# C Exercises: Find the equilibrium index of an array

## C Array: Exercise-86 with Solution

Write a program in C to find the equilibrium index of an array.

Expected Output:

The given array is:

0 -4 7 -4 -2 6 -3 0

The equilibrium index found at : 7 5 0

The problem involves writing a C program to find the equilibrium index of an array, where the sum of elements on the left is equal to the sum of elements on the right. The program should iterate through the array, computing the sums dynamically to identify all such indices. The output will display the array and the positions of all equilibrium indices found.

**Sample Solution:**

**C Code:**

```
#include <stdio.h>
// Function to find equilibrium indices in the array
void findEquiIndex(int arr1[], int n)
{
int sumofleft[n]; // Initialize an array to store sums of elements to the left of each index
sumofleft[0] = 0; // Initialize the sum at index 0 as 0 (no elements to the left)
// Calculate the cumulative sum of elements to the left of each index
for (int i = 1; i < n; i++)
{
sumofleft[i] = sumofleft[i - 1] + arr1[i - 1]; // Sum of elements to the left of the current index
}
int sumofright = 0; // Initialize the sum of elements to the right as 0
// Traverse the array from right to left to check for equilibrium indices
for (int i = n - 1; i >= 0; i--)
{
if (sumofleft[i] == sumofright) // If the sum to the left matches the sum to the right
{
printf("%d ", i); // Print the index as equilibrium index
}
sumofright += arr1[i]; // Add the current element to the sum of elements to the right
}
}
int main()
{
int arr1[] = { 0, -4, 7, -4, -2, 6, -3, 0 };
int n = sizeof(arr1) / sizeof(arr1[0]);
int i;
// Print the original array
printf("The given array is: \n");
for (i = 0; i < n; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
// Find and print equilibrium indices
printf("The equilibrium index found at : ");
findEquiIndex(arr1, n);
return 0;
}
```

Output:

The given array is: 0 -4 7 -4 -2 6 -3 0 The equilibrium index found at : 7 5 0

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