pair
(2,4)has the minimum sum of 6. After replacement,nums = [5,6].
The array nums became non-decreasing in two operations.
Example 2:
Input: nums = [1,2,2]
Output: 0
Explanation:
The array nums is already sorted.
Constraints:
1 <= nums.length <= 50-1000 <= nums[i] <= 1000
Solution
class Solution {
fun minimumPairRemoval(nums: IntArray): Int {
var nums = nums
var operations = 0
while (!isNonDecreasing(nums)) {
var minSum = Int.Companion.MAX_VALUE
var index = 0
// Find the leftmost pair with minimum sum
for (i in 0..<nums.size - 1) {
val sum = nums[i] + nums[i + 1]
if (sum < minSum) {
minSum = sum
index = i
}
}
// Merge the pair at index
val newNums = IntArray(nums.size - 1)
var j = 0
var i = 0
while (i < nums.size) {
if (i == index) {
newNums[j++] = nums[i] + nums[i + 1]
// Skip the next one since it's merged
i++
} else {
newNums[j++] = nums[i]
}
i++
}
nums = newNums
operations++
}
return operations
}
private fun isNonDecreasing(nums: IntArray): Boolean {
for (i in 1..<nums.size) {
if (nums[i] < nums[i - 1]) {
return false
}
}
return true
}
}