Skip to content
Related Articles

Related Articles

Improve Article

Minimum letters to be removed to make all occurrences of a given letter continuous

  • Last Updated : 11 Jun, 2021

Given a string str and a character K, the task is to find the minimum number of elements that should be removed from this sequence so that all occurrences of the given character K become continuous. 

Examples: 

Input: str = “ababababa”, K = ‘a’ 
Output:
Explanation: 
All the occurrences of the character ‘b’ should be removed in order to make the occurrences of ‘a’ continuous. 

Input: str = “kprkkoinkopt”, K = ‘k’ 
Output:
 

Approach: The idea is to find the number of characters other than K in between the first and last occurrence of the given character K. In order to do this, the following steps are followed:  



  1. Find the first occurrence of the character K.
  2. Find the last occurrence of the character K.
  3. Iterate between the indices and find the number of characters present in between those indices other than K. This is the required answer.

Below is the implementation of the above approach:  

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number of
// deletions required to make the occurrences
// of the given character K continuous
int noOfDeletions(string str, char k)
{
    int ans = 0, cnt = 0, pos = 0;
 
    // Find the first occurrence of the given letter
    while (pos < str.length() && str[pos] != k) {
        pos++;
    }
 
    int i = pos;
 
    // Iterate from the first occurrence
    // till the end of the sequence
    while (i < str.length()) {
 
        // Find the index from where the occurrence
        // of the character is not continuous
        while (i < str.length() && str[i] == k) {
            i = i + 1;
        }
 
        // Update the answer with the number of
        // elements between non-consecutive occurrences
        // of the given letter
        ans = ans + cnt;
        cnt = 0;
        while (i < str.length() && str[i] != k) {
            i = i + 1;
 
            // Update the count for all letters
            // which are not equal to the given letter
            cnt = cnt + 1;
        }
    }
 
    // Return the count
    return ans;
}
 
// Driver code
int main()
{
    string str1 = "ababababa";
    char k1 = 'a';
    // Calling the function
    cout << noOfDeletions(str1, k1) << endl;
 
    string str2 = "kprkkoinkopt";
    char k2 = 'k';
    // Calling the function
    cout << noOfDeletions(str2, k2) << endl;
}

Java




// Java implementation of the above approach
import java.util.*;
 
class GFG{
  
// Function to find the minimum number of
// deletions required to make the occurrences
// of the given character K continuous
static int noOfDeletions(String str, char k)
{
    int ans = 0, cnt = 0, pos = 0;
  
    // Find the first occurrence of the given letter
    while (pos < str.length() && str.charAt(pos) != k) {
        pos++;
    }
  
    int i = pos;
  
    // Iterate from the first occurrence
    // till the end of the sequence
    while (i < str.length()) {
  
        // Find the index from where the occurrence
        // of the character is not continuous
        while (i < str.length() && str.charAt(i) == k) {
            i = i + 1;
        }
  
        // Update the answer with the number of
        // elements between non-consecutive occurrences
        // of the given letter
        ans = ans + cnt;
        cnt = 0;
        while (i < str.length() && str.charAt(i) != k) {
            i = i + 1;
  
            // Update the count for all letters
            // which are not equal to the given letter
            cnt = cnt + 1;
        }
    }
  
    // Return the count
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    String str1 = "ababababa";
    char k1 = 'a';
 
    // Calling the function
    System.out.print(noOfDeletions(str1, k1) +"\n");
  
    String str2 = "kprkkoinkopt";
    char k2 = 'k';
 
    // Calling the function
    System.out.print(noOfDeletions(str2, k2) +"\n");
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the above approach
 
# Function to find the minimum number of
# deletions required to make the occurrences
# of the given character K continuous
def noOfDeletions(string, k) :
 
    ans = 0; cnt = 0; pos = 0;
 
    # Find the first occurrence of the given letter
    while (pos < len(string) and string[pos] != k) :
        pos += 1;
 
    i = pos;
 
    # Iterate from the first occurrence
    # till the end of the sequence
    while (i < len(string)) :
 
        # Find the index from where the occurrence
        # of the character is not continuous
        while (i < len(string) and string[i] == k) :
            i = i + 1;
         
        # Update the answer with the number of
        # elements between non-consecutive occurrences
        # of the given letter
        ans = ans + cnt;
        cnt = 0;
        while (i < len(string) and string[i] != k) :
            i = i + 1;
 
            # Update the count for all letters
            # which are not equal to the given letter
            cnt = cnt + 1;
             
    # Return the count
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    str1 = "ababababa";
    k1 = 'a';
     
    # Calling the function
    print(noOfDeletions(str1, k1));
     
    str2 = "kprkkoinkopt";
    k2 = 'k';
     
    # Calling the function
    print(noOfDeletions(str2, k2));
     
# This code is contributed by AnkitRai01

C#




// C# implementation of the above approach
using System;
 
class GFG{
   
// Function to find the minimum number of
// deletions required to make the occurrences
// of the given character K continuous
static int noOfDeletions(String str, char k)
{
    int ans = 0, cnt = 0, pos = 0;
   
    // Find the first occurrence of the given letter
    while (pos < str.Length && str[pos] != k) {
        pos++;
    }
   
    int i = pos;
   
    // Iterate from the first occurrence
    // till the end of the sequence
    while (i < str.Length) {
   
        // Find the index from where the occurrence
        // of the character is not continuous
        while (i < str.Length && str[i] == k) {
            i = i + 1;
        }
   
        // Update the answer with the number of
        // elements between non-consecutive occurrences
        // of the given letter
        ans = ans + cnt;
        cnt = 0;
        while (i < str.Length && str[i] != k) {
            i = i + 1;
   
            // Update the count for all letters
            // which are not equal to the given letter
            cnt = cnt + 1;
        }
    }
   
    // Return the count
    return ans;
}
   
// Driver code
public static void Main(String[] args)
{
    String str1 = "ababababa";
    char k1 = 'a';
  
    // Calling the function
    Console.Write(noOfDeletions(str1, k1) +"\n");
   
    String str2 = "kprkkoinkopt";
    char k2 = 'k';
  
    // Calling the function
    Console.Write(noOfDeletions(str2, k2) +"\n");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to find the minimum number of
// deletions required to make the occurrences
// of the given character K continuous
function noOfDeletions(str, k)
{
    var ans = 0, cnt = 0, pos = 0;
 
    // Find the first occurrence of the given letter
    while (pos < str.length && str[pos] != k)
    {
        pos++;
    }
 
    var i = pos;
 
    // Iterate from the first occurrence
    // till the end of the sequence
    while (i < str.length)
    {
 
        // Find the index from where the occurrence
        // of the character is not continuous
        while (i < str.length && str[i] == k)
        {
            i = i + 1;
        }
 
        // Update the answer with the number of
        // elements between non-consecutive occurrences
        // of the given letter
        ans = ans + cnt;
        cnt = 0;
        while (i < str.length && str[i] != k)
        {
            i = i + 1;
 
            // Update the count for all letters
            // which are not equal to the given letter
            cnt = cnt + 1;
        }
    }
 
    // Return the count
    return ans;
}
 
// Driver code
var str1 = "ababababa";
var k1 = 'a';
 
// Calling the function
document.write( noOfDeletions(str1, k1) + "<br>");
var str2 = "kprkkoinkopt";
var k2 = 'k';
 
// Calling the function
document.write( noOfDeletions(str2, k2));
 
// This code is contributed by rrrtnx
 
</script>
Output: 
4
5

 

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




My Personal Notes arrow_drop_up
Recommended Articles
Page :