Given a binary array arr[] of 1’s and 0’s of length N. The task is to find the number of elements that 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=3Input : N = 2 , arr=[1, 1]
Output : 0
Approach:
- Run a loop for all the elements of the 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 <bits/stdc++.h>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 codeint 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 approachclass 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 codepublic 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 approachdef 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 codearr = [0, 0, 0, 0, 0, 1, 0]print(distinct(arr))# This code is contributed by Mohit Kumar |
C#
// C# implementation of// the above approachusing 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 codepublic 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 |
Javascript
<script>// JavaScript implementation of// the above approachfunction distinct(arr, n){ let count = 0; // if array has only one element, return 1 if (n == 1) return 1; for ( let 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 let arr = [0, 0, 0, 0, 0, 1, 0]; let n = arr.length; document.write(distinct(arr, n));</script> |
Output
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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