805. Split Array With Same Average

class Solution {
    public boolean splitArraySameAverage(int[] A) {
        // for a previous i numbers in A [0, i)
        // countB = m, countC = i - m
        //  B sumB      C sumC 
        // We could have same
        // For each coming A[i] -> we have 2 choices
        // put into B -> sumB = sumB + A[i], countB = countB + 1, sumC = sumC, countC = countC
        // put into C -> sumB = sumB, countB = countB, sumC = sumC + A[i], countC = countC + 1
       
        // split into 2 groups -> avgB = sumB / countB, avgC = sumC / countC
        // sumB / countB = sumC / countC -> sumB * countC = sumC * countB -> sumB * (count - countB) = (sum - sumB) * countB
        // avg = avgB = avgC
        // sumB * count = sum * countB -> sumC * count = sum * countC
        int count = A.length;
        int sum = 0;
        for(int a : A) sum += a;
        // Can we find a subset of A whose average equals to the average of A
        // knapsack problem
       
       
       
       
       
    }
}