Medium
Given an integer array nums
, your goal is to make all elements in nums
equal. To complete one operation, follow these steps:
nums
. Let its index be i
(0-indexed) and its value be largest
. If there are multiple elements with the largest value, pick the smallest i
.nums
strictly smaller than largest
. Let its value be nextLargest
.nums[i]
to nextLargest
.Return the number of operations to make all elements in nums
equal.
Example 1:
Input: nums = [5,1,3]
Output: 3
Explanation: It takes 3 operations to make all elements in nums equal:
largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].
Example 2:
Input: nums = [1,1,1]
Output: 0
Explanation: All elements in nums are already equal.
Example 3:
Input: nums = [1,1,2,2,3]
Output: 4
Explanation: It takes 4 operations to make all elements in nums equal:
largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].
Constraints:
1 <= nums.length <= 5 * 104
1 <= nums[i] <= 5 * 104
class Solution {
fun reductionOperations(nums: IntArray): Int {
val arr = IntArray(100001)
for (i in nums) {
arr[i]++
}
var `val` = 0
var curr = 0
for (i in 100000 downTo 0) {
if (arr[i] != 0) {
`val` += curr
curr += arr[i]
}
}
return `val`
}
}