Given a string of lowercase characters, the task is to print the string in a manner such that a character comes first in string displays first with all its occurrences in string.
Examples:
Input : str = "neveropen" Output: ggeeeekkssfor Explanation: In the given string 'g' comes first and occurs 2 times so it is printed first Then 'e' comes in this string and 4 times so it gets printed. Similarly remaining string is printed. Input : str = "occurrence" output : occcurreen Input : str = "cdab" Output : cdab
This problem is a string version of following problem for array of integers. Group multiple occurrence of array elements ordered by first occurrences Since given strings have only 26 possible characters, it is easier to implement for strings.
Implementation:
- Count the occurrence of all the characters in given string using an array of size 26.
- Then start traversing the string. Print every character its count times.
C++
// C++ program to print all occurrences of every character // together. # include<bits/stdc++.h> using namespace std; // Since only lower case characters are there const int MAX_CHAR = 26; // Function to print the string void printGrouped(string str) { int n = str.length(); // Initialize counts of all characters as 0 int count[MAX_CHAR] = {0}; // Count occurrences of all characters in string for (int i = 0 ; i < n ; i++) count[str[i]-'a']++; // Starts traversing the string for (int i = 0; i < n ; i++) { // Print the character till its count in // hash array while (count[str[i]-'a']--) cout << str[i]; // Make this character's count value as 0. count[str[i]-'a'] = 0; } } // Driver code int main() { string str = "neveropen"; printGrouped(str); return 0; } |
Java
// C++ program to print all occurrences of every character // together. import java.io.*; import java.util.*; class GFG { // Since only lower case characters are there static int MAX_CHAR = 26; // Function to print the string static void printGrouped(String str) { int n = str.length(); // Initialize counts of all characters as 0 int count[] = new int[MAX_CHAR]; Arrays.fill(count,0); // Count occurrences of all characters in string for (int i = 0 ; i < n ; i++) count[str.charAt(i)-'a']++; // Starts traversing the string for (int i = 0; i < n ; i++) { // Print the character till its count in // hash array while (count[str.charAt(i)-'a']-->0) System.out.print(str.charAt(i)); // Make this character's count value as 0. count[str.charAt(i)-'a'] = 0; } } // Driver Code public static void main(String[] args) { String str = "neveropen"; printGrouped(str); } } // This code is contributed by shruti456rawal |
Python3
# Python3 program to print all occurrences # of every character together. # Since only lower case characters are there MAX_CHAR = 26 # Function to print the string def printGrouped(string): n = len(string) # Initialize counts of all characters as 0 count = [0] * MAX_CHAR # Count occurrences of all characters in string for i in range(n): count[ord(string[i]) - ord("a")] += 1 # Starts traversing the string for i in range(n): # Print the character till its count in # hash array while count[ord(string[i]) - ord("a")]: print(string[i], end = "") count[ord(string[i]) - ord("a")] -= 1 # Make this character's count value as 0. count[ord(string[i]) - ord("a")] = 0 # Driver code if __name__ == "__main__": string = "neveropen" printGrouped(string) # This code is contributed by # sanjeev2552 |
C#
// C# program to print all // occurrences of every // character together. using System; class GFG { // Since only lower case // characters are there static int MAX_CHAR = 26; // Method to print // the string static void printGrouped(String str) { int n = str.Length; // Initialize counts of // all characters as 0 int []count = new int[MAX_CHAR]; // Count occurrences of // all characters in string for (int i = 0 ; i < n ; i++) count[str[i] - 'a']++; // Starts traversing // the string for (int i = 0; i < n ; i++) { // Print the character // till its count in // hash array while (count[str[i] - 'a'] != 0) { Console.Write(str[i]); count[str[i] - 'a']--; } // Make this character's // count value as 0. count[str[i] - 'a'] = 0; } } // Driver code public static void Main() { string str = "neveropen"; printGrouped(str); } } // This code is contributed by Sam007 |
Javascript
// javascript program to print all occurrences of every character // together. var MAX_CHAR = 26; // Function to print the string function printGrouped(str) { var n = str.length; // Initialize counts of all characters as 0 var count = Array(MAX_CHAR).fill(0); // Count occurrences of all characters in string for (var i=0; i < n; i++) { count[str.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]++; } // Starts traversing the string for (var i=0; i < n; i++) { // Print the character till its count in // hash array while (count[str.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]-- > 0) { console.log(str.charAt(i)); } // Make this character's count value as 0. count[str.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)] = 0; } } // Driver Code var str = "neveropen"; printGrouped(str); // This code is contributed by Aarti_Rathi |
Output
ggeeeekkssfor
Time complexity : O(n)
Auxiliary Space : O(1)
This article is contributed by Sahil Chhabra. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
