Java Program for Equilibrium index of an array

Equilibrium index of an array is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in an array A: 

Example : 

Input: A[] = {-7, 1, 5, 2, -4, 3, 0} 
Output: 3 
3 is an equilibrium index, because: 
A[0] + A[1] + A[2] = A[4] + A[5] + A[6]

Input: A[] = {1, 2, 3} 
Output: -1 

Write a function int equilibrium(int[] arr, int n); that given a sequence arr[] of size n, returns an equilibrium index (if any) or -1 if no equilibrium indexes exist. 

Method 1 (Simple but inefficient) 
Use two loops. Outer loop iterates through all the element and inner loop finds out whether the current index picked by the outer loop is equilibrium index or not. Time complexity of this solution is O(n^2). 

// Java program to find equilibrium
// index of an array

class EquilibriumIndex {
    int equilibrium(int arr[], int n)
    {
        int i, j;
        int leftsum, rightsum;

        /* Check for indexes one by one until 
           an equilibrium index is found */
        for (i = 0; i < n; ++i) {

            /* get left sum */
            leftsum = 0;  
            for (j = 0; j < i; j++)
                leftsum += arr[j];

            /* get right sum */
            rightsum = 0;
            for (j = i + 1; j < n; j++)
                rightsum += arr[j];

            /* if leftsum and rightsum are same, 
               then we are done */
            if (leftsum == rightsum)
                return i;
        }

        /* return -1 if no equilibrium index is found */
        return -1;
    }
    // Driver code
    public static void main(String[] args)
    {
        EquilibriumIndex equi = new EquilibriumIndex();
        int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.length;
        System.out.println(equi.equilibrium(arr, arr_size));
    }
}

// This code has been contributed by Mayank Jaiswal

3

Time Complexity: O(n^2)

Method 2 (Tricky and Efficient) 
The idea is to get the total sum of the array first. Then Iterate through the array and keep updating the left sum which is initialized as zero. In the loop, we can get the right sum by subtracting the elements one by one. Thanks to Sambasiva for suggesting this solution and providing code for this.

1) Initialize leftsum  as 0
2) Get the total sum of the array as sum
3) Iterate through the array and for each index i, do following.
    a)  Update sum to get the right sum.  
           sum = sum - arr[i] 
       // sum is now right sum
    b) If leftsum is equal to sum, then return current index. 
       // update leftsum for next iteration.
    c) leftsum = leftsum + arr[i]
4) return -1 
// If we come out of loop without returning then
// there is no equilibrium index

The image below shows the dry run of the above approach: 

Below is the implementation of the above approach: 

// Java program to find equilibrium
// index of an array

class EquilibriumIndex {
    int equilibrium(int arr[], int n)
    {
        int sum = 0; // initialize sum of whole array
        int leftsum = 0; // initialize leftsum

        /* Find sum of the whole array */
        for (int i = 0; i < n; ++i)
            sum += arr[i];

        for (int i = 0; i < n; ++i) {
            sum -= arr[i]; // sum is now right sum for index i

            if (leftsum == sum)
                return i;

            leftsum += arr[i];
        }

        /* If no equilibrium index found, then return 0 */
        return -1;
    }

   // Driver code
    public static void main(String[] args)
    {
        EquilibriumIndex equi = new EquilibriumIndex();
        int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
        int arr_size = arr.length;
        System.out.println("First equilibrium index is " + 
                          equi.equilibrium(arr, arr_size));
    }
}

// This code has been contributed by Mayank Jaiswal

First equilibrium index is 3

Output: 
First equilibrium index is 3

Time Complexity: O(n)

Method 3 :

This is a quite simple and straightforward method. The idea is to take the prefix sum of the array twice. Once from the front end of array and another from the back end of array.

After taking both prefix sums run a loop and check for some i if both the prefix sum from one array is equal to prefix sum from the second array then that point can be considered as the Equilibrium point.

// Java program to find equilibrium
// index of an array
class GFG{

static int equilibrium(int a[], int n)
{ 
    if (n == 1)
        return (0);
    
    int[] front = new int[n];
    int[] back = new int[n];

    // Taking the prefixsum from front end array
    for (int i = 0; i < n; i++)
    {
        if (i != 0)
        {
            front[i] = front[i - 1] + a[i];
        }
        else 
        {
            front[i] = a[i];
        }
    }
  
    // Taking the prefixsum from back end of array
    for (int i = n - 1; i > 0; i--) 
    {
        if (i <= n - 2) 
        {
            back[i] = back[i + 1] + a[i];
        }
        else 
        {
            back[i] = a[i];
        }
    }
    
    // Checking for equilibrium index by
    //comparing front and back sums 
    for(int i = 0; i < n; i++) 
    {
        if (front[i] == back[i])
        {
            return i;
        }
    }
    
    // If no equilibrium index found,then return -1
    return -1;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { -7, 1, 5, 2, -4, 3, 0 };
    int arr_size = arr.length;
    
    System.out.println("First Point of equilibrium " +
                       "is at index " + 
                       equilibrium(arr, arr_size));
}
}

// This code is contributed by Lovish Aggarwal

First Point of equilibrium is at index 3

Time Complexity: O(N)

Space Complexity: O(N)

Please refer complete article on Equilibrium index of an array for more details!
 

kartik

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