1664. Ways to Make a Fair Array

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1664. Ways to Make a Fair Array

Description

You are given an integer arraynums. 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:

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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:

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

Example 3:

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

Constraints:

  • 1 <= nums.length <= 10^5
  • 1 <= nums[i] <= 10^4

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int waysToMakeFair(vector<int>& nums) {
        vector<int> preSumOdd = {0}, preSumEven = {0};
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            if (i % 2) {
                preSumOdd.push_back(preSumOdd.back() + nums[i]);
                preSumEven.push_back(preSumEven.back());
            } else {
                preSumEven.push_back(preSumEven.back() + nums[i]);
                preSumOdd.push_back(preSumOdd.back());
            }
        }
        int res = 0, total_odd = (n + 1) / 2, total_even = n - total_odd;
        for (int i = 0; i < n; i++) {
            int previousEven = preSumEven[i], previousOdd = preSumOdd[i];
            int nextEven = preSumOdd[n] - preSumOdd[i + 1], nextOdd = preSumEven[n] - preSumEven[i + 1];
            if (previousEven + nextEven == previousOdd + nextOdd) {
                res++;
            }

        }
        return res;
    }
};