Problem:

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

Problem Analysis:

• Initialization:

  • Store the length of the array in a variable (n).
  • Initialize a variable (max_val) to track the maximum valid value.

• Single Pass Update:

  • Iterate through the array once.
  • Update each element in the array to be the minimum of its current value and the previous element plus one.

• Maximum Value Tracking:

  • During the iteration, track the maximum valid value encountered in the max_val variable.

• Result:

  • Return the maximum valid value (max_val) as the result.

• Time Complexity:

  • O(N * log(N)) due to the sort operation.
    • Sorting the array has a time complexity of O(N * log(N)), where N is the length of the array.
  • The subsequent single pass through the array has a linear time complexity of O(N).
  • Overall, the dominant factor is the sort operation.

• Space Complexity:

  • O(1) (constant space).
    • The algorithm uses a constant amount of extra space, regardless of the input size.
    • Sorting is typically in-place and doesn't require additional space proportional to the input size.

Solutions:

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

        max_val = 1
        for i in range(1, n):
            max_val = min(arr[i], max_val + 1)

        return max_val

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