LeetCode #2449 — HARD

Minimum Number of Operations to Make Arrays Similar

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given two positive integer arrays nums and target, of the same length.

In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and:

set nums[i] = nums[i] + 2 and
set nums[j] = nums[j] - 2.

Two arrays are considered to be similar if the frequency of each element is the same.

Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

Example 1:

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

Step 01

Brute Force Baseline

Problem summary: You are given two positive integer arrays nums and target, of the same length. In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and: set nums[i] = nums[i] + 2 and set nums[j] = nums[j] - 2. Two arrays are considered to be similar if the frequency of each element is the same. Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[8,12,6]
[2,14,10]

Example 2

[1,2,5]
[4,1,3]

Example 3

[1,1,1,1,1]
[1,1,1,1,1]

Core Insight

What unlocks the optimal approach

Solve for even and odd numbers separately.
Greedily match smallest even element from nums to smallest even element from target, then similarly next smallest element and so on.
Similarly, match odd elements too.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
class Solution {
    public long makeSimilar(int[] nums, int[] target) {
        Arrays.sort(nums);
        Arrays.sort(target);
        List<Integer> a1 = new ArrayList<>();
        List<Integer> a2 = new ArrayList<>();
        List<Integer> b1 = new ArrayList<>();
        List<Integer> b2 = new ArrayList<>();
        for (int v : nums) {
            if (v % 2 == 0) {
                a1.add(v);
            } else {
                a2.add(v);
            }
        }
        for (int v : target) {
            if (v % 2 == 0) {
                b1.add(v);
            } else {
                b2.add(v);
            }
        }
        long ans = 0;
        for (int i = 0; i < a1.size(); ++i) {
            ans += Math.abs(a1.get(i) - b1.get(i));
        }
        for (int i = 0; i < a2.size(); ++i) {
            ans += Math.abs(a2.get(i) - b2.get(i));
        }
        return ans / 4;
    }
}

// Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
func makeSimilar(nums []int, target []int) int64 {
	sort.Ints(nums)
	sort.Ints(target)
	a1, a2, b1, b2 := []int{}, []int{}, []int{}, []int{}
	for _, v := range nums {
		if v%2 == 0 {
			a1 = append(a1, v)
		} else {
			a2 = append(a2, v)
		}
	}
	for _, v := range target {
		if v%2 == 0 {
			b1 = append(b1, v)
		} else {
			b2 = append(b2, v)
		}
	}
	ans := 0
	for i := 0; i < len(a1); i++ {
		ans += abs(a1[i] - b1[i])
	}
	for i := 0; i < len(a2); i++ {
		ans += abs(a2[i] - b2[i])
	}
	return int64(ans / 4)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
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

// Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
// class Solution {
//     public long makeSimilar(int[] nums, int[] target) {
//         Arrays.sort(nums);
//         Arrays.sort(target);
//         List<Integer> a1 = new ArrayList<>();
//         List<Integer> a2 = new ArrayList<>();
//         List<Integer> b1 = new ArrayList<>();
//         List<Integer> b2 = new ArrayList<>();
//         for (int v : nums) {
//             if (v % 2 == 0) {
//                 a1.add(v);
//             } else {
//                 a2.add(v);
//             }
//         }
//         for (int v : target) {
//             if (v % 2 == 0) {
//                 b1.add(v);
//             } else {
//                 b2.add(v);
//             }
//         }
//         long ans = 0;
//         for (int i = 0; i < a1.size(); ++i) {
//             ans += Math.abs(a1.get(i) - b1.get(i));
//         }
//         for (int i = 0; i < a2.size(); ++i) {
//             ans += Math.abs(a2.get(i) - b2.get(i));
//         }
//         return ans / 4;
//     }
// }

// Accepted solution for LeetCode #2449: Minimum Number of Operations to Make Arrays Similar
function makeSimilar(nums: number[], target: number[]): number {
    nums.sort((a, b) => a - b);
    target.sort((a, b) => a - b);

    const a1: number[] = [];
    const a2: number[] = [];
    const b1: number[] = [];
    const b2: number[] = [];

    for (const v of nums) {
        if (v % 2 === 0) {
            a1.push(v);
        } else {
            a2.push(v);
        }
    }

    for (const v of target) {
        if (v % 2 === 0) {
            b1.push(v);
        } else {
            b2.push(v);
        }
    }

    let ans = 0;
    for (let i = 0; i < a1.length; ++i) {
        ans += Math.abs(a1[i] - b1[i]);
    }

    for (let i = 0; i < a2.length; ++i) {
        ans += Math.abs(a2[i] - b2[i]);
    }

    return ans / 4;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.