Given a sequence arr[] of size n, Write a function int equilibrium(int[] arr, int n) that returns an equilibrium index (if any) or -1 if no equilibrium index exists.
The 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.
Examples:
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
Naive approach: To solve the problem follow the below idea:
Use two loops. The 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
Steps to solve the problem:
1. iterate through i=1 to n:
*declare a leftsum variable to zero.
*iterate through 0 till i and add arr[i] to leftsum.
*declare a rightsum variable to zero.
*iterate through i+1 till n and add arr[i] to rightsum.
*check if leftsum is equal to rightsum than return arr[i].
2. return -1 in case of no point.
Below is the implementation of the above approach:
C++
// C++ program to find equilibrium// index of an array#include <bits/stdc++.h>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 codeint main(){ int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; int arr_size = sizeof(arr) / sizeof(arr[0]); // Function call 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 <stdio.h>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 codeint main(){ int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; int arr_size = sizeof(arr) / sizeof(arr[0]); // Function call printf("%d", equilibrium(arr, arr_size)); getchar(); return 0;} |
Java
// Java program to find equilibrium// index of an arrayclass 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; // Function call System.out.println(equi.equilibrium(arr, arr_size)); }}// This code has been contributed by Mayank Jaiswal |
Python3
# Python3 program to find equilibrium# index of an array# function to find the equilibrium indexdef 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 codeif __name__ == "__main__": arr = [-7, 1, 5, 2, -4, 3, 0] # Function call print(equilibrium(arr))# This code is contributed by Abhishek Sharama |
C#
// C# program to find equilibrium// index of an arrayusing 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; // Function call Console.Write(equilibrium(arr, arr_size)); }}// This code is contributed by Sam007 |
PHP
<?php // PHP program to find equilibrium // index of an array function equilibrium($arr, $n) { $i; $j; $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 $arr = array( -7, 1, 5, 2, -4, 3, 0 ); $arr_size = sizeof($arr); // Function callecho equilibrium($arr, $arr_size); // This code is contributed // by akt_mit?> |
Javascript
<script>// JavaScript Program to find equilibrium// index of an arrayfunction equilibrium(arr, n){ var i, j; var 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(let j = 0; j < i; j++) leftsum += arr[j]; /*get right sum*/ rightsum = 0; for(let 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 var arr = new Array(-7,1,5,2,-4,3,0); n = arr.length; document.write(equilibrium(arr,n)); // This code is contributed by simranarora5sos </script> |
Output
3
Time Complexity: O(N2)
Auxiliary Space: O(1)
Equilibrium index of an Array using Array-Sum:
To solve the problem follow the below idea:
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.
Follow the given steps to solve the problem:
- Initialize leftsum as 0
- Get the total sum of the array as sum
- Iterate through the array and for each index i, do the following:
- Update the sum to get the right sum.
- sum = sum – arr[i]
- The sum is now the right sum
- If leftsum is equal to the sum, then return the current index.
- update left sum for the next iteration.
