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

Solution 1

class Solution {
    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {
        int l = 0;
        int r = changeIndices.length + 1;
        while (l < r) {
            final int m = (l + r) / 2;
            if (canMark(nums, changeIndices, m)) {
                r = m;
            } else {
                l = m + 1;
            }
        }
        return l <= changeIndices.length ? l : -1;
    }

    private boolean canMark(int[] nums, int[] changeIndices, int second) {
        int numMarked = 0;
        int decrement = 0;
        // indexToLastSecond[i] := the last second to mark the index i
        int[] indexToLastSecond = new int[nums.length];
        Arrays.fill(indexToLastSecond, -1);

        for (int i = 0; i < second; ++i) {
            indexToLastSecond[changeIndices[i] - 1] = i;
        }

        for (int i = 0; i < second; ++i) {
            // Convert to 0-indexed.
            final int index = changeIndices[i] - 1;
            if (i == indexToLastSecond[index]) {
                // Reach the last occurrence of the number.
                // So, the current second will be used to mark the index.
                if (nums[index] > decrement) {
                    // The decrement is less than the number to be marked.
                    return false;
                }
                decrement -= nums[index];
                ++numMarked;
            } else {
                ++decrement;
            }
        }
        return numMarked == nums.length;
    }
}

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;
    }
};

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

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
}

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;
}

3048 - Earliest Second to Mark Indices I

3048. Earliest Second to Mark Indices I

Description

Solutions

Solution 1

All Problems

All Solutions

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3048. Earliest Second to Mark Indices I

Description

Solutions

Solution 1

All Problems

All Solutions