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3048. Earliest Second to Mark Indices I

  • Time: $O(m\log m)$
  • Space: $O(m)$
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class Solution {
 public:
  int earliestSecondToMarkIndices(vector<int>& nums,
                                  vector<int>& changeIndices) {
    int l = 0;
    int r = changeIndices.size() + 1;

    while (l < r) {
      const int m = (l + r) / 2;
      if (canMark(nums, changeIndices, m))
        r = m;
      else
        l = m + 1;
    }

    return l <= changeIndices.size() ? l : -1;
  }

 private:
  // Returns true if all indices of `nums` can be marked within `second`.
  bool canMark(const vector<int>& nums, const vector<int>& changeIndices,
               int second) {
    int numMarked = 0;
    int decrement = 0;
    // indexToLastSecond[i] := the last second to mark the index i
    vector<int> indexToLastSecond(nums.size(), -1);

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

    for (int i = 0; i < second; ++i) {
      const int index = changeIndices[i] - 1;  // Convert to 0-indexed.
      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.size();
  }
};

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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;
  }

  // Returns true if all indices of `nums` can be marked within `second`.
  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) {
      final int index = changeIndices[i] - 1; // Convert to 0-indexed.
      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;
  }
}

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class Solution:
  def earliestSecondToMarkIndices(
      self,
      nums: list[int],
      changeIndices: list[int],
  ) -> int:
    def canMark(second: int) -> bool:
      """
      Returns True if all indices of `nums` can be marked within `second`.
      """
      numMarked = 0
      decrement = 0
      indexToLastSecond = {}

      for i in range(second):
        indexToLastSecond[changeIndices[i] - 1] = i

      for i in range(second):
        index = changeIndices[i] - 1  # Convert to 0-indexed
        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 += 1
        else:
          decrement += 1

      return numMarked == len(nums)

    l = 1
    r = len(changeIndices) + 1
    ans = bisect.bisect_left(range(l, r), True, key=canMark) + l
    return ans if ans <= len(changeIndices) else -1