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 + A + A = A + A + A

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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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).

C++

 // C++ program to find equilibrium // index of an array #include using namespace std;    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 int main() {     int arr[] = { -7, 1, 5, 2, -4, 3, 0 };     int arr_size = sizeof(arr) / sizeof(arr);     cout << equilibrium(arr, arr_size);     return 0; }    // This code is contributed  // by Akanksha Rai(Abby_akku)

C

 // C program to find equilibrium // index of an array    #include    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 int main() {     int arr[] = { -7, 1, 5, 2, -4, 3, 0 };     int arr_size = sizeof(arr) / sizeof(arr);     printf("%d", equilibrium(arr, arr_size));        getchar();     return 0; }

Java

 // 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

Python3

 # Python program to find equilibrium  # index of an array    # function to find the equilibrium index def equilibrium(arr):     leftsum = 0     rightsum = 0     n = len(arr)        # Check for indexes one by one      # until an equilibrium index is found     for i in range(n):         leftsum = 0         rightsum = 0                # get left sum         for j in range(i):             leftsum += arr[j]                    # get right sum         for j in range(i + 1, n):             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 arr = [-7, 1, 5, 2, -4, 3, 0] print (equilibrium(arr))    # This code is contributed by Abhishek Sharama

C#

 // C# program to find equilibrium // index of an array    using System;    class GFG {     static 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) {                // initialize left sum             // for current index i             leftsum = 0;                // initialize right sum             // for current index i             rightsum = 0;                /* get left sum */             for (j = 0; j < i; j++)                 leftsum += arr[j];                /* get right sum */             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()     {         int[] arr = { -7, 1, 5, 2, -4, 3, 0 };         int arr_size = arr.Length;            Console.Write(equilibrium(arr, arr_size));     } }    // This code is contributed by Sam007

PHP



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:

C++

 // C++ program to find equilibrium  // index of an array  #include using namespace std;    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  int main()  {      int arr[] = { -7, 1, 5, 2, -4, 3, 0 };      int arr_size = sizeof(arr) / sizeof(arr);      cout << "First equilibrium index is " << equilibrium(arr, arr_size);      return 0;  }     // This is code is contributed by rathbhupendra

C

 // C program to find equilibrium // index of an array    #include    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 int main() {     int arr[] = { -7, 1, 5, 2, -4, 3, 0 };     int arr_size = sizeof(arr) / sizeof(arr);     printf("First equilibrium index is %d",                   equilibrium(arr, arr_size));        getchar();     return 0; }

Java

 // 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

Python3

 # Python program to find the equilibrium # index of an array    # function to find the equilibrium index def equilibrium(arr):        # finding the sum of whole array     total_sum = sum(arr)     leftsum = 0     for i, num in enumerate(arr):                    # total_sum is now right sum         # for index i         total_sum -= num                    if leftsum == total_sum:             return i         leftsum += num               # If no equilibrium index found,        # then return -1     return -1        # Driver code arr = [-7, 1, 5, 2, -4, 3, 0] print ('First equilibrium index is ',        equilibrium(arr))    # This code is contributed by Abhishek Sharma

C#

 // C# program to find the equilibrium //  index of an array    using System;    class GFG {     static int equilibrium(int[] arr, int n)     {         // initialize sum of whole array         int sum = 0;            // initialize leftsum         int leftsum = 0;            /* Find sum of the whole array */         for (int i = 0; i < n; ++i)             sum += arr[i];            for (int i = 0; i < n; ++i) {                // sum is now right sum             // for index i             sum -= arr[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()     {         int[] arr = { -7, 1, 5, 2, -4, 3, 0 };         int arr_size = arr.Length;            Console.Write("First equilibrium index is " +                            equilibrium(arr, arr_size));     } } // This code is contributed by Sam007

PHP



Output:

First equilibrium index is 3

Time Complexity: O(n)

As pointed out by Sameer, we can remove the return statement and add a print statement to print all equilibrium indexes instead of returning only one.

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.

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