How to find an element in array such that sum of left array is equal to sum of right array

Question

here is the input and output the code should generate:

• Input : 1 4 2 5 0
• Output : 2

Explanation : If 2 is the partition, then: 1+ 4=5(LHS) and 5+0=5(RHS) should print 2, but in my case I'm not getting the output.

I tried my code with the logic that.. the n-th element would be present somewhere in between the array. So considering the sum of element of arr till n-th element will be LHS and from n-th element it would be till the end of the arr which is RHS. Also i added a while loop which increments the pointing element n.

#include <iostream.h>
#include <conio.h>
void main(){
   int arr[5],arr2[5],i,j=0,k=0,n=1;
   cout<<"input_arr\n";
   for(i=0;i<5;i++)
    cin>>arr[i];

   while(n<=5)
{
   for(i=n;i<5;n++)
      j+=arr[i];   //sum of right side arr from n-th   element
   for(i=0;i<n;i++)
    k+=arr2[i];  //sum of left side arr till n-th element
   if(k==j)
    cout<<"\n"<<arr[n];  // prints the equilbrium element
   else
         n++;
}
    getch();
}

Answer 1

First off, you want to reset j, k every iteration with this method, if I understand correctly.

Second, the second for loop increments n instead of i, for some reason.

Third, you're also declaring arr2, not initializing it, and adding values in arr2 to k for some reason.

You're including the median as well. Try n = 2.

j = arr[2] + arr[3]+arr[4] = 2 + 5 + 0 = 7
k = arr[0] + arr[1] = 1 + 4 = 0

proposed changes to while loop (to maintain the same method):

while (n <= 5) {
    j = 0; k = 0;
    for (i = n+1; i < 5; i++) {
        j+= arr[i];
    }
    for (i = 0; i < n; i++) {
        k+= arr[i]
    }
    if (k == j) {
        cout << "\n" << arr[n];
        break; //you should end the for loop here
    }
    n++;
}