- leftsum = leftsum + arr[i]
- Return -1
- If we come out of the 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 <bits/stdc++.h>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 codeint main(){ int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; int arr_size = sizeof(arr) / sizeof(arr[0]); // Function call 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 <stdio.h>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 codeint main(){ int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; int arr_size = sizeof(arr) / sizeof(arr[0]); // Function call printf("First equilibrium index is %d", equilibrium(arr, arr_size)); getchar(); return 0;} |
Java
// Java program to find equilibrium// index of an arrayclass 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; // Function call 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 indexdef 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 codeif __name__ == "__main__": arr = [-7, 1, 5, 2, -4, 3, 0] # Function call print('First equilibrium index is ', equilibrium(arr))# This code is contributed by Abhishek Sharma |
C#
// C# program to find the equilibrium// index of an arrayusing 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; // Function call Console.Write("First equilibrium index is " + equilibrium(arr, arr_size)); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to find equilibrium // index of an array function equilibrium($arr, $n) { $sum = 0; // initialize sum of // whole array $leftsum = 0; // initialize leftsum /* Find sum of the whole array */ for ($i = 0; $i < $n; ++$i) $sum += $arr[$i]; for ($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$arr = array( -7, 1, 5, 2, -4, 3, 0 ); $arr_size = sizeof($arr); // Function callecho "First equilibrium index is ", equilibrium($arr, $arr_size); // This code is contributed by ajit?> |
Javascript
<script>// program to find equilibrium// index of an array function equilibrium(arr, n){ sum = 0; // initialize sum of whole array leftsum = 0; // initialize leftsum /* Find sum of the whole array */ for (let i = 0; i < n; ++i) sum += arr[i]; for (let 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 codearr =new Array(-7, 1, 5, 2, -4, 3, 0);n=arr.length;document.write("First equilibrium index is " + equilibrium(arr, n));// This code is contributed by simranarora5sos</script> |
Output
First equilibrium index is 3
Time Complexity: O(N)
Auxiliary Space: O(1)
Equilibrium index of an array using Prefix-Sum:
To solve the problem follow the below idea:
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 the array and another from the back end of the 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
Follow the given steps to solve the problem:
- Declare two arrays to store the prefix sum of the array from the front and the back
- Run a loop from 1 to N and check if at any point prefix sum of the array from the front is equal to the prefix sum of the array from the back
- If any such index is found then return that index
- Else return -1
Below is the implementation of the above approach:
C++
// C++ program to find equilibrium index of an array#include <bits/stdc++.h>using namespace std;int equilibrium(int a[], int n){ if (n == 1) return (0); int forward[n] = { 0 }; int rev[n] = { 0 }; // Taking the prefixsum from front end array //As We Know that in prefixsum from front end the first prefixsum will be equal to first element in the array //As well as from the back end the first prefixsum will be equal to last element in the array forward[0]=a[0]; rev[n-1]=a[n-1]; for (int i = 1; i < n; i++) { forward[i] = forward[i - 1] + a[i]; } // Taking the prefixsum from back end of array for (int i = n - 2; i >= 0; i--) { rev[i] = rev[i + 1] + a[i]; } // Checking if forward prefix sum // is equal to rev prefix // sum for (int i = 0; i < n; i++) { if (forward[i] == rev[i]) { return i; } } return -1; // If You want all the points // of equilibrium create // vector and push all equilibrium // points in it and // return the vector}// Driver codeint main(){ int arr[] = { -7, 1, 5, 2, -4, 3, 0 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << "First Point of equilibrium is at index " << equilibrium(arr, n) << "\n"; return 0;} |
Java
// Java program to find equilibrium// index of an arrayimport java.io.*;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; // Function call System.out.println("First Point of equilibrium " + "is at index " + equilibrium(arr, arr_size)); }}// This code is contributed by Lovish Aggarwal |
Python3
# Python3 program to find the equilibrium# index of an array# Function to find the equilibrium indexdef equilibrium(arr): left_sum = [] right_sum = [] # Iterate from 0 to len(arr) for i in range(len(arr)): # If i is not 0 if(i): left_sum.append(left_sum[i-1]+arr[i]) right_sum.append(right_sum[i-1]+arr[len(arr)-1-i]) else: left_sum.append(arr[i]) right_sum.append(arr[len(arr)-1]) # Iterate from 0 to len(arr) for i in range(len(arr)): if(left_sum[i] == right_sum[len(arr) - 1 - i]): return(i) # If no equilibrium index found,then return -1 return -1# Driver codeif __name__ == "__main__": arr = [-7, 1, 5, 2, -4, 3, 0] # Function call print('First equilibrium index is ', equilibrium(arr))# This code is contributed by Lokesh Sharma |
C#
// C# program to find equilibrium// index of an arrayusing System;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; // Function call Console.WriteLine("First Point of equilibrium " + "is at index " + equilibrium(arr, arr_size)); }}// This code is contributed by ukasp |
Javascript
<script>// Program to find equilibrium index of an array function equilibrium(a, n){ if (n == 1) return (0); var forward = new Array(0); var rev = new Array(0); // Taking the prefixsum from front end array for (let i = 0; i < n; i++) { if (i) { forward[i] = forward[i - 1] + a[i]; } else { forward[i] = a[i]; } } // Taking the prefixsum from back end of array for (let i = n - 1; i > 0; i--) { if (i <= n - 2) { rev[i] = rev[i + 1] + a[i]; } else { rev[i] = a[i]; } } // Checking if forward prefix sum // is equal to rev prefix // sum for (let i = 0; i < n; i++) { if (forward[i] == rev[i]) { return i; } } return -1; // If You want all the points // of equilibrium create // vector and push all equilibrium // points in it and // return the vector} // Driver code arr = new Array(-7, 1, 5, 2, -4, 3, 0); n = arr.length; document.write("First Point of equilibrium is at index " + equilibrium(arr, n) + "\n"); // This code is contributed by simranarora5sos</script> |
Output
First Point of equilibrium is at index 3
Time Complexity: O(N)
Auxiliary Space: O(N)
Equilibrium index of an array using two pointers:
The given code is trying to find the equilibrium index of an array, where an equilibrium index is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes.
