Related Articles
Distinct adjacent elements in a binary array
• Last Updated : 01 Aug, 2019

Given a binary array arr[] of 1’s and 0’s of length N. The task is to find the number of elements which are different with respect to their neighbors.
Note:At least one of the neighbors should be distinct.

Examples:

Input : N = 4 , arr=[1, 0, 1, 1]
Output : 3
arr[0]=1 is distinct since it’s neighbor arr[1]=0 is different.
arr[1]=0 is also distinct, as it has two different neighbors i.e, arr[2]=1 & arr[0]=1.
arr[2]=1 has same neighbor in arr[3]=1 but has different neighbor in arr[1]=0. So it’s distinct.
But arr[3]=1 is not distinct as it’s neighbor arr[2]=1 is the same.
So total distinct elements are 1+1+1+0=3

Input : N = 2 , arr=[1, 1]
Output : 0

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

Approach:

• Run a loop for all the elements of list and compare every element with its previous and next neighbors. Increment count by 1 if the element is distinct.
• The first element has to be compared only with its next neighbor and similarly the last element has to be compared only with its previous element.
• The remaining elements have two neighbors . If anyone of two neighbors is different then it is considered distinct.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of``// the above approach``#include `` ` `using` `namespace` `std;`` ` `int` `distinct(``int` `arr[], ``int` `n)``{``    ``int` `count = 0;`` ` `    ``// if array has only one element, return 1``    ``if` `(n == 1)``        ``return` `1;`` ` `    ``for` `( ``int` `i = 0; i < n - 1; i++)``    ``{``  ` `        ``// For first element compare``        ``// with only next element``        ``if``(i == 0)``        ``{``            ``if``(arr[i] != arr[i + 1])``                ``count += 1;``        ``}`` ` `        ``// For remaining elements compare with``        ``// both prev and next elements``        ``else``        ``{``            ``if``(arr[i] != arr[i + 1] || ``               ``arr[i] != arr[i - 1])``                ``count += 1;``        ``}``    ``}``     ` `    ``// For last element compare``    ``// with only prev element``    ``if``(arr[n - 1] != arr[n - 2])``        ``count += 1;`` ` `    ``return` `count;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = {0, 0, 0, 0, 0, 1, 0};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << distinct(arr, n);``     ` `    ``return` `0;``}`

## Java

 `// Java implementation of``// the above approach``class` `GFG ``{``static` `int` `distinct(``int` `[]arr, ``int` `n)``{``    ``int` `count = ``0``;`` ` `    ``// if array has only one element, ``    ``// return 1``    ``if` `(n == ``1``)``        ``return` `1``;`` ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++)``    ``{`` ` `        ``// For first element compare``        ``// with only next element``        ``if``(i == ``0``)``        ``{``            ``if``(arr[i] != arr[i + ``1``])``                ``count += ``1``;``        ``}`` ` `        ``// For remaining elements compare with``        ``// both prev and next elements``        ``else``        ``{``            ``if``(arr[i] != arr[i + ``1``] || ``               ``arr[i] != arr[i - ``1``])``                ``count += ``1``;``        ``}``    ``}``     ` `    ``// For last element compare``    ``// with only prev element``    ``if``(arr[n - ``1``] != arr[n - ``2``])``        ``count += ``1``;`` ` `    ``return` `count;``}`` ` `// Driver code``public` `static` `void` `main(String[] args) ``{``    ``int` `arr[] = {``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0``};``    ``int` `n = arr.length;``    ``System.out.println(distinct(arr, n));``}``}`` ` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of ``# the above approach``def` `distinct(arr):``    ``count ``=` `0`` ` `    ``# if array has only one element, return 1``    ``if` `len``(arr) ``=``=` `1``:``        ``return` `1``     ` `    ``for` `i ``in` `range``(``0``, ``len``(arr) ``-` `1``):`` ` `        ``# For first element compare``        ``# with only next element``        ``if``(i ``=``=` `0``):``            ``if``(arr[i] !``=` `arr[i ``+` `1``]):``                ``count ``+``=` `1`` ` `        ``# For remaining elements compare with``        ``# both prev and next elements``        ``elif``(i > ``0` `& i < ``len``(arr) ``-` `1``):``            ``if``(arr[i] !``=` `arr[i ``+` `1``] ``or` `               ``arr[i] !``=` `arr[i ``-` `1``]):``                ``count ``+``=` `1`` ` `    ``# For last element compare ``    ``# with only prev element``    ``if``(arr[``len``(arr) ``-` `1``] !``=` `arr[``len``(arr) ``-` `2``]):``        ``count ``+``=` `1``    ``return` `count`` ` `# Driver code``arr ``=` `[``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0``]`` ` `print``(distinct(arr))`` ` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of``// the above approach``using` `System;``     ` `class` `GFG ``{``static` `int` `distinct(``int` `[]arr, ``int` `n)``{``    ``int` `count = 0;`` ` `    ``// if array has only one element, ``    ``// return 1``    ``if` `(n == 1)``        ``return` `1;`` ` `    ``for` `(``int` `i = 0; i < n - 1; i++)``    ``{`` ` `        ``// For first element compare``        ``// with only next element``        ``if``(i == 0)``        ``{``            ``if``(arr[i] != arr[i + 1])``                ``count += 1;``        ``}`` ` `        ``// For remaining elements compare with``        ``// both prev and next elements``        ``else``        ``{``            ``if``(arr[i] != arr[i + 1] || ``               ``arr[i] != arr[i - 1])``                ``count += 1;``        ``}``    ``}``     ` `    ``// For last element compare``    ``// with only prev element``    ``if``(arr[n - 1] != arr[n - 2])``        ``count += 1;`` ` `    ``return` `count;``}`` ` `// Driver code``public` `static` `void` `Main(String[] args) ``{``    ``int` `[]arr = {0, 0, 0, 0, 0, 1, 0};``    ``int` `n = arr.Length;``    ``Console.WriteLine(distinct(arr, n));``}``}`` ` `// This code is contributed by Princi Singh`
Output:
```3
```

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up