LeetCode #1664 — MEDIUM

Ways to Make a Fair Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums. You can choose exactly one index (0-indexed) and remove the element. Notice that the index of the elements may change after the removal.

For example, if nums = [6,1,7,4,1]:

Choosing to remove index 1 results in nums = [6,7,4,1].
Choosing to remove index 2 results in nums = [6,1,4,1].
Choosing to remove index 4 results in nums = [6,1,7,4].

An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values.

Return the number of indices that you could choose such that after the removal, nums is fair.

Example 1:

Input: nums = [2,1,6,4]
Output: 1
Explanation:
Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair.
Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair.
Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair.
Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair.
There is 1 index that you can remove to make nums fair.

Example 2:

Input: nums = [1,1,1]
Output: 3
Explanation: You can remove any index and the remaining array is fair.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: You cannot make a fair array after removing any index.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. You can choose exactly one index (0-indexed) and remove the element. Notice that the index of the elements may change after the removal. For example, if nums = [6,1,7,4,1]: Choosing to remove index 1 results in nums = [6,7,4,1]. Choosing to remove index 2 results in nums = [6,1,4,1]. Choosing to remove index 4 results in nums = [6,1,7,4]. An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values. Return the number of indices that you could choose such that after the removal, nums is fair.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,1,6,4]

Example 2

[1,1,1]

Example 3

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

The parity of the indices after the removed element changes.
Calculate prefix sums for even and odd indices separately to calculate for each index in O(1).

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

Upper-end input sizes

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 #1664: Ways to Make a Fair Array
class Solution {
    public int waysToMakeFair(int[] nums) {
        int s1 = 0, s2 = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            s1 += i % 2 == 0 ? nums[i] : 0;
            s2 += i % 2 == 1 ? nums[i] : 0;
        }
        int t1 = 0, t2 = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            ans += i % 2 == 0 && t2 + s1 - t1 - v == t1 + s2 - t2 ? 1 : 0;
            ans += i % 2 == 1 && t2 + s1 - t1 == t1 + s2 - t2 - v ? 1 : 0;
            t1 += i % 2 == 0 ? v : 0;
            t2 += i % 2 == 1 ? v : 0;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1664: Ways to Make a Fair Array
func waysToMakeFair(nums []int) (ans int) {
	var s1, s2, t1, t2 int
	for i, v := range nums {
		if i%2 == 0 {
			s1 += v
		} else {
			s2 += v
		}
	}
	for i, v := range nums {
		if i%2 == 0 && t2+s1-t1-v == t1+s2-t2 {
			ans++
		}
		if i%2 == 1 && t2+s1-t1 == t1+s2-t2-v {
			ans++
		}
		if i%2 == 0 {
			t1 += v
		} else {
			t2 += v
		}
	}
	return
}

# Accepted solution for LeetCode #1664: Ways to Make a Fair Array
class Solution:
    def waysToMakeFair(self, nums: List[int]) -> int:
        s1, s2 = sum(nums[::2]), sum(nums[1::2])
        ans = t1 = t2 = 0
        for i, v in enumerate(nums):
            ans += i % 2 == 0 and t2 + s1 - t1 - v == t1 + s2 - t2
            ans += i % 2 == 1 and t2 + s1 - t1 == t1 + s2 - t2 - v
            t1 += v if i % 2 == 0 else 0
            t2 += v if i % 2 == 1 else 0
        return ans

// Accepted solution for LeetCode #1664: Ways to Make a Fair Array
struct Solution;

impl Solution {
    fn ways_to_make_fair(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut left_odd = vec![0; n];
        let mut left_even = vec![0; n];
        let mut right_odd = vec![0; n];
        let mut right_even = vec![0; n];
        let mut odd_sum = 0;
        let mut even_sum = 0;
        for i in 0..n {
            left_odd[i] = odd_sum;
            left_even[i] = even_sum;
            if i % 2 == 0 {
                even_sum += nums[i];
            } else {
                odd_sum += nums[i];
            }
        }
        odd_sum = 0;
        even_sum = 0;
        for i in (0..n).rev() {
            right_odd[i] = odd_sum;
            right_even[i] = even_sum;
            if i % 2 == 0 {
                even_sum += nums[i];
            } else {
                odd_sum += nums[i];
            }
        }
        let mut res = 0;
        for i in 0..n {
            if left_odd[i] + right_even[i] == left_even[i] + right_odd[i] {
                res += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![2, 1, 6, 4];
    let res = 1;
    assert_eq!(Solution::ways_to_make_fair(nums), res);
    let nums = vec![1, 1, 1];
    let res = 3;
    assert_eq!(Solution::ways_to_make_fair(nums), res);
    let nums = vec![1, 2, 3];
    let res = 0;
    assert_eq!(Solution::ways_to_make_fair(nums), res);
}

// Accepted solution for LeetCode #1664: Ways to Make a Fair Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1664: Ways to Make a Fair Array
// class Solution {
//     public int waysToMakeFair(int[] nums) {
//         int s1 = 0, s2 = 0;
//         int n = nums.length;
//         for (int i = 0; i < n; ++i) {
//             s1 += i % 2 == 0 ? nums[i] : 0;
//             s2 += i % 2 == 1 ? nums[i] : 0;
//         }
//         int t1 = 0, t2 = 0;
//         int ans = 0;
//         for (int i = 0; i < n; ++i) {
//             int v = nums[i];
//             ans += i % 2 == 0 && t2 + s1 - t1 - v == t1 + s2 - t2 ? 1 : 0;
//             ans += i % 2 == 1 && t2 + s1 - t1 == t1 + s2 - t2 - v ? 1 : 0;
//             t1 += i % 2 == 0 ? v : 0;
//             t2 += i % 2 == 1 ? v : 0;
//         }
//         return ans;
//     }
// }

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.