#1846

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Element After Decreasing and Rearranging

Determine the maximum value in an array after decreasing elements and rearranging using a greedy invariant approach.

Problem Statement

Given an array of positive integers, you may perform operations to decrease elements or rearrange them so that the array satisfies specific adjacency constraints. Each element can be reduced any number of times, and the goal is to adjust the array to meet the rules while keeping all values positive.

You must return the maximum element possible in the array after performing these operations optimally. The challenge is to determine how to rearrange and decrease values using a greedy approach to maintain the invariant that each element is at most one greater than its predecessor.

Examples

Example 1

Input: arr = [2,2,1,2,1]

Output: 2

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

One possible way to satisfy the conditions is by doing the following:

Rearrange arr so it becomes [1,100,1000].
Decrease the value of the second element to 2.
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

The array already satisfies the conditions, and the largest element is 5.

Constraints

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

Solution Approach

Sort and Initialize

Start by sorting the array to simplify the greedy adjustments. Initialize the first element as 1 since the smallest value must remain positive.

Iterate with Greedy Adjustment

Traverse the sorted array, setting each element to the minimum of its current value or the previous element plus one. This preserves the invariant while minimizing the maximum element.

Return Maximum Value

After processing all elements, the last element in the adjusted array represents the largest possible value. Return this value as the result.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

Sorting takes O(n log n) but can be simplified to O(n) with counting sort if values are bounded. Iterating and adjusting is O(n). The space complexity is O(n) if a separate array is used for adjustment, or O(1) in-place.

What Interviewers Usually Probe

Look for sorting the array before applying operations.
Check if each element can be at most one greater than its previous.
Validate the maximum value after greedy adjustments.

Common Pitfalls or Variants

Common pitfalls

Forgetting to sort before applying greedy decreases.
Not enforcing the invariant that arr[i] <= arr[i-1] + 1.
Returning the initial max instead of the adjusted maximum element.

Follow-up variants

Find maximum after only decreasing elements without rearranging.
Return the array configuration instead of just the maximum element.
Apply the same approach for arrays with negative numbers allowed.

FAQ

What is the main pattern used in Maximum Element After Decreasing and Rearranging?

The key pattern is greedy choice plus invariant validation, ensuring each element is at most one greater than its predecessor after sorting.

Can we avoid sorting the array?

Sorting is necessary to enforce the greedy invariant efficiently; without sorting, determining the correct maximum becomes complex.

What is the time complexity of the optimal solution?

The solution runs in O(n) for the adjustment step and O(n log n) if standard sorting is used, with O(n) space for storing adjusted values.

Does rearranging affect the final maximum value?

Yes, rearranging allows the elements to be positioned optimally so that the invariant holds and the maximum element is minimized.

What are common mistakes in this problem?

Common mistakes include not sorting, failing to enforce the invariant, and returning the original maximum instead of the adjusted maximum value.

terminal

Solution

Solution 1: Sorting + Greedy Algorithm

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

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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)

Continue Practicing

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