The code uses two pointers, left and right, to keep track of the sum of elements to the left and right of the pivot index. It starts by initializing the left pointer to 0, pivot to 0, and right pointer to the sum of all elements in the array minus the first element.
The code then enters a while loop where the pivot index is incremented until the left and right pointers are equal, or until the pivot index is the last index in the array.
In each iteration of the loop, the pivot index is incremented, and the right pointer is decremented by the value of the element at the current pivot index. The left pointer is incremented by the value of the element at the previous pivot index.
The code then checks if the left pointer is equal to the right pointer. If it is, the current pivot index is returned as the equilibrium index. If the pivot index is the last index in the array and the left pointer is still not equal to the right pointer, the function returns -1, indicating that no equilibrium index was found.
C++
#include <iostream>#include <vector>using namespace std;int pivotIndex(vector<int>& nums) { int left = 0, pivot = 0, right = 0; for (int i = 1; i < nums.size(); i++) { right += nums[i]; } while (pivot < nums.size() - 1 && right != left) { pivot++; right -= nums[pivot]; left += nums[pivot - 1]; } return (left == right) ? pivot : -1;}int main() { vector<int> nums = {1, 7, 3, 6, 5, 6}; int result = pivotIndex(nums); cout <<"First Point of equilibrium is at index "<< result << endl; return 0;} |
Java
import java.util.List;class Solution { public int pivotIndex(List<Integer> nums) { int left = 0, pivot = 0, right = 0; for (int i = 1; i < nums.size(); i++) { right += nums.get(i); } while (pivot < nums.size() - 1 && right != left) { pivot++; right -= nums.get(pivot); left += nums.get(pivot - 1); } return (left == right) ? pivot : -1; } public static void main(String[] args) { List<Integer> nums = List.of(1, 7, 3, 6, 5, 6); int result = new Solution().pivotIndex(nums); System.out.println(result); }} |
Python3
from typing import Listdef pivotIndex(nums: List[int]) -> int: left, pivot, right = 0, 0, sum(nums)-nums[0] while pivot < len(nums)-1 and right != left: pivot += 1 right -= nums[pivot] left += nums[pivot-1] return pivot if left == right else -1def main(): # test the function with an example array nums = [1, 7, 3, 6, 5, 6] result = pivotIndex(nums) print(result)if __name__ == "__main__": main() |
Javascript
function pivotIndex(nums) { let left = 0, pivot = 0, right = nums.slice(1).reduce((a, b) => a + b, 0); while (pivot < nums.length - 1 && right !== left) { pivot++; right -= nums[pivot]; left += nums[pivot - 1]; } return (left === right) ? pivot : -1;}console.log(pivotIndex([1, 7, 3, 6, 5, 6])); |
C#
using System;using System.Linq;class Solution { public static int PivotIndex(int[] nums) { int left = 0, pivot = 0, right = nums.Skip(1).Sum(); while (pivot < nums.Length - 1 && right != left) { pivot++; right -= nums[pivot]; left += nums[pivot - 1]; } return (left == right) ? pivot : -1; } public static void Main(string[] args) { int[] nums = {1, 7, 3, 6, 5, 6}; int result = PivotIndex(nums); Console.WriteLine(result); }} |
Output
First Point of equilibrium is at index 3
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
