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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;
};

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