Question

1846. Maximum Element After Decreasing and Rearranging

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Description

You are given an array of positive integers arr. Perform some operations (possibly none) on arr so that it satisfies these conditions:

• The value of the first element in arr must be 1.
• The absolute difference between any 2 adjacent elements must be less than or equal to 1. In other words, abs(arr[i] - arr[i - 1]) <= 1 for each i where 1 <= i < arr.length (0-indexed). abs(x) is the absolute value of x.

There are 2 types of operations that you can perform any number of times:

• Decrease the value of any element of arr to a smaller positive integer.
• Rearrange the elements of arr to be in any order.

Return _the maximum possible value of an element in _arr_ after performing the operations to satisfy the conditions_.

 

Example 1:

Input: arr = [2,2,1,2,1]
Output: 2
Explanation: 
We can satisfy the conditions by rearranging arr so it becomes [1,2,2,2,1].
The largest element in arr is 2.

Example 2:

Input: arr = [100,1,1000]
Output: 3
Explanation: 
One possible way to satisfy the conditions is by doing the following:
1. Rearrange arr so it becomes [1,100,1000].
2. Decrease the value of the second element to 2.
3. Decrease the value of the third element to 3.
Now arr = [1,2,3], which satisfies the conditions.
The largest element in arr is 3.

Example 3:

Input: arr = [1,2,3,4,5]
Output: 5
Explanation: The array already satisfies the conditions, and the largest element is 5.

 

Constraints:

• 1 <= arr.length <= 105
• 1 <= arr[i] <= 109

Solutions

Solution 1: Sorting + Greedy Algorithm

First, we sort the array and then set the first element of the array to $1$.

Next, we start traversing the array from the second element. If the difference between the current element and the previous one is more than $1$, we greedily reduce the current element to the previous element plus $1$.

Finally, we return the maximum element in the array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.

class Solution:
  def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int:
    arr.sort()
    arr[0] = 1
    for i in range(1, len(arr)):
      d = max(0, arr[i] - arr[i - 1] - 1)
      arr[i] -= d
    return max(arr)

class Solution {
  public int maximumElementAfterDecrementingAndRearranging(int[] arr) {
    Arrays.sort(arr);
    arr[0] = 1;
    int ans = 1;
    for (int i = 1; i < arr.length; ++i) {
      int d = Math.max(0, arr[i] - arr[i - 1] - 1);
      arr[i] -= d;
      ans = Math.max(ans, arr[i]);
    }
    return ans;
  }
}

class Solution {
public:
  int maximumElementAfterDecrementingAndRearranging(vector<int>& arr) {
    sort(arr.begin(), arr.end());
    arr[0] = 1;
    int ans = 1;
    for (int i = 1; i < arr.size(); ++i) {
      int d = max(0, arr[i] - arr[i - 1] - 1);
      arr[i] -= d;
      ans = max(ans, arr[i]);
    }
    return ans;
  }
};

func maximumElementAfterDecrementingAndRearranging(arr []int) int {
  sort.Ints(arr)
  ans := 1
  arr[0] = 1
  for i := 1; i < len(arr); i++ {
    d := max(0, arr[i]-arr[i-1]-1)
    arr[i] -= d
    ans = max(ans, arr[i])
  }
  return ans
}

function maximumElementAfterDecrementingAndRearranging(arr: number[]): number {
  arr.sort((a, b) => a - b);
  arr[0] = 1;
  let ans = 1;
  for (let i = 1; i < arr.length; ++i) {
    const d = Math.max(0, arr[i] - arr[i - 1] - 1);
    arr[i] -= d;
    ans = Math.max(ans, arr[i]);
  }
  return ans;
}

public class Solution {
  public int MaximumElementAfterDecrementingAndRearranging(int[] arr) {
    Array.Sort(arr);
    int n = arr.Length;
    arr[0] = 1;
    for (int i = 1; i < n; ++i) {
      arr[i] = Math.Min(arr[i], arr[i - 1] + 1);
    }
    return arr[n - 1];
  }
}