# Find the only different element in an array

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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,

1. 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.
2. 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 ` `using` `namespace` `std; ` ` `  `// Function to find minimum range ` `// increments to sort an array ` `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 Code ` `int` `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 array ` `static` `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 Code ` `public` `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 Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``arr ``=` `[``10``, ``20``, ``20``, ``20``, ``20``]  ` `    ``n ``=` `len``(arr)  ` `    ``print``(findTheOnlyDifferent(arr, n))  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## 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 array ` `static` `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 Code ` `public` `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

 ` `

Output:

```0
```

Time Complexity: O(n)
Auxiliary Space: O(1)

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