LeetCode in Kotlin

3048. Earliest Second to Mark Indices I

Medium

You are given two 1-indexed integer arrays, nums and, changeIndices, having lengths n and m, respectively.

Initially, all indices in nums are unmarked. Your task is to mark all indices in nums.

In each second, s, in order from 1 to m (inclusive), you can perform one of the following operations:

• Choose an index i in the range [1, n] and decrement nums[i] by 1.
• If nums[changeIndices[s]] is equal to 0, mark the index changeIndices[s].
• Do nothing.

Return an integer denoting the earliest second in the range [1, m] when all indices in nums can be marked by choosing operations optimally, or -1 if it is impossible.

Example 1:

Input: nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]

Output: 8

Explanation: In this example, we have 8 seconds. The following operations can be performed to mark all indices:

Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0].

Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0].

Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0].

Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0].

Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0.

Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0.

Second 7: Do nothing.

Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0.

Now all indices have been marked.

It can be shown that it is not possible to mark all indices earlier than the 8th second.

Hence, the answer is 8.

Example 2:

Input: nums = [1,3], changeIndices = [1,1,1,2,1,1,1]

Output: 6

Explanation: In this example, we have 7 seconds. The following operations can be performed to mark all indices:

Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2].

Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1].

Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0].

Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0.

Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0].

Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0.

Now all indices have been marked.

It can be shown that it is not possible to mark all indices earlier than the 6th second.

Hence, the answer is 6.

Example 3:

Input: nums = [0,1], changeIndices = [2,2,2]

Output: -1

Explanation: In this example, it is impossible to mark all indices because index 1 isn’t in changeIndices. Hence, the answer is -1.

Constraints:

• 1 <= n == nums.length <= 2000
• 0 <= nums[i] <= 109
• 1 <= m == changeIndices.length <= 2000
• 1 <= changeIndices[i] <= n

Solution

class Solution {
    fun earliestSecondToMarkIndices(nums: IntArray, changeIndices: IntArray): Int {
        val n = nums.size
        if (nums.isEmpty() || changeIndices.isEmpty()) {
            return 0
        }
        val last = IntArray(n)
        last.fill(-1)
        for (i in changeIndices.indices) {
            changeIndices[i] -= 1
        }
        var low = 0
        var high = changeIndices.size - 1
        while (low < high) {
            val mid = low + (high - low) / 2
            if (isPossible(mid, nums, changeIndices, last)) {
                high = mid
            } else {
                low = mid + 1
            }
        }
        return if (isPossible(low, nums, changeIndices, last)) low + 1 else -1
    }

    private fun isPossible(s: Int, nums: IntArray, changeIndices: IntArray, last: IntArray): Boolean {
        val n = nums.size
        last.fill(-1)
        for (i in 0..s) {
            last[changeIndices[i]] = i
        }
        var marked = 0
        var operations = 0
        for (i in 0..s) {
            if (i == last[changeIndices[i]]) {
                if (nums[changeIndices[i]] > operations) {
                    return false
                }
                operations -= nums[changeIndices[i]]
                marked++
            } else {
                operations++
            }
        }
        return marked == n
    }
}

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