How do you solve Minimum Moves to Make Array Complementary on LeetCode?

Minimum Moves to Make Array Complementary (LeetCode #1674) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to pair elements and their sums is crucial for this problem.

What is the brute force solution for Minimum Moves to Make Array Complementary?

The brute force approach for Minimum Moves to Make Array Complementary is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible target sum for the pairs in the array. For each target sum, it checks how many moves are needed to make each pair of elements add up to that sum. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Moves to Make Array Complementary?

Minimum Moves to Make Array Complementary uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Moves to Make Array Complementary?

Always start with a brute force solution to understand the problem better. Explain your thought process clearly, especially when transitioning to an optimal solution. Practice identifying patterns in problems to help recognize when to use specific data structures.

What are common mistakes when solving Minimum Moves to Make Array Complementary?

Not considering the range of possible target sums correctly. Failing to account for edge cases where no moves are needed.

#1674

Minimum Moves to Make Array Complementary

Medium

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a frequency array to track how many modifications are needed for each potential target sum. This reduces the complexity by avoiding repeated calculations for each target sum.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency array to count the number of moves needed for each possible sum from 2 to 2 * limit.
2Step 2: For each pair (nums[i], nums[n-1-i]), update the frequency array based on how many moves are needed to achieve each target sum.
3Step 3: Calculate the cumulative moves needed for each target sum and find the minimum.

solution.py18 lines

1# Full working Python code
2
3def minMoves(nums, limit):
4    n = len(nums)
5    moves = [0] * (2 * limit + 2)
6    for i in range(n // 2):
7        a, b = nums[i], nums[n - 1 - i]
8        low = max(2, a + b)
9        high = min(2 * limit, a + b)
10        moves[low] += 1
11        moves[high + 1] -= 1
12    min_moves = float('inf')
13    current_moves = 0
14    for target in range(2, 2 * limit + 1):
15        current_moves += moves[target]
16        min_moves = min(min_moves, current_moves)
17    return min_moves
18

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the array a constant number of times, and the space complexity is O(n) due to the frequency array used to track modifications.

1Understanding how to pair elements and their sums is crucial for this problem.
2Using a frequency array can significantly optimize the solution by reducing the number of calculations.

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