1846. Maximum Element After Decreasing and Rearranging

• Time: $O(\texttt{sort})$
• Space: $O(\texttt{sort})$

C++JavaPython

class Solution {
 public:
  int maximumElementAfterDecrementingAndRearranging(vector<int>& arr) {
    ranges::sort(arr);
    arr[0] = 1;

    for (int i = 1; i < arr.size(); ++i)
      arr[i] = min(arr[i], arr[i - 1] + 1);

    return arr.back();
  }
};

class Solution {
  public int maximumElementAfterDecrementingAndRearranging(int[] arr) {
    Arrays.sort(arr);
    arr[0] = 1;

    for (int i = 1; i < arr.length; ++i)
      arr[i] = Math.min(arr[i], arr[i - 1] + 1);

    return arr[arr.length - 1];
  }
}

class Solution:
  def maximumElementAfterDecrementingAndRearranging(
      self, arr: list[int],
  ) -> int:
    arr.sort()
    arr[0] = 1

    for i in range(1, len(arr)):
      arr[i] = min(arr[i], arr[i - 1] + 1)

    return arr[-1]