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1664. Ways to Make a Fair Array

Difficulty Topics

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

Solution

ways-to-make-a-fair-array.py
class Solution:
    def waysToMakeFair(self, nums: List[int]) -> int:
        ro = re = 0
        for i,x in enumerate(nums):
            if i & 1: ro += x
            else: re += x

        lo = le = res = 0
        for i,x in enumerate(nums):
            if i & 1: ro -= x
            else: re -= x

            res += (lo + re) == (le + ro)

            if i & 1: lo += x
            else: le += x

        return res