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

The brute force approach for Minimum Operations to Make All Array Elements Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. For each query, we will calculate the total number of operations needed to make every element in the array equal to the query value. This is done by iterating through the entire array for each query. The optimal Optimal Solution approach improves this to O(n log n + m log n).

What algorithm or data structure is used to solve Minimum Operations to Make All Array Elements Equal?

Minimum Operations to Make All Array Elements Equal uses the following concepts: Array, Binary Search, Sorting, Prefix Sum. The recommended patterns to study are: Prefix Sum, Sorting, Binary Search.

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

Always clarify the problem statement and constraints before diving into coding. Think about the time complexity of your approach and be ready to discuss potential optimizations. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Minimum Operations to Make All Array Elements Equal?

Not resetting the array state between queries, leading to incorrect calculations. Failing to account for edge cases, such as when all elements are already equal to the query.

#2602

Minimum Operations to Make All Array Elements Equal

Medium

ArrayBinary SearchSortingPrefix SumPrefix SumSortingBinary Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n + m log n)Space O(n)

By sorting the nums array and using prefix sums, we can efficiently calculate the total operations needed for each query in linear time after the initial sort.

⚙️

Algorithm

5 steps

1Step 1: Sort the nums array.
2Step 2: Create a prefix sum array where each element at index i is the sum of the first i elements of nums.
3Step 3: For each query, use binary search to find the position where the query would fit in the sorted nums array.
4Step 4: Calculate the total operations using the prefix sums for elements less than and greater than the query.
5Step 5: Store the result for each query and return the result array.

solution.py11 lines

1def minOperations(nums, queries):
2    nums.sort()
3    prefix_sum = [0] * (len(nums) + 1)
4    for i in range(1, len(nums) + 1):
5        prefix_sum[i] = prefix_sum[i - 1] + nums[i - 1]
6    result = []
7    for query in queries:
8        idx = bisect.bisect_left(nums, query)
9        total_operations = query * idx - prefix_sum[idx] + (prefix_sum[len(nums)] - prefix_sum[idx]) - query * (len(nums) - idx)
10        result.append(total_operations)
11    return result

ℹ

Complexity note: The time complexity is O(n log n) for sorting the array and O(m log n) for processing each query using binary search, making it much more efficient than the brute force approach.

1Using prefix sums allows for quick calculations of cumulative values, reducing the need for repeated summation.
2Sorting the array enables efficient searching and comparison, which is crucial for optimizing the solution.

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