Maximum Element After Decreasing and Rearranging

To satisfy the given conditions, we first sort the array arr to make sure the differences between adjacent elements can be handled more easily. Then we iterate through the sorted array and adjust the values as needed to ensure that the absolute difference between any two adjacent elements is less than or equal to 1.

Here’s the code:

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class Solution:
    def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int:
        arr.sort()

        # Set the first element to 1, as the problem specifies
        arr[0] = 1

        # Iterate through the rest of the sorted array, adjusting values as needed
        for i in range(1, len(arr)):
            # If the difference between the current element and the previous element is greater than 1,
            # we set the current element to one more than the previous element
            if arr[i] - arr[i - 1] > 1:
                arr[i] = arr[i - 1] + 1

        # The largest element will be the last element in the adjusted array
        return arr[-1]

Explanation:

• We first sort the array to allow easier handling of adjacent elements.
• The first element must be 1, so we set it explicitly.
• We then iterate through the rest of the sorted array, checking the difference between adjacent elements. If the difference is greater than 1, we set the current element to one more than the previous element, as this will satisfy the given conditions.
• Finally, we return the last element in the adjusted array, as it will be the largest element.