Given an array of integers where all elements are same except one element, find the only different element in the array. It may be assumed that the size of the array is at least two.
Examples:
Input : arr[] = {10, 10, 10, 20, 10, 10}
Output : 3
arr[3] is the only different element.
Input : arr[] = {30, 10, 30, 30, 30}
Output : 1
arr[1] is the only different element.
A simple solution is to traverse the array. For every element, check if it is different from others or not. The time complexity of this solution would be O(n*n)
A better solution is to use hashing. We count frequencies of all elements. The hash table will have two elements. We print the element with value (or frequency) equal to 1. This solution works in O(n) time but requires O(n) extra space.
An efficient solution is to first check the first three elements. There can be two cases,
- Two elements are same, i.e., one is different according to given conditions in the question. In this case, the different element is among the first three, so we return the different element.
- All three elements are same. In this case, the different element lies in the remaining array. So we traverse the array from the fourth element and simply check if the value of the current element is different from previous or not.
Below is the implementation of the above idea.
C++
// C++ program to find the only different// element in an array.#include <bits/stdc++.h>using namespace std;// Function to find minimum range// increments to sort an arrayint findTheOnlyDifferent(int arr[], int n){ // Array size must be at least two. if (n == 1) return -1; // If there are two elements, then we // can return any one element as different if (n == 2) return 0; // Check if the different element is among // first three if (arr[0] == arr[1] && arr[0] != arr[2]) return 2; if (arr[0] == arr[2] && arr[0] != arr[1]) return 1; if (arr[1] == arr[2] && arr[0] != arr[1]) return 0; // If we reach here, then first three elements // must be same (assuming that only one element // is different. for (int i = 3; i < n; i++) if (arr[i] != arr[i - 1]) return i; return -1;}// Driver Codeint main(){ int arr[] = { 10, 20, 20, 20, 20 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findTheOnlyDifferent(arr, n); return 0;} |
Java
// Java program to find the only different// element in an array.public class GFG{// Function to find minimum range// increments to sort an arraystatic int findTheOnlyDifferent(int[] arr, int n){ // Array size must be at least two. if (n == 1) return -1; // If there are two elements, then we // can return any one element as different if (n == 2) return 0; // Check if the different element is among // first three if (arr[0] == arr[1] && arr[0] != arr[2]) return 2; if (arr[0] == arr[2] && arr[0] != arr[1]) return 1; if (arr[1] == arr[2] && arr[0] != arr[1]) return 0; // If we reach here, then first three elements // must be same (assuming that only one element // is different. for (int i = 3; i < n; i++) if (arr[i] != arr[i - 1]) return i; return -1;}// Driver Codepublic static void main(String args[]){ int[] arr = { 10, 20, 20, 20, 20 }; int n = arr.length; System.out.println(findTheOnlyDifferent(arr, n));}// This code is contributed// by Ryuga} |
Python3
# Python3 program to find the only # different element in an array. # Function to find minimum range # increments to sort an array def findTheOnlyDifferent(arr, n): # Array size must be at least two. if n == 1: return -1 # If there are two elements, # then we can return any one # element as different if n == 2: return 0 # Check if the different element # is among first three if arr[0] == arr[1] and arr[0] != arr[2]: return 2 if arr[0] == arr[2] and arr[0] != arr[1]: return 1 if arr[1] == arr[2] and arr[0] != arr[1]: return 0 # If we reach here, then first # three elements must be same # (assuming that only one element # is different. for i in range(3, n): if arr[i] != arr[i - 1]: return i return -1# Driver Codeif __name__ == "__main__": arr = [10, 20, 20, 20, 20] n = len(arr) print(findTheOnlyDifferent(arr, n)) # This code is contributed # by Rituraj Jain |
Javascript
// JavaScript program to find the only different// element in an array.// Function to find minimum range// increments to sort an arrayfunction findTheOnlyDifferent(arr, n) { // Array size must be at least two. if (n === 1) return -1; // If there are two elements, then we // can return any one element as different if (n === 2) return 0; // Check if the different element is among // first three if (arr[0] === arr[1] && arr[0] !== arr[2]) return 2; if (arr[0] === arr[2] && arr[0] !== arr[1]) return 1; if (arr[1] === arr[2] && arr[0] !== arr[1]) return 0; // If we reach here, then first three elements // must be same (assuming that only one element // is different. for (let i = 3; i < n; i++) if (arr[i] !== arr[i - 1]) return i; return -1;}// Driver Codeconst arr = [10, 20, 20, 20, 20];const n = arr.length;console.log(findTheOnlyDifferent(arr, n)); |
