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Leetcode #1674: Minimum Moves to Make Array Complementary

In this guide, we solve Leetcode #1674 Minimum Moves to Make Array Complementary in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Hash Table, Prefix Sum

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input: nums = [1,2,4,3], limit = 4 Output: 1 Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed). nums[0] + nums[3] = 1 + 3 = 4. nums[1] + nums[2] = 2 + 2 = 4. nums[2] + nums[1] = 2 + 2 = 4. nums[3] + nums[0] = 3 + 1 = 4. Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.

Python Solution

class Solution: def minMoves(self, nums: List[int], limit: int) -> int: d = [0] * (2 * limit + 2) n = len(nums) for i in range(n // 2): x, y = nums[i], nums[-i - 1] if x > y: x, y = y, x d[2] += 2 d[x + 1] -= 2 d[x + 1] += 1 d[x + y] -= 1 d[x + y + 1] += 1 d[y + limit + 1] -= 1 d[y + limit + 1] += 2 return min(accumulate(d[2:]))

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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