# 805. Split Array With Same Average

## Description

You are given an integer array nums.

You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B).

Return true if it is possible to achieve that and false otherwise.

Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr.

Example 1:

Input: nums = [1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.


Example 2:

Input: nums = [3,1]
Output: false


Constraints:

• 1 <= nums.length <= 30
• 0 <= nums[i] <= 104

## Solutions

• class Solution {
public boolean splitArraySameAverage(int[] nums) {
int n = nums.length;
if (n == 1) {
return false;
}
int s = Arrays.stream(nums).sum();
for (int i = 0; i < n; ++i) {
nums[i] = nums[i] * n - s;
}
int m = n >> 1;
Set<Integer> vis = new HashSet<>();
for (int i = 1; i < 1 << m; ++i) {
int t = 0;
for (int j = 0; j < m; ++j) {
if (((i >> j) & 1) == 1) {
t += nums[j];
}
}
if (t == 0) {
return true;
}
vis.add(t);
}
for (int i = 1; i < 1 << (n - m); ++i) {
int t = 0;
for (int j = 0; j < (n - m); ++j) {
if (((i >> j) & 1) == 1) {
t += nums[m + j];
}
}
if (t == 0 || (i != (1 << (n - m)) - 1) && vis.contains(-t)) {
return true;
}
}
return false;
}
}

• class Solution {
public:
bool splitArraySameAverage(vector<int>& nums) {
int n = nums.size();
if (n == 1) return false;
int s = accumulate(nums.begin(), nums.end(), 0);
for (int& v : nums) v = v * n - s;
int m = n >> 1;
unordered_set<int> vis;
for (int i = 1; i < 1 << m; ++i) {
int t = 0;
for (int j = 0; j < m; ++j)
if (i >> j & 1) t += nums[j];
if (t == 0) return true;
vis.insert(t);
}
for (int i = 1; i < 1 << (n - m); ++i) {
int t = 0;
for (int j = 0; j < (n - m); ++j)
if (i >> j & 1) t += nums[m + j];
if (t == 0 || (i != (1 << (n - m)) - 1 && vis.count(-t))) return true;
}
return false;
}
};

• class Solution:
def splitArraySameAverage(self, nums: List[int]) -> bool:
n = len(nums)
if n == 1:
return False
s = sum(nums)
for i, v in enumerate(nums):
nums[i] = v * n - s
m = n >> 1
vis = set()
for i in range(1, 1 << m):
t = sum(v for j, v in enumerate(nums[:m]) if i >> j & 1)
if t == 0:
return True
vis.add(t)
for i in range(1, 1 << (n - m)):
t = sum(v for j, v in enumerate(nums[m:]) if i >> j & 1)
if t == 0 or (i != (1 << (n - m)) - 1 and -t in vis):
return True
return False


• func splitArraySameAverage(nums []int) bool {
n := len(nums)
if n == 1 {
return false
}
s := 0
for _, v := range nums {
s += v
}
for i, v := range nums {
nums[i] = v*n - s
}
m := n >> 1
vis := map[int]bool{}
for i := 1; i < 1<<m; i++ {
t := 0
for j, v := range nums[:m] {
if (i >> j & 1) == 1 {
t += v
}
}
if t == 0 {
return true
}
vis[t] = true
}
for i := 1; i < 1<<(n-m); i++ {
t := 0
for j, v := range nums[m:] {
if (i >> j & 1) == 1 {
t += v
}
}
if t == 0 || (i != (1<<(n-m))-1 && vis[-t]) {
return true
}
}
return false
}