Easy
You are given an integer array nums
, an integer k
, and an integer multiplier
.
You need to perform k
operations on nums
. In each operation:
x
in nums
. If there are multiple occurrences of the minimum value, select the one that appears first.x
with x * multiplier
.Return an integer array denoting the final state of nums
after performing all k
operations.
Example 1:
Input: nums = [2,1,3,5,6], k = 5, multiplier = 2
Output: [8,4,6,5,6]
Explanation:
Operation | Result |
---|---|
After operation 1 | [2, 2, 3, 5, 6] |
After operation 2 | [4, 2, 3, 5, 6] |
After operation 3 | [4, 4, 3, 5, 6] |
After operation 4 | [4, 4, 6, 5, 6] |
After operation 5 | [8, 4, 6, 5, 6] |
Example 2:
Input: nums = [1,2], k = 3, multiplier = 4
Output: [16,8]
Explanation:
Operation | Result |
---|---|
After operation 1 | [4, 2] |
After operation 2 | [4, 8] |
After operation 3 | [16, 8] |
Constraints:
1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5
@Suppress("NAME_SHADOWING")
class Solution {
fun getFinalState(nums: IntArray, k: Int, multiplier: Int): IntArray {
var k = k
while (k-- > 0) {
var min = nums[0]
var index = 0
for (i in nums.indices) {
if (min > nums[i]) {
min = nums[i]
index = i
}
}
nums[index] = nums[index] * multiplier
}
return nums
}
}