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3048. Earliest Second to Mark Indices I
Description
You are given two 1indexed 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 decrementnums[i]
by1
.  If
nums[changeIndices[s]]
is equal to0
, mark the indexchangeIndices[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] <= 10^{9}
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 0indexed. 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[i1] { return false } decrement = nums[i1] 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; }