LeetCode in Kotlin

3086. Minimum Moves to Pick K Ones

Hard

You are given a binary array nums of length n, a positive integer k and a non-negative integer maxChanges.

Alice plays a game, where the goal is for Alice to pick up k ones from nums using the minimum number of moves. When the game starts, Alice picks up any index aliceIndex in the range [0, n - 1] and stands there. If nums[aliceIndex] == 1 , Alice picks up the one and nums[aliceIndex] becomes 0(this does not count as a move). After this, Alice can make any number of moves (including zero) where in each move Alice must perform exactly one of the following actions:

• Select any index j != aliceIndex such that nums[j] == 0 and set nums[j] = 1. This action can be performed at most maxChanges times.
• Select any two adjacent indices x and y (|x - y| == 1) such that nums[x] == 1, nums[y] == 0, then swap their values (set nums[y] = 1 and nums[x] = 0). If y == aliceIndex, Alice picks up the one after this move and nums[y] becomes 0.

Return the minimum number of moves required by Alice to pick exactly k ones.

Example 1:

Input: nums = [1,1,0,0,0,1,1,0,0,1], k = 3, maxChanges = 1

Output: 3

Explanation: Alice can pick up 3 ones in 3 moves, if Alice performs the following actions in each move when standing at aliceIndex == 1:

• At the start of the game Alice picks up the one and nums[1] becomes 0. nums becomes [1,**1**,1,0,0,1,1,0,0,1].
• Select j == 2 and perform an action of the first type. nums becomes [1,**0**,1,0,0,1,1,0,0,1]
• Select x == 2 and y == 1, and perform an action of the second type. nums becomes [1,**1**,0,0,0,1,1,0,0,1]. As y == aliceIndex, Alice picks up the one and nums becomes [1,**0**,0,0,0,1,1,0,0,1].
• Select x == 0 and y == 1, and perform an action of the second type. nums becomes [0,**1**,0,0,0,1,1,0,0,1]. As y == aliceIndex, Alice picks up the one and nums becomes [0,**0**,0,0,0,1,1,0,0,1].

Note that it may be possible for Alice to pick up 3 ones using some other sequence of 3 moves.

Example 2:

Input: nums = [0,0,0,0], k = 2, maxChanges = 3

Output: 4

Explanation: Alice can pick up 2 ones in 4 moves, if Alice performs the following actions in each move when standing at aliceIndex == 0:

• Select j == 1 and perform an action of the first type. nums becomes [**0**,1,0,0].
• Select x == 1 and y == 0, and perform an action of the second type. nums becomes [**1**,0,0,0]. As y == aliceIndex, Alice picks up the one and nums becomes [**0**,0,0,0].
• Select j == 1 again and perform an action of the first type. nums becomes [**0**,1,0,0].
• Select x == 1 and y == 0 again, and perform an action of the second type. nums becomes [**1**,0,0,0]. As y == aliceIndex, Alice picks up the one and nums becomes [**0**,0,0,0].

Constraints:

• 2 <= n <= 105
• 0 <= nums[i] <= 1
• 1 <= k <= 105
• 0 <= maxChanges <= 105
• maxChanges + sum(nums) >= k

Solution

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun minimumMoves(nums: IntArray, k: Int, maxChanges: Int): Long {
        var maxAdjLen = 0
        val n = nums.size
        var numOne = 0
        var l = 0
        var r = 0
        while (r < n) {
            if (nums[r] != 1) {
                maxAdjLen = max(maxAdjLen, (r - l))
                l = r + 1
            } else {
                numOne++
            }
            r++
        }
        maxAdjLen = min(3, max(maxAdjLen, (r - l)))
        if (maxAdjLen + maxChanges >= k) {
            return if (maxAdjLen >= k) {
                k - 1L
            } else {
                max(0, (maxAdjLen - 1)) + (k - maxAdjLen) * 2L
            }
        }
        val ones = IntArray(numOne)
        var ind = 0
        for (i in 0 until n) {
            if (nums[i] == 1) {
                ones[ind++] = i
            }
        }
        val preSum = LongArray(ones.size + 1)
        for (i in 1 until preSum.size) {
            preSum[i] = preSum[i - 1] + ones[i - 1]
        }
        val target = k - maxChanges
        l = 0
        var res = Long.MAX_VALUE
        while (l <= ones.size - target) {
            r = l + target - 1
            val mid = (l + r) / 2
            val median = ones[mid]
            val sum1 = preSum[mid + 1] - preSum[l]
            val sum2 = preSum[r + 1] - preSum[mid + 1]
            val area1 = (mid - l + 1).toLong() * median
            val area2 = (r - mid).toLong() * median
            val curRes = area1 - sum1 + sum2 - area2
            res = min(res.toDouble(), curRes.toDouble()).toLong()
            l++
        }
        res += 2L * maxChanges
        return res
    }
}

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