#2449

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Number of Operations to Make Arrays Similar

Determine the minimum operations to make two arrays similar by adjusting pairs while respecting element frequencies and parity.

array greedy sorting

Problem Statement

You are given two positive integer arrays nums and target of equal length. In one operation, you may select two distinct indices i and j and increment one element while decrementing the other by 2 units each. The goal is to make nums similar to target in the fewest operations.

Two arrays are considered similar if they contain the same frequency of each element. Your task is to compute the minimum number of such operations required to transform nums into a similar array matching target while following parity and value constraints.

Examples

Example 1

Input: nums = [8,12,6], target = [2,14,10]

Output: 2

It is possible to make nums similar to target in two operations:

Choose i = 0 and j = 2, nums = [10,12,4].
Choose i = 1 and j = 2, nums = [10,14,2]. It can be shown that 2 is the minimum number of operations needed.

Example 2

Input: nums = [1,2,5], target = [4,1,3]

Output: 1

We can make nums similar to target in one operation:

Choose i = 1 and j = 2, nums = [1,4,3].

Example 3

Input: nums = [1,1,1,1,1], target = [1,1,1,1,1]

Output: 0

The array nums is already similiar to target.

Constraints

n == nums.length == target.length
1 <= n <= 105
1 <= nums[i], target[i] <= 106
It is possible to make nums similar to target.

Solution Approach

Sort and Split by Parity

Sort both nums and target arrays and separate even and odd numbers. This ensures parity is maintained, preventing invalid operations and aligning values for greedy adjustments.

Compute Differences

Calculate the differences between corresponding elements of the sorted parity-separated arrays. Each difference represents the total adjustment needed, and pairing largest positive with largest negative differences ensures minimal operations.

Apply Greedy Operations

Perform operations by transferring 2 units between mismatched pairs, reducing the total differences iteratively. Continue until all differences are zero, counting each move to determine the minimum number of operations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Sorting the arrays takes O(n log n) time. Computing differences and applying greedy pair operations is O(n). Space complexity is O(n) for parity-separated arrays and difference tracking.

What Interviewers Usually Probe

Checks if you consider parity separately to avoid invalid operations.
Observes whether you identify that sum invariants guarantee feasibility of transformations.
Watches if you optimize operations using sorted differences rather than brute force.

Common Pitfalls or Variants

Common pitfalls

Failing to separate even and odd numbers, causing impossible operations.
Miscounting operations by not pairing differences optimally, leading to non-minimal results.
Assuming identical sums of arrays is sufficient without verifying frequency alignment.

Follow-up variants

Allowing operations of ±k instead of ±2, requiring adjustment to the greedy pairing calculation.
Arrays with negative numbers, introducing additional parity and difference handling complexity.
Determining the minimum operations under a restricted set of indices instead of any two elements.

FAQ

What is the key strategy for Minimum Number of Operations to Make Arrays Similar?

The main strategy is to separate even and odd numbers, sort them, compute differences, and use greedy pair adjustments to minimize operations.

Do I need to match sums of nums and target before operations?

Matching sums is necessary but not sufficient; you must also align frequencies and parity to ensure valid operations.

Can I perform multiple operations on the same index?

Yes, indices can be used repeatedly as long as each operation follows the ±2 adjustment rule and parity is preserved.

What is the complexity of this approach?

Time complexity is O(n log n) for sorting and O(n) for difference computations and greedy operations. Space complexity is O(n) for temporary arrays.

Why split arrays by parity in this problem pattern?

Splitting by parity ensures all operations are valid, prevents impossible transformations, and aligns with the greedy invariant approach of this problem.

terminal

Solution

Solution 1: Odd-Even Classification + Sorting

Notice that, because each operation will only increase or decrease the value of an element by $2$, the parity of the element will not change.

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class Solution:
    def makeSimilar(self, nums: List[int], target: List[int]) -> int:
        nums.sort(key=lambda x: (x & 1, x))
        target.sort(key=lambda x: (x & 1, x))
        return sum(abs(a - b) for a, b in zip(nums, target)) // 4

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