Split Array With Same Average in C++

Suppose we have one array A, we must move every element of A to either list B or list C. (These lists B and C are initially empty.) We have to check whether after such a move, it is possible that the average value of B is equal to the average value of C, and B and C are both non-empty.

So if the input is like − [1,2,3,4,5,6,7,8,9,10], then the result will be true,

To solve this, we will follow these steps −

• n := size of A, total := 0
• for initialize i := 0, when i < n, update (increase i by 1), do −
  • total := total + A[i]
• isPossible := false, m := n / 2
• for initialize i := 1, when i <= m and not isPossible is non-zero, update (increase i by 1), do −
  • if total * i mod n is same as 0, then −
    • isPossible := true
• if not isPossible is non-zero, then −
  • return false
• Define one 2D array dp of size (total + 1) x (n / 2) + 1)
• dp[0, 0] := true
• for initialize i := 0, when i < n, update (increase i by 1), do −
  • x := A[i]
  • for initialize j := total, when j >= x, update (decrease j by 1), do −
    • for initialize l := 1, when l <= (n / 2), update (increase l by 1), do −
      • dp[j, l] := dp[j, l] OR dp[j - x, l - 1]
• for initialize i := 1, when i <= (n / 2), update (increase i by 1), do −
  • if (total * i) mod n is same as 0 and dp[total * i / n, i] is non-zero, then −
    • return true
• return false

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool splitArraySameAverage(vector<int>& A) {
      int n = A.size();
      int total = 0 ;
      for(int i = 0; i < n; i++) total += A[i];
      bool isPossible = false;
      int m = n / 2;
      for (int i = 1; i <= m && !isPossible; ++i)
      if (total*i%n == 0) isPossible = true;
      if (!isPossible) return false;
      vector < vector <bool> > dp(total + 1, vector <bool>((n / 2) + 1));
      dp[0][0] = true;
      for(int i = 0; i < n; i++){
         int x = A[i];
         for(int j = total; j >= x; j--){
            for(int l = 1; l <= (n / 2); l++){
               dp[j][l] = dp[j][l] || dp[j - x][l - 1];
            }
         }
      }
      for(int i = 1 ; i <= (n / 2); i++){
         if((total * i) % n == 0 && dp[total * i / n][i]) return true;
      }
      return false;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,2,3,4,5,6,7,8,9,10};
   cout << (ob.splitArraySameAverage(v));
}

Input

{1,2,3,4,5,6,7,8,9,10}

Output

1

Arnab Chakraborty

Published on 02-Jun-2020 14:49:31