<p>You are given an integer array <code>nums</code>.</p>

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

<p>Return <code>true</code> if it is possible to achieve that and <code>false</code> otherwise.</p>

<p><strong>Note</strong> that for an array <code>arr</code>, <code>average(arr)</code> is the sum of all the elements of <code>arr</code> over the length of <code>arr</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 30</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>


class Solution {
public:
    bool splitArraySameAverage(vector<int>& nums) {
        int n = nums.size();
        int totalSum = std::accumulate(nums.begin(), nums.end(), 0);

        // Subset A must have k elements with sum k* totalSum/n.
        for (int k =1; k<=n /2; ++k){
            if ((totalSum *k)% n == 0){
                int targetSum = (totalSum *k)/n;
                if (canPartition(nums, k, targetSum,0)){
                    return true;
                }
            }
        }
        return false;
    }
    private:
        bool canPartition(const std::vector<int>& nums, int k, int targetSum,int start){
            if (k==0) return targetSum ==0;
            if (targetSum<0) return false;

            for (int i = start; i<nums.size(); i++){
                // if there is redundant number ignore
                if (i >start && nums[i] == nums[i-1]) continue;
                if (canPartition(nums, k-1,targetSum-nums[i], i+1)){
                    return true;
                }
            }
            return false;
        }
};
12801