1664. Ways to Make a Fair Array LeetCode Solution

Last updated: February 24, 2025

In this guide, you will get 1664. Ways to Make a Fair Array LeetCode Solution with the best time and space complexity. The solution to Ways to Make a Fair Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Ways to Make a Fair Array solution in C++
Ways to Make a Fair Array solution in Java
Ways to Make a Fair Array solution in Python
Additional Resources

1664. Ways to Make a Fair Array LeetCode Solution image

Problem Statement of Ways to Make a Fair Array

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

1664. Ways to Make a Fair Array LeetCode Solution in C++

class Solution {
 public:
  int waysToMakeFair(vector<int>& nums) {
    const int n = nums.size();
    int ans = 0;
    vector<int> even(n + 1);  // the sum of even-indexed nums[0..i)
    vector<int> odd(n + 1);   // the sum of odd-indexed nums[0..i)

    for (int i = 1; i <= n; ++i) {
      odd[i] = odd[i - 1];
      even[i] = even[i - 1];
      if (i % 2 == 0)
        even[i] += nums[i - 1];
      else
        odd[i] += nums[i - 1];
    }

    const int sum = even.back() + odd.back();

    for (int i = 0; i < n; ++i) {
      const int evenSum = even[i] + odd.back() - odd[i + 1];
      const int oddSum = sum - nums[i] - evenSum;
      if (evenSum == oddSum)
        ++ans;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

1664. Ways to Make a Fair Array LeetCode Solution in Java

class Solution {
  public int waysToMakeFair(int[] nums) {
    final int n = nums.length;
    int ans = 0;
    int[] even = new int[n + 1]; // the sum of even-indexed nums[0..i)
    int[] odd = new int[n + 1];  // the sum of odd-indexed nums[0..i)

    for (int i = 1; i <= n; ++i) {
      odd[i] = odd[i - 1];
      even[i] = even[i - 1];
      if (i % 2 == 0)
        even[i] += nums[i - 1];
      else
        odd[i] += nums[i - 1];
    }

    final int sum = even[n] + odd[n];

    for (int i = 0; i < n; ++i) {
      final int evenSum = even[i] + odd[n] - odd[i + 1];
      final int oddSum = sum - nums[i] - evenSum;
      if (evenSum == oddSum)
        ++ans;
    }

    return ans;
  }
}
// code provided by PROGIEZ

1664. Ways to Make a Fair Array LeetCode Solution in Python

class Solution {
 public:
  int waysToMakeFair(vector<int>& nums) {
    const int n = nums.size();
    int ans = 0;
    // l[0] := the sum of even-indexed nums[0..i)
    // l[1] := the sum of odd-indexed nums[0..i)
    // r[0] := the sum of even-indexed nums[i + 1..n)
    // r[1] := the sum of odd-indexed nums[i + 1..n)
    vector<int> l(2);
    vector<int> r(2);

    for (int i = 0; i < n; ++i)
      r[i % 2] += nums[i];

    for (int i = 0; i < n; ++i) {
      r[i % 2] -= nums[i];
      if (l[0] + r[1] == l[1] + r[0])
        ++ans;
      l[i % 2] += nums[i];
    }

    return ans;
  }
};
 # code by PROGIEZ

Additional Resources

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