C#
// C# program to find the only different// element in an array.using System;class GFG{// Function to find minimum range// increments to sort an arraystatic int findTheOnlyDifferent(int[] arr, int n){ // Array size must be at least two. if (n == 1) return -1; // If there are two elements, then we // can return any one element as different if (n == 2) return 0; // Check if the different element is among // first three if (arr[0] == arr[1] && arr[0] != arr[2]) return 2; if (arr[0] == arr[2] && arr[0] != arr[1]) return 1; if (arr[1] == arr[2] && arr[0] != arr[1]) return 0; // If we reach here, then first three elements // must be same (assuming that only one element // is different. for (int i = 3; i < n; i++) if (arr[i] != arr[i - 1]) return i; return -1;}// Driver Codepublic static void Main(){ int[] arr = { 10, 20, 20, 20, 20 }; int n = arr.Length; Console.Write(findTheOnlyDifferent(arr, n));}}// This code is contributed// by Akanksha Rai |
PHP
<?php// PHP program to find the only different// element in an array.// Function to find minimum range// increments to sort an arrayfunction findTheOnlyDifferent($arr, $n){ // Array size must be at least two. if ($n == 1) return -1; // If there are two elements, then we // can return any one element as different if ($n == 2) return 0; // Check if the different element is among // first three if ($arr[0] == $arr[1] && $arr[0] != $arr[2]) return 2; if ($arr[0] == $arr[2] && $arr[0] != $arr[1]) return 1; if ($arr[1] == $arr[2] && $arr[0] != $arr[1]) return 0; // If we reach here, then first three // elements must be same (assuming that // only one element is different. for ($i = 3; $i < $n; $i++) if ($arr[$i] != $arr[$i - 1]) return $i; return -1;}// Driver Code$arr = array(10, 20, 20, 20, 20);$n = sizeof($arr);echo findTheOnlyDifferent($arr, $n);// This code is contributed by ajit?> |
Output
0
Time Complexity: O(n)
Auxiliary Space: O(1)
SECOND APPROACH : Using STL
Another approach to this problem is by using vector method . Like we will simply copy all the element in another array and then we will simply compare the first element of new sorted array with previous one and the index which is not same will be our answer .
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>using namespace std;void solve(){ int arr[] = { 10, 20, 20, 20, 20 }; int n = sizeof(arr) / sizeof(arr[0]); int arr2[n]; for(int i=0;i<n;i++) { arr2[i]=arr[i]; } sort(arr2,arr2+n); for (int i = 0; i < n; i++) { if (arr[i] != arr2[1]) { cout << i << "\n"; } } }int main() { solve(); return 0;} |
Java
// Java program to find the only different// element in an array.import java.util.Arrays;// Function to find minimum range// increments to sort an arraypublic class Main { public static void main(String[] args) { solve(); } static void solve() { int[] arr = {10, 20, 20, 20, 20}; int n = arr.length; int[] arr2 = Arrays.copyOf(arr, n); Arrays.sort(arr2); for (int i = 0; i < n; i++) { if (arr[i] != arr2[1]) { System.out.println(i); } } }} |
Python3
# Python3 program to find the only different# Function to solve the problemdef solve(): # Initialize an array arr = [10, 20, 20, 20, 20] n = len(arr) # Create a copy of the array and sort it arr2 = arr.copy() arr2.sort() # Loop through the original array and print the indices # of elements that are not equal to the second smallest element for i in range(n): if arr[i] != arr2[1]: print(i)# Call the solve functionsolve() |
Javascript
// Function to find the second smallest element index in an arrayfunction findSecondSmallestIndex(arr) { let n = arr.length; // Create a copy of the original array let arr2 = [...arr]; // Sort the copied array arr2.sort((a, b) => a - b); // Loop through the original array for (let i = 0; i < n; i++) { // If the current element is not equal to the second smallest element if (arr[i] !== arr2[1]) { // Return the index of the current element return i; } }}// Test the functionlet arr = [10, 20, 20, 20, 20];let index = findSecondSmallestIndex(arr);console.log(` ${index}`); |
C#
using System;class MainClass { public static void Main(string[] args) { // Initialize an array int[] arr = { 10, 20, 20, 20, 20 }; int n = arr.Length; // Create a copy of the array and sort it int[] arr2 = new int[n]; Array.Copy(arr, arr2, n); Array.Sort(arr2); // Loop through the original array and print the // indices of elements that are not equal to the // second smallest element for (int i = 0; i < n; i++) { if (arr[i] != arr2[1]) { Console.WriteLine(i); } } }} |
Output
0
Time Complexity: O(n)
Auxiliary Space: O(1)
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