What is the time complexity of Reduction Operations to Make the Array Elements Equal?

The optimal solution for Reduction Operations to Make the Array Elements Equal runs in O(n log n) time and uses O(1) space. The complexity is O(n log n) due to the sorting step, which is more efficient than the brute force approach. The subsequent operations count is O(n).

What is the brute force solution for Reduction Operations to Make the Array Elements Equal?

The brute force approach for Reduction Operations to Make the Array Elements Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach repeatedly finds the largest element and reduces it to the next largest element until all elements are equal. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Reduction Operations to Make the Array Elements Equal?

Reduction Operations to Make the Array Elements Equal uses the following concepts: Array, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Reduction Operations to Make the Array Elements Equal?

Always consider sorting as a first step when dealing with order-related problems. Think about how to reduce the problem size with each operation. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Reduction Operations to Make the Array Elements Equal?

Not considering edge cases where all elements are already equal. Failing to optimize the search for the next largest element.

#1887

Reduction Operations to Make the Array Elements Equal

Medium

ArraySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

By sorting the array first, we can efficiently determine how many operations are needed to reduce all elements to the smallest value without repeatedly searching for the largest and next largest values.

⚙️

Algorithm

3 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Initialize a variable to count operations and iterate through the array from the second last element to the first.
3Step 3: For each element, calculate the difference between it and the next element, and add this difference to the operations count.

solution.py6 lines

1def reductionOperations(nums):
2    nums.sort()
3    operations = 0
4    for i in range(len(nums) - 1):
5        operations += nums[i + 1] - nums[i]
6    return operations

ℹ

Complexity note: The complexity is O(n log n) due to the sorting step, which is more efficient than the brute force approach. The subsequent operations count is O(n).

1Sorting the array allows us to efficiently calculate the number of operations needed.
2The difference between consecutive elements in a sorted array directly gives the number of operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.