How do you solve Minimum Operations to Equalize Subarrays on LeetCode?

Minimum Operations to Equalize Subarrays (LeetCode #3762) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: All elements must have the same remainder when divided by k to be equalized.

What is the time complexity of Minimum Operations to Equalize Subarrays?

The optimal solution for Minimum Operations to Equalize Subarrays runs in O(n log n) time and uses O(n) space. Sorting the subarray takes O(n log n), and checking remainders takes O(n), leading to overall efficient performance.

What is the brute force solution for Minimum Operations to Equalize Subarrays?

The brute force approach for Minimum Operations to Equalize Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible operations for each query by trying to equalize the subarray elements. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Equalize Subarrays?

Minimum Operations to Equalize Subarrays uses the following concepts: Array, Math, Binary Search, Segment Tree. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Equalize Subarrays?

Clarify the constraints and edge cases before diving into coding. Discuss your thought process and approach to the problem. Practice similar problems to strengthen your understanding of patterns.

What are common mistakes when solving Minimum Operations to Equalize Subarrays?

Overlooking the requirement for the same remainder when divided by k. Not considering edge cases where subarrays are of length 1.

#3762

Minimum Operations to Equalize Subarrays

Hard

ArrayMathBinary SearchSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

To equalize elements, they must share the same remainder when divided by k. We can transform the problem to finding the median of the adjusted values.

⚙️

Algorithm

3 steps

1Step 1: For each query, check if all elements have the same remainder when divided by k. If not, return -1.
2Step 2: Transform the elements by dividing by k and sort them.
3Step 3: Calculate the number of operations needed to make all elements equal to the median.

solution.py13 lines

1def min_operations(nums, k, queries):
2    results = []
3    for l, r in queries:
4        subarray = nums[l:r+1]
5        remainders = [num % k for num in subarray]
6        if len(set(remainders)) > 1:
7            results.append(-1)
8            continue
9        transformed = sorted(num // k for num in subarray)
10        median = transformed[len(transformed) // 2]
11        ops = sum(abs(num // k - median) for num in subarray)
12        results.append(ops)
13    return results

ℹ

Complexity note: Sorting the subarray takes O(n log n), and checking remainders takes O(n), leading to overall efficient performance.

1All elements must have the same remainder when divided by k to be equalized.
2Using the median minimizes the total operations needed to equalize values.

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