# 1664. Ways to Make a Fair Array

## Description

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

## Solutions

• 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;
}
}

• class Solution {
public:
int waysToMakeFair(vector<int>& nums) {
int s1 = 0, s2 = 0;
int n = nums.size();
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;
ans += i % 2 == 1 && t2 + s1 - t1 == t1 + s2 - t2 - v;
t1 += i % 2 == 0 ? v : 0;
t2 += i % 2 == 1 ? v : 0;
}
return ans;
}
};

• 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


• 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
}

• /**
* @param {number[]} nums
* @return {number}
*/
var waysToMakeFair = function (nums) {
let [s1, s2, t1, t2] = [0, 0, 0, 0];
const n = nums.length;
for (let i = 0; i < n; ++i) {
if (i % 2 == 0) {
s1 += nums[i];
} else {
s2 += nums[i];
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
const v = nums[i];
ans += i % 2 == 0 && t2 + s1 - t1 - v == t1 + s2 - t2;
ans += i % 2 == 1 && t2 + s1 - t1 == t1 + s2 - t2 - v;
t1 += i % 2 == 0 ? v : 0;
t2 += i % 2 == 1 ? v : 0;
}
return ans;
};