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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 total * i mod n is same as 0, then −
- 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 l := 1, when l <= (n / 2), update (increase l by 1), do −
- 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
- if (total * i) mod n is same as 0 and dp[total * i / n, i] is non-zero, then −
- return false
Let us see the following implementation to get better understanding −
Example
#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
Updated on: 02-Jun-2020
244 Views
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