What is the time complexity of Minimum Number of Operations to Make Arrays Similar?

The optimal solution for Minimum Number of Operations to Make Arrays Similar runs in O(n log n) time and uses O(1) space. The sorting step dominates the time complexity, making it O(n log n), while we only use a constant amount of extra space for counters.

What is the brute force solution for Minimum Number of Operations to Make Arrays Similar?

The brute force approach for Minimum Number of Operations to Make Arrays Similar is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible pair of indices to adjust the values in the nums array to match the target array. This method is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make Arrays Similar?

Minimum Number of Operations to Make Arrays Similar uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting, Two Pointers.

What are the interview tips for Minimum Number of Operations to Make Arrays Similar?

Always clarify the problem constraints and ensure you understand the operations allowed. Start with a brute force solution to understand the problem before optimizing. Discuss your thought process and reasoning behind each step during the interview.

What are common mistakes when solving Minimum Number of Operations to Make Arrays Similar?

Not considering the need to balance both excess and deficit properly. Failing to sort the arrays before processing, which can lead to incorrect calculations.

#2449

Minimum Number of Operations to Make Arrays Similar

Hard

ArrayGreedySortingGreedySortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal approach sorts both arrays and calculates the difference in values. By focusing on the excess and deficit of values, we can efficiently determine how many operations are needed to balance the two arrays.

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Algorithm

3 steps

1Step 1: Sort both nums and target arrays.
2Step 2: Separate the excess and deficit values for both arrays based on their differences.
3Step 3: Calculate the total excess and deficit, and divide by 2 to find the number of operations needed.

solution.py14 lines

1# Full working Python code
2
3def min_operations(nums, target):
4    nums.sort()
5    target.sort()
6    excess = 0
7    deficit = 0
8    for n, t in zip(nums, target):
9        if n > t:
10            excess += (n - t) // 2
11        elif n < t:
12            deficit += (t - n) // 2
13    return max(excess, deficit)
14

ℹ

Complexity note: The sorting step dominates the time complexity, making it O(n log n), while we only use a constant amount of extra space for counters.

1Separating excess and deficit values helps in understanding how many operations are needed.
2Sorting both arrays allows for a structured way to compare and balance values.

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