In this tutorial, we’re going to explore a unique array manipulation challenge presented by LeetCode problem “1887. Reduction Operations to Make the Array Elements Equal”. This problem is a test of understanding array operations and optimization techniques.

Problem Statement

The task is to transform an integer array nums into an array where all elements are equal. In each operation, we can perform the following steps:

1. Find the largest element and its index.
2. Find the next largest element, strictly smaller than the largest.
3. Reduce the largest element to the next largest element.

Our goal is to determine the number of such operations required to make all elements in nums equal.

Examples

1. Input: nums = [5,1,3] Output: 3 Explanation: It takes 3 steps, reducing the largest elements step by step until all elements are 1.
2. Input: nums = [1,1,1] Output: 0 Explanation: All elements are already equal, so no operations are needed.
3. Input: nums = [1,1,2,2,3] Output: 4 Explanation: Sequentially reducing the largest elements leads to 4 operations until all elements are 1.

Solution Approach

Our approach is to sort the array and count the number of operations required to make elements equal. We exploit the fact that reducing a number to its next largest number can be done in bulk for numbers of the same value.

Step-by-Step Solution

Sort the Array:

• Sort nums in ascending order to efficiently compare adjacent elements.

Initialize the Counter:

• Set an operation counter ans to zero.

Iterate Backwards:

• Traverse the array from the end to the start.
• For each unique element, count how many steps it takes to reduce this and all larger elements to this value.

Return the Result:

• The ans variable will contain the total number of operations required.

Implementation

Here’s the Python code for the solution:

class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        nums.sort()  # Sorting the array

        size = len(nums)
        ans = 0

        # Iterate through the array from the end to start
        for i in range(size - 1, 0, -1):
            # If the current element is different from the next,
            # increment the answer by the number of elements remaining
            if nums[i - 1] != nums[i]:
                ans += size - i

        return ans

Complexity Analysis

• Time Complexity: O(n log n) due to sorting.
• Space Complexity: O(1) as no extra space is used apart from variables.

Conclusion

This problem demonstrates an efficient way to transform an array into a uniform array with minimal operations. By carefully analyzing the sorted array and performing bulk operations, we achieve the desired outcome efficiently.