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

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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 = "zambiatek"
         char = e, 
         count = 2
Output : ksforzambiatek

Asked in: Oracle

Implementation:

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.
Print the string after the index till the string gets traversed in the loop.
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 = "zambiatek for zambiatek"; 
    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 = "zambiatek for zambiatek"; 
        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 = "zambiatek for zambiatek"
    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 = "zambiatek for zambiatek"; 
        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 = "zambiatek for zambiatek"; 
printString(str, 'e', 2); 
  
// This code is contributed by Amit Katiyar  
  
</script> 

Output

ks for zambiatek

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

This article is contributed by Sahil Chhabra. If you like zambiatek and would like to contribute, you can also write an article using write.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek main page and help other Geeks.

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