How do you solve Ways to Make a Fair Array on LeetCode?

Ways to Make a Fair Array (LeetCode #1664) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The parity of indices changes after removing an element, affecting the sums.

What is the brute force solution for Ways to Make a Fair Array?

The brute force approach for Ways to Make a Fair Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each index of the array, removing the element at that index, and then calculating the sums of the remaining elements at even and odd indices to determine if the array is fair. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Ways to Make a Fair Array?

Ways to Make a Fair Array uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Array.

What are the interview tips for Ways to Make a Fair Array?

Always consider edge cases, such as arrays with all identical elements. Explain your thought process clearly while coding; it shows your understanding. Practice writing both brute-force and optimal solutions to strengthen your problem-solving skills.

What are common mistakes when solving Ways to Make a Fair Array?

Not considering how the index changes after removal, leading to incorrect sum calculations. Failing to optimize the solution, resulting in unnecessary time complexity.

#1664

Ways to Make a Fair Array

Medium

ArrayPrefix SumPrefix SumArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses prefix sums to keep track of the even and odd indexed sums as we iterate through the array, allowing us to calculate the sums in constant time after the initial setup.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total sums of even and odd indexed elements in the original array.
2Step 2: Initialize variables to keep track of the current even and odd sums as we iterate through the array.
3Step 3: For each index, calculate the new even and odd sums if the current index is removed, using the prefix sums.
4Step 4: If the adjusted sums are equal, increment the counter.
5Step 5: Return the counter after checking all indices.

solution.py20 lines

1# Full working Python code
2
3def ways_to_make_fair(nums):
4    total_even = sum(nums[i] for i in range(0, len(nums), 2))
5    total_odd = sum(nums[i] for i in range(1, len(nums), 2))
6    current_even = 0
7    current_odd = 0
8    count = 0
9    for i in range(len(nums)):
10        if i % 2 == 0:
11            total_even -= nums[i]
12        else:
13            total_odd -= nums[i]
14        if current_even + total_odd == current_odd + total_even:
15            count += 1
16        if i % 2 == 0:
17            current_even += nums[i]
18        else:
19            current_odd += nums[i]
20    return count

ℹ

Complexity note: The optimal solution runs in linear time because we only traverse the array a constant number of times, maintaining a running total of even and odd indexed sums.

1The parity of indices changes after removing an element, affecting the sums.
2Using prefix sums allows for efficient recalculation of sums after each removal.

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