LeetCode in Kotlin

3266. Final Array State After K Multiplication Operations II

Hard

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:

• Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
• Replace the selected minimum value x with x * multiplier.

After the k operations, apply modulo 109 + 7 to every value in nums.

Return an integer array denoting the final state of nums after performing all k operations and then applying the modulo.

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]
After applying modulo	[8, 4, 6, 5, 6]

Example 2:

Input: nums = [100000,2000], k = 2, multiplier = 1000000

Output: [999999307,999999993]

Explanation:

Operation	Result
After operation 1	[100000, 2000000000]
After operation 2	[100000000000, 2000000000]
After applying modulo	[999999307, 999999993]

Constraints:

• 1 <= nums.length <= 104
• 1 <= nums[i] <= 109
• 1 <= k <= 109
• 1 <= multiplier <= 106

Solution

import java.util.PriorityQueue
import kotlin.math.max

@Suppress("NAME_SHADOWING")
class Solution {
    fun getFinalState(nums: IntArray, k: Int, multiplier: Int): IntArray {
        var k = k
        if (multiplier == 1) {
            return nums
        }
        val n = nums.size
        var mx = 0
        for (x in nums) {
            mx = max(mx, x)
        }
        val a = LongArray(n)
        var left = k
        var shouldExit = false
        run {
            var i = 0
            while (i < n && !shouldExit) {
                var x = nums[i].toLong()
                while (x < mx) {
                    x *= multiplier.toLong()
                    if (--left < 0) {
                        shouldExit = true
                        break
                    }
                }
                a[i] = x
                i++
            }
        }
        if (left < 0) {
            val pq =
                PriorityQueue { p: LongArray, q: LongArray ->
                    if (p[0] != q[0]
                    ) java.lang.Long.compare(p[0], q[0])
                    else java.lang.Long.compare(p[1], q[1])
                }
            for (i in 0 until n) {
                pq.offer(longArrayOf(nums[i].toLong(), i.toLong()))
            }
            while (k-- > 0) {
                val p = pq.poll()
                p[0] *= multiplier.toLong()
                pq.offer(p)
            }
            while (pq.isNotEmpty()) {
                val p = pq.poll()
                nums[p[1].toInt()] = (p[0] % MOD).toInt()
            }
            return nums
        }

        val ids: Array<Int> = Array(n) { i: Int -> i }
        ids.sortWith { i: Int?, j: Int? -> java.lang.Long.compare(a[i!!], a[j!!]) }
        k = left
        val pow1 = pow(multiplier.toLong(), k / n)
        val pow2 = pow1 * multiplier % MOD
        for (i in 0 until n) {
            val j = ids[i]
            nums[j] = (a[j] % MOD * (if (i < k % n) pow2 else pow1) % MOD).toInt()
        }
        return nums
    }

    private fun pow(x: Long, n: Int): Long {
        var x = x
        var n = n
        var res: Long = 1
        while (n > 0) {
            if (n % 2 > 0) {
                res = res * x % MOD
            }
            x = x * x % MOD
            n /= 2
        }
        return res
    }

    companion object {
        private const val MOD = 1000000007
    }
}

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