What is the time complexity of Minimum Operations to Make Elements Within K Subarrays Equal?

The optimal solution for Minimum Operations to Make Elements Within K Subarrays Equal runs in O(n log n) time and uses O(n) space. Sorting each window takes O(x log x) and we do this for n-x+1 windows, leading to O(n log n).

What is the brute force solution for Minimum Operations to Make Elements Within K Subarrays Equal?

The brute force approach for Minimum Operations to Make Elements Within K Subarrays Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible combinations of subarrays of size x and calculate the operations needed to make them equal. This approach 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 Make Elements Within K Subarrays Equal?

Minimum Operations to Make Elements Within K Subarrays Equal uses the following concepts: Array, Hash Table, Math, Dynamic Programming, Sliding Window, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make Elements Within K Subarrays Equal?

Explain your thought process clearly. Discuss edge cases, like small arrays. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Operations to Make Elements Within K Subarrays Equal?

Not considering non-overlapping subarrays. Forgetting to sort the window for median calculation.

#3505

Minimum Operations to Make Elements Within K Subarrays Equal

Hard

ArrayHash TableMathDynamic ProgrammingSliding WindowHeap (Priority Queue)Hash 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)

Utilize the sliding window technique to efficiently calculate the operations needed for each window of size x. The median minimizes the total operations.

⚙️

Algorithm

3 steps

1Step 1: Use a sliding window of size x to calculate the median for each window.
2Step 2: For each window, compute the operations needed to make all elements equal to the median.
3Step 3: Track the minimum operations for at least k non-overlapping windows.

solution.py10 lines

1import numpy as np
2
3def min_operations(nums, x, k):
4    n = len(nums)
5    ops = []
6    for i in range(n - x + 1):
7        median = int(np.median(nums[i:i+x]))
8        ops.append(sum(abs(nums[j] - median) for j in range(i, i + x)))
9    ops.sort()
10    return sum(ops[:k])

ℹ

Complexity note: Sorting each window takes O(x log x) and we do this for n-x+1 windows, leading to O(n log n).

1Using the median minimizes operations for equalizing numbers.
2Sliding windows allow efficient calculation of overlapping subarrays.

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