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1846. Maximum Element After Decreasing and Rearranging

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 {
    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;
    }
};

  • 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)


  • 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];
    }
}

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