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3048. Earliest Second to Mark Indices I

Description

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

Solutions

We notice that if we can mark all indices within $t$ seconds, then we can also mark all indices within $t' \geq t$ seconds. Therefore, we can use binary search to find the earliest seconds.

We define the left and right boundaries of binary search as $l = 1$ and $r = m + 1$, where $m$ is the length of the array changeIndices. For each $t = \frac{l + r}{2}$, we check whether we can mark all indices within $t$ seconds. If we can, we move the right boundary to $t$, otherwise we move the left boundary to $t + 1$. Finally, we judge whether the left boundary is greater than $m$, if it is, return $-1$, otherwise return the left boundary.

The key to the problem is how to judge whether we can mark all indices within $t$ seconds. We can use an array $last$ to record the latest time each index needs to be marked, use a variable $decrement$ to record the current number of times that can be reduced, and use a variable $marked$ to record the number of indices that have been marked.

We traverse the first $t$ elements of the array changeIndices, for each element $i$, if $last[i] = s$, then we need to check whether $decrement$ is greater than or equal to $nums[i - 1]$, if it is, we subtract $nums[i - 1]$ from $decrement$, and add one to $marked$; otherwise, we return False. If $last[i] \neq s$, then we can temporarily not mark the index, so we add one to $decrement$. Finally, we check whether $marked$ is equal to $n$, if it is, we return True, otherwise return False.

The time complexity is $O(m \times \log m)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of nums and changeIndices respectively.

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class Solution:
    def earliestSecondToMarkIndices(
        self, nums: List[int], changeIndices: List[int]
    ) -> int:
        def check(t: int) -> bool:
            decrement = 0
            marked = 0
            last = {i: s for s, i in enumerate(changeIndices[:t])}
            for s, i in enumerate(changeIndices[:t]):
                if last[i] == s:
                    if decrement < nums[i - 1]:
                        return False
                    decrement -= nums[i - 1]
                    marked += 1
                else:
                    decrement += 1
            return marked == len(nums)

        m = len(changeIndices)
        l = bisect_left(range(1, m + 2), True, key=check) + 1
        return -1 if l > m else l

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class Solution {
    private int[] nums;
    private int[] changeIndices;

    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {
        this.nums = nums;
        this.changeIndices = changeIndices;
        int m = changeIndices.length;
        int l = 1, r = m + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l > m ? -1 : l;
    }

    private boolean check(int t) {
        int[] last = new int[nums.length + 1];
        for (int s = 0; s < t; ++s) {
            last[changeIndices[s]] = s;
        }
        int decrement = 0;
        int marked = 0;
        for (int s = 0; s < t; ++s) {
            int i = changeIndices[s];
            if (last[i] == s) {
                if (decrement < nums[i - 1]) {
                    return false;
                }
                decrement -= nums[i - 1];
                ++marked;
            } else {
                ++decrement;
            }
        }
        return marked == nums.length;
    }
}

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class Solution {
public:
    int earliestSecondToMarkIndices(vector<int>& nums, vector<int>& changeIndices) {
        int n = nums.size();
        int last[n + 1];
        auto check = [&](int t) {
            memset(last, 0, sizeof(last));
            for (int s = 0; s < t; ++s) {
                last[changeIndices[s]] = s;
            }
            int decrement = 0, marked = 0;
            for (int s = 0; s < t; ++s) {
                int i = changeIndices[s];
                if (last[i] == s) {
                    if (decrement < nums[i - 1]) {
                        return false;
                    }
                    decrement -= nums[i - 1];
                    ++marked;
                } else {
                    ++decrement;
                }
            }
            return marked == n;
        };

        int m = changeIndices.size();
        int l = 1, r = m + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l > m ? -1 : l;
    }
};

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func earliestSecondToMarkIndices(nums []int, changeIndices []int) int {
    n, m := len(nums), len(changeIndices)
    l := sort.Search(m+1, func(t int) bool {
        last := make([]int, n+1)
        for s, i := range changeIndices[:t] {
            last[i] = s
        }
        decrement, marked := 0, 0
        for s, i := range changeIndices[:t] {
            if last[i] == s {
                if decrement < nums[i-1] {
                    return false
                }
                decrement -= nums[i-1]
                marked++
            } else {
                decrement++
            }
        }
        return marked == n
    })
    if l > m {
        return -1
    }
    return l
}

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function earliestSecondToMarkIndices(nums: number[], changeIndices: number[]): number {
    const [n, m] = [nums.length, changeIndices.length];
    let [l, r] = [1, m + 1];
    const check = (t: number): boolean => {
        const last: number[] = Array(n + 1).fill(0);
        for (let s = 0; s < t; ++s) {
            last[changeIndices[s]] = s;
        }
        let [decrement, marked] = [0, 0];
        for (let s = 0; s < t; ++s) {
            const i = changeIndices[s];
            if (last[i] === s) {
                if (decrement < nums[i - 1]) {
                    return false;
                }
                decrement -= nums[i - 1];
                ++marked;
            } else {
                ++decrement;
            }
        }
        return marked === n;
    };
    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l > m ? -1 : l;
}

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