# C Exercises: Check if an array can be splitted in such a position that, the sum of left side of the splitting is equal to the sum of the right side

## C Array: Exercise-99 with Solution

Write a program in C to return true if an array can be split in such a way that the left side of the splitting is equal to the sum of the right side.

**Sample Solution:**

**C Code:**

```
#include<stdio.h>
#include <stdbool.h>
// Function to check if the array can be split at a position where the sum of both sides are equal
bool canBalance(int arr1[], int n)
{
int l = n;
// Iterate through each element of the array
for (int i = 0; i < l; i++)
{
int rhs = 0, lhs = 0;
// Nested loop to calculate the sums of left-hand side (lhs) and right-hand side (rhs)
for (int k = 0; k < l; k++)
{
if (k > i)
{
lhs += arr1[k]; // Add elements to lhs if the index is greater than the current position (i)
}
else
{
rhs += arr1[k]; // Add elements to rhs if the index is less than or equal to the current position (i)
}
}
// Check if the sum of both sides (rhs and lhs) is equal
if (rhs == lhs)
{
return true; // If found, return true
}
}
return false; // Return false if no such position is found
}
int main()
{
int arr1[] = {1, 3, 3, 8, 4, 3, 2, 3, 3};
int arr_size = sizeof(arr1) / sizeof(arr1[0]);
int i;
bool bl;
// Print the original array
printf("The given array is: ");
for (i = 0; i < arr_size; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
// Check if the array can be split into two parts with equal sum and print the result
bl = canBalance(arr1, arr_size);
if (bl == true)
{
printf("The array can be split at a position where the sum of both sides are equal. ");
}
else
{
printf("The array cannot be split at any position where the sum of both sides are equal. ");
}
return 0;
}
```

Sample Output:

The given array is : 1 3 3 8 4 3 2 3 3 The array can be split in a position where the sum of both side are equal.

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