Given a string S, remove all the consecutive duplicates. Note that this problem is different from Recursively remove all adjacent duplicates. Here we keep one character and remove all subsequent same characters.
Examples:
Input : aaaaabbbbbb Output : ab Input : geeksforgeeks Output : geksforgeks Input : aabccba Output : abcba
Recursive Solution:
The above problem can be solved using recursion.
- If the string is empty, return.
- Else compare the adjacent characters of the string. If they are same then shift the characters one by one to the left. Call recursion on string S
- If they not same then call recursion from S+1 string.
The recursion tree for the string S = aabcca is shown below.
aabcca S = aabcca
/
abcca S = abcca
/
bcca S = abcca
/
cca S = abcca
/
ca S = abca
/
a S = abca (Output String)
/
empty string
Below is the implementation of the above approach:
// Recursive Program to remove consecutive // duplicates from string S. #include <bits/stdc++.h> using namespace std; // A recursive function that removes // consecutive duplicates from string S void removeDuplicates(char* S) { // When string is empty, return if (S[0] == '\0') return; // if the adjacent characters are same if (S[0] == S[1]) { // Shift character by one to left int i = 0; while (S[i] != '\0') { S[i] = S[i + 1]; i++; } // Check on Updated String S removeDuplicates(S); } // If the adjacent characters are not same // Check from S+1 string address removeDuplicates(S + 1); } // Driver Program int main() { char S1[] = "geeksforgeeks"; removeDuplicates(S1); cout << S1 << endl; char S2[] = "aabcca"; removeDuplicates(S2); cout << S2 << endl; return 0; } |
Output:
geksforgeks abca
The worst case time complexity of the above solution is O(n2). The worst case happens when all characters are same.
Iterative Solution :
The idea is to keep track of two indexes, index of current character in str and index of next distinct character in str. Whenever we see same character, we only increment current character index. We see different character, we increment index of distinct character.
C++
// C++ program to remove consecutive // duplicates from a given string. #include <bits/stdc++.h> using namespace std; // A iterative function that removes // consecutive duplicates from string S void removeDuplicates(char S[]){ int n = strlen(S); // We don't need to do anything for // empty or single character string. if (n < 2) return; // j is used to store index is result // string (or index of current distinct // character) int j = 0; // Traversing string for (int i=1; i<n; i++) { // If current character S[i] // is different from S[j] if (S[j] != S[i]) { j++; S[j] = S[i]; } } // Putting string termination // character. j++; S[j] = '\0'; } // Driver Program int main() { char S1[] = "geeksforgeeks"; removeDuplicates(S1); cout << S1 << endl; char S2[] = "aabcca"; removeDuplicates(S2); cout << S2 << endl; return 0; } |
Java
// Java program to remove consecutive // duplicates from a given string. import java.util.*; class GFG { // A iterative function that removes // consecutive duplicates from string S static void removeDuplicates(char[] S) { int n = S.length; // We don't need to do anything for // empty or single character string. if (n < 2) { return; } // j is used to store index is result // string (or index of current distinct // character) int j = 0; // Traversing string for (int i = 1; i < n; i++) { // If current character S[i] // is different from S[j] if (S[j] != S[i]) { j++; S[j] = S[i]; } } System.out.println(Arrays.copyOfRange(S, 0, j + 1)); } // Driver code public static void main(String[] args) { char S1[] = "geeksforgeeks".toCharArray(); removeDuplicates(S1); char S2[] = "aabcca".toCharArray(); removeDuplicates(S2); } } /* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 program to remove consecutive # duplicates from a given string. # A iterative function that removes # consecutive duplicates from string S def removeDuplicates(S): n = len(S) # We don't need to do anything for # empty or single character string. if (n < 2) : return # j is used to store index is result # string (or index of current distinct # character) j = 0 # Traversing string for i in range(n): # If current character S[i] # is different from S[j] if (S[j] != S[i]): j += 1 S[j] = S[i] # Putting string termination # character. j += 1 S = S[:j] return S # Driver Code if __name__ == '__main__': S1 = "geeksforgeeks" S1 = list(S1.rstrip()) S1 = removeDuplicates(S1) print(*S1, sep = "") S2 = "aabcca" S2 = list(S2.rstrip()) S2 = removeDuplicates(S2) print(*S2, sep = "") # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to remove consecutive // duplicates from a given string. using System; class GFG { // A iterative function that removes // consecutive duplicates from string S static void removeDuplicates(char[] S) { int n = S.Length; // We don't need to do anything for // empty or single character string. if (n < 2) { return; } // j is used to store index is result // string (or index of current distinct // character) int j = 0; // Traversing string for (int i = 1; i < n; i++) { // If current character S[i] // is different from S[j] if (S[j] != S[i]) { j++; S[j] = S[i]; } } char []A = new char[j+1]; Array.Copy(S,0, A, 0, j + 1); Console.WriteLine(A); } // Driver code public static void Main(String[] args) { char []S1 = "geeksforgeeks".ToCharArray(); removeDuplicates(S1); char []S2 = "aabcca".ToCharArray(); removeDuplicates(S2); } } // This code is contributed by 29AjayKumar |
Output:
geksforgeks abca
Time Complexity : O(n)
Auxiliary Space : O(1)
This article is contributed by Ankur Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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