Given a string, the task is to find the number of characters with the same adjacent characters.
Note: First and last character will always be counted as they will have only one adjacent character.
Examples:
Input: str = “eneveropenk”
Output: 4
Characters with same adjacent characters are e, g, s, kInput: str = “eeeeeeee”
Output: 8
Characters with same adjacent characters are e, e, e, e, e, e, e, e
Approach:
- If the length of the string is less than 3 then return the length of the string.
- Initialize the count with 2 as first and last will always be counted.
- Start traversing the string.
- Check if the previous and next characters of the current character are the same.
- Increment count, if yes.
- Return count.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to count the characters// with same adjacent charactersint countChar(string str){ int n = str.length(); // if length is less than 3 // then return length as there // will be only two characters if (n <= 2) return n; int count = 2; // Traverse the string for (int i = 1; i < n - 1; i++) // Increment the count if the previous // and next character is same if (str[i - 1] == str[i + 1]) count++; // Return count return count;}// Driver codeint main(){ string str = "eneveropenk"; cout << countChar(str); return 0;} |
Java
// Java implementation of the above approach class GFG{ // Function to count the characters // with same adjacent characters static int countChar(String str) { int n = str.length(); // if length is less than 3 // then return length as there // will be only two characters if (n <= 2) return n; int count = 2; // Traverse the string for (int i = 1; i < n - 1; i++) // Increment the count if the previous // and next character is same if (str.charAt(i - 1) == str.charAt(i + 1)) count++; // Return count return count; } // Driver code public static void main(String []args) { String str = "eneveropenk"; System.out.println(countChar(str)); }}// This code is contributed// by ihritik |
Python3
# Python 3 implementation of above approach# Function to count the characters# with same adjacent charactersdef countChar(str): n = len(str) # if length is less than 3 # then return length as there # will be only two characters if (n <= 2): return n count = 2 # Traverse the string for i in range(1, n - 1): # Increment the count if the previous # and next character is same if (str[i - 1] == str[i + 1]): count += 1 # Return count return count# Driver codeif __name__ == '__main__': str = "eneveropenk" print(countChar(str)) # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of above approachusing System;class GFG{// Function to count the characters// with same adjacent charactersstatic int countChar(String str){ int n = str.Length; // if length is less than 3 // then return length as there // will be only two characters if (n <= 2) return n; int count = 2; // Traverse the string for (int i = 1; i < n - 1; i++) // Increment the count if the previous // and next character is same if (str[i - 1] == str[i + 1]) count++; // Return count return count;}// Driver codepublic static void Main(){ String str = "eneveropenk"; Console.WriteLine(countChar(str));}}// This code is contributed// by Subhadeep |
PHP
<?php// PHP implementation of above approach // Function to count the characters // with same adjacent characters function countChar($str) { $n = strlen($str); // if length is less than 3 // then return length as there // will be only two characters if ($n <= 2) return $n; $count = 2; // Traverse the string for ($i = 1; $i < $n - 1; $i++) // Increment the count if the previous // and next character is same if ($str[$i - 1] == $str[$i + 1]) $count++; // Return count return $count; } // Driver code $str = "eneveropenk"; echo countChar($str); // This code is contributed by Sach?> |
Javascript
<script>// Javascript implementation of above approach// Function to count the characters// with same adjacent charactersfunction countChar(str){ var n = str.length; // if length is less than 3 // then return length as there // will be only two characters if (n <= 2) return n; var count = 2; // Traverse the string for (var i = 1; i < n - 1; i++) // Increment the count if the previous // and next character is same if (str[i - 1] == str[i + 1]) count++; // Return count return count;}// Driver codevar str = "eneveropenk";document.write( countChar(str));</script> |
Output
4
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Example in c:
Approach:
Initialize a counter variable to zero, which will keep track of the number of characters with the same neighbors.
Initialize three variables, prev, curr, and next, to the first three characters of the string.
Traverse the string from the fourth character to the last character:
a. Update the prev, curr, and next variables to the previous, current, and next characters, respectively.
b. If the curr character is the same as both the prev and next characters, increment the counter variable.
Return the counter variable.
C++
#include <iostream>#include <cstring>using namespace std;int count_same_neighbors(char *str) { int len = strlen(str); int count = 0; char prev = str[0]; char curr = str[1]; char next = str[2]; // Traverse the string and count characters with the same neighbors for (int i = 3; i < len; i++) { prev = curr; curr = next; next = str[i]; if (prev == curr && curr == next) { count++; } } // Return the count of characters with the same neighbors return count;}int main() { char str[] = "aabbccddddee"; int count = count_same_neighbors(str); cout << "Number of characters with the same neighbors: " << count << endl; return 0;} |
C
#include <stdio.h>#include <string.h>int count_same_neighbors(char *str) { int len = strlen(str); int count = 0; char prev = str[0]; char curr = str[1]; char next = str[2]; // Traverse the string and count characters with the same neighbors for (int i = 3; i < len; i++) { prev = curr; curr = next; next = str[i]; if (prev == curr && curr == next) { count++; } } // Return the count of characters with the same neighbors return count;}int main() { char str[] = "aabbccddddee"; int count = count_same_neighbors(str); printf("Number of characters with the same neighbors: %d\n", count); return 0;} |
Java
public class Main { public static int countSameNeighbors(String str) { int len = str.length(); int count = 0; char prev = str.charAt(0); char curr = str.charAt(1); char next = str.charAt(2); // Traverse the string and count characters with the // same neighbors for (int i = 3; i < len; i++) { prev = curr; curr = next; next = str.charAt(i); if (prev == curr && curr == next) { count++; } } // Return the count of characters with the same // neighbors return count; } public static void main(String[] args) { String str = "aabbccddddee"; int count = countSameNeighbors(str); System.out.println( "Number of characters with the same neighbors: " + count); }} |
Python3
def count_same_neighbors(str): length = len(str) count = 0 prev = str[0] curr = str[1] next = str[2] # Traverse the string and count characters with the same neighbors for i in range(3, length): prev = curr curr = next next = str[i] if prev == curr and curr == next: count += 1 # Return the count of characters with the same neighbors return countstr = "aabbccddddee"count = count_same_neighbors(str)print("Number of characters with the same neighbors: ", end = "")print(count) |
C#
using System;class Program { static int CountSameNeighbors(string str) { int len = str.Length; int count = 0; char prev = str[0]; char curr = str[1]; char next = str[2]; // Traverse the string and count characters with the // same neighbors for (int i = 3; i < len; i++) { prev = curr; curr = next; next = str[i]; if (prev == curr && curr == next) { count++; } } // Return the count of characters with the same // neighbors return count; } static void Main(string[] args) { string str = "aabbccddddee"; int count = CountSameNeighbors(str); Console.WriteLine( "Number of characters with the same neighbors: " + count); }} |
Javascript
function count_same_neighbors(str) { let len = str.length; let count = 0; let prev = str[0]; let curr = str[1]; let next = str[2]; // Traverse the string and count characters with the same neighbors for (let i = 3; i < len; i++) { prev = curr; curr = next; next = str[i]; if (prev == curr && curr == next) { count++; } } // Return the count of characters with the same neighbors return count;}let str = "aabbccddddee";let count = count_same_neighbors(str);console.log("Number of characters with the same neighbors: " + count); |
Output
Number of characters with the same neighbors: 2
time complexity of O(n)
space complexity of O(1)
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