package LeetCode_1664
/**
* 1664. Ways to Make a Fair Array
* https://leetcode.com/problems/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. 1 <= nums.length <= 105
2. 1 <= nums[i] <= 104
* */
class Solution {
/*
* solution: increasing and decreasing even, odd; then compare two side if equals,
* Time:O(n), Space:O(1)
* */
fun waysToMakeFair(nums: IntArray): Int {
var even = 0
var odd = 0
for (i in nums.indices) {
if (i % 2 == 0) {
even += nums[i]
} else {
odd += nums[i]
}
}
var even2 = 0
var odd2 = 0
var result = 0
for (i in nums.indices) {
//decreasing
if (i % 2 == 0) {
even -= nums[i]
} else {
odd -= nums[i]
}
//compare two side
if (even + odd2 == odd + even2) {
result++
}
//increasing
if (i % 2 == 0) {
even2 += nums[i]
} else {
odd2 += nums[i]
}
}
return result
}
}