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Print the string after the specified character has occurred given no. of times

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  • Difficulty Level : Easy
  • Last Updated : 14 Jul, 2022
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Given a string, a character, and a count, the task is to print the string after the specified character has occurred count number of times. Print “Empty string” in case of any unsatisfying conditions. (Given character is not present, or present but less than given count, or given count completes on last index). If given count is 0, then given character doesn’t matter, just print the whole string.

Examples: 

Input  :  str = "This is demo string" 
          char = i,    
          count = 3
Output :  ng

Input :  str = "geeksforgeeks"
         char = e, 
         count = 2
Output : ksforgeeks

Asked in: Oracle

Recommended Practice

Implementation: 

  1. Start traversing the string.
    • Increment occ_count if current char is equal to given char.
    • Get out of the loop, if occ_count becomes equal to given count.
  2. Print the string after the index till the string gets traversed in the loop. 
  3. If index has reached the last, then print “Empty string”.

Implementation:

C++




// C++ program for above implementation
#include <iostream>
using namespace std;
 
// Function to print the string
void printString(string str, char ch, int count)
{
    int occ = 0, i;
 
    // If given count is 0
    // print the given string and return
    if (count == 0) {
        cout << str;
        return;
    }
 
    // Start traversing the string
    for (i = 0; i < str.length(); i++) {
 
        // Increment occ if current char is equal
        // to given character
        if (str[i] == ch)
            occ++;
 
        // Break the loop if given character has
        // been occurred given no. of times
        if (occ == count)
            break;
    }
 
    // Print the string after the occurrence
    // of given character given no. of times
    if (i < str.length() - 1)
        cout << str.substr(i + 1, str.length() - (i + 1));
 
    // Otherwise string is empty
    else
        cout << "Empty string";
}
 
// Drivers code
int main()
{
    string str = "geeks for geeks";
    printString(str, 'e', 2);
    return 0;
}

Java




// Java program for above implementation
 
public class GFG
{
    // Method to print the string
    static void printString(String str, char ch, int count)
    {
        int occ = 0, i;
      
        // If given count is 0
        // print the given string and return
        if (count == 0) {
            System.out.println(str);
            return;
        }
      
        // Start traversing the string
        for (i = 0; i < str.length(); i++) {
      
            // Increment occ if current char is equal
            // to given character
            if (str.charAt(i) == ch)
                occ++;
      
            // Break the loop if given character has
            // been occurred given no. of times
            if (occ == count)
                break;
        }
      
        // Print the string after the occurrence
        // of given character given no. of times
        if (i < str.length() - 1)
            System.out.println(str.substring(i + 1));
      
        // Otherwise string is empty
        else
            System.out.println("Empty string");
    }
     
    // Driver Method
    public static void main(String[] args)
    {
        String str = "geeks for geeks";
        printString(str, 'e', 2);
    }
}

Python3




# Python3 program for above implementation
 
# Function to print the string
def printString(str, ch, count):
    occ, i = 0, 0
 
    # If given count is 0
    # print the given string and return
    if (count == 0):
        print(str)
 
    # Start traversing the string
    for i in range(len(str)):
 
        # Increment occ if current char
        # is equal to given character
        if (str[i] == ch):
            occ += 1
 
        # Break the loop if given character has
        # been occurred given no. of times
        if (occ == count):
            break
 
    # Print the string after the occurrence
    # of given character given no. of times
    if (i < len(str)- 1):
        print(str[i + 1: len(str) - i + 2])
 
    # Otherwise string is empty
    else:
        print("Empty string")
 
# Driver code
if __name__ == '__main__':
    str = "geeks for geeks"
    printString(str, 'e', 2)
 
# This code is contributed
# by 29AjayKumar

C#




// C# program for above implementation
using System;
 
public class GFG {
     
    // Method to print the string
    static public void printString(string str,
                           char ch, int count)
    {
        int occ = 0, i;
     
        // If given count is 0
        // print the given string
        // and return
        if (count == 0) {
            Console.WriteLine(str);
            return;
        }
     
        // Start traversing the string
        for (i = 0; i < str.Length; i++)
        {
     
            // Increment occ if current
            // char is equal to given
            // character
            if (str[i] == ch)
                occ++;
     
            // Break the loop if given
            // character has been occurred
            // given no. of times
            if (occ == count)
                break;
        }
     
        // Print the string after the
        // occurrence of given character
        // given no. of times
        if (i < str.Length - 1)
            Console.WriteLine(str.Substring(i + 1));
     
        // Otherwise string is empty
        else
            Console.WriteLine("Empty string");
    }
     
    // Driver Method
    static public void Main()
    {
        string str = "geeks for geeks";
        printString(str, 'e', 2);
    }
}
 
// This code is contributed by vt_m.

Javascript




<script>
 
// JavaScript program for above implementation
 
// Method to print the string
function printString(str, ch , count)
{
    var occ = 0, i;
  
    // If given count is 0
    // print the given string and return
    if (count == 0) {
        document.write(str);
        return;
    }
  
    // Start traversing the string
    for (i = 0; i < str.length; i++) {
  
        // Increment occ if current char is equal
        // to given character
        if (str.charAt(i) == ch)
            occ++;
  
        // Break the loop if given character has
        // been occurred given no. of times
        if (occ == count)
            break;
    }
  
    // Print the string after the occurrence
    // of given character given no. of times
    if (i < str.length - 1)
        document.write(str.substring(i + 1));
  
    // Otherwise string is empty
    else
        document.write("Empty string");
}
 
// Driver Method
var str = "geeks for geeks";
printString(str, 'e', 2);
 
// This code is contributed by Amit Katiyar
 
</script>

Output

ks for geeks

Time complexity : O(n) 
Auxiliary Space : O(1)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.


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