LeetCode in Kotlin
1300. Sum of Mutated Array Closest to Target
Medium
Given an integer array arr and a target value target, return the integer value such that when we change all the integers larger than value in the given array to be equal to value, the sum of the array gets as close as possible (in absolute difference) to target.
In case of a tie, return the minimum such integer.
Notice that the answer is not neccesarilly a number from arr.
Example 1:
Input: arr = [4,9,3], target = 10
Output: 3
Explanation: When using 3 arr converts to [3, 3, 3] which sums 9 and that’s the optimal answer.
Example 2:
Input: arr = [2,3,5], target = 10
Output: 5
Example 3:
Input: arr = [60864,25176,27249,21296,20204], target = 56803
Output: 11361
Constraints:
1 <= arr.length <= 1041 <= arr[i], target <= 105
Solution
class Solution {
fun findBestValue(arr: IntArray, target: Int): Int {
arr.sort()
val n = arr.size
var lo = 0
var hi = arr[n - 1]
var min = Int.MAX_VALUE
var ans = -1
while (lo <= hi) {
val mid = (lo + hi) / 2
val m = check(mid, arr, target)
val l = check(mid - 1, arr, target)
val r = check(mid + 1, arr, target)
if (m < min || m == min && ans > mid) {
min = m
ans = mid
} else if (l <= r) {
hi = mid - 1
} else {
lo = mid + 1
}
}
return ans
}
fun check(v: Int, arr: IntArray, target: Int): Int {
var sum = 0
for (i in arr.indices) {
sum += if (arr[i] >= v) {
return Math.abs(sum + (arr.size - i) * v - target)
} else {
arr[i]
}
}
return Math.abs(sum - target)
}
}