Check if a String can be converted to Pangram in K changes

Given a String str containing only lowercase English alphabets and an integer K. The task is to check that whether the string can be converted to a Pangram by performing at most K changes. In one change we can remove any existing character and add a new character.

Pangram: A pangram is a sentence containing every letter in the English Alphabet.

Note: Given that length of string is greater than 26 always and in one operation we have to remove an existing element to add a new element.

Examples: 

Input : str = "qwqqwqeqqwdsdadsdasadsfsdsdsdasasas"
        K = 4
Output : False
Explanation : Making just 4 modifications in this string, 
it can't be changed to a pangram. 

Input : str = "qwqqwqeqqwdsdadsdasadsfsdsdsdasasas"
        K = 24
Output : True
Explanation : By making 19 modifications in the string, 
it can be changed to a pangram.

Approach: 

1. Traverse the string character by character to keep track of all the characters present in the array using a boolean visit array.
2. Using a variable count, traverse the visit array to keep count of the missing characters.
3. If count value is less than or equal to K, print True.
4. Else print False.

Below is the implementation of above approach:

// C++ program to check if a
// String can be converted 
// to Pangram by atmost k modifications 
#include<bits/stdc++.h>
using namespace std;

// Function to find if string 
// can be converted to Pangram
// by atmost k modifications 
bool isPangram(string S, int k) 
{ 
    if (S.length() < 26)
        return false;

    // visit array to keep track 
    // of all the characters 
    // present in the array 
    int visited[26]; 

    for(int i = 0; i < S.length(); i++) 
        visited[S[i] - 'a'] = true;

    // A variable to keep count 
    // of characters missing
    // in the string 
    int count = 0;

    for(int i = 0; i < 26; i++) 
    {
        if (!visited[i]) 
            count += 1;
    }

    // Comparison of count 
    // with given value K 
    if(count <= k )
        return true;
    return false;
}
        
// Driver Code
int main()
{

    string S = "thequickquickfoxmumpsoverthelazydog";
    int k = 15;
    
    // function calling 
    isPangram(S, k) ? cout<< "true" :
                      cout<< "false";

    return 0;
}

// This code is contributed by ChitraNayal

// Java Program to check if a String can be 
// converted to Pangram by atmost k modifications

public class GFG {

    // Function to find if string can be converted 
    // to Pangram by atmost k modifications
    static boolean isPangram(String S, int k)
    {
        if (S.length() < 26)
            return false;

        // visit array to keep track of all
        // the characters present in the array
        boolean[] visited = new boolean[26];

        for (int i = 0; i < S.length(); i++) {
            visited[S.charAt(i) - 'a'] = true;
        }

        // A variable to keep count of 
        // characters missing in the string
        int count = 0;

        for (int i = 0; i < 26; i++) {
            if (!visited[i])
                count++;
        }
        
        // Comparison of count with given value K
        if (count <= k)
            return true;
        return false;
    }
    
    // Driver code
    public static void main(String[] args)
    {
        String S = "thequickquickfoxmumpsoverthelazydog";
        
        int k = 15;
        
        System.out.print(isPangram(S, k));
    }
}

# Python 3 program to check 
# if a String can be converted 
# to Pangram by atmost k modifications 

# Function to find if string 
# can be converted to Pangram
# by atmost k modifications 
def isPangram(S, k) :

    if len(S) < 26 :
        return False

    # visit array to keep track 
    # of all the characters 
    # present in the array 
    visited = [0] * 26

    for char in S :
        visited[ord(char) - ord('a')] = True

    # A variable to keep count 
    # of characters missing
    # in the string 
    count = 0

    for i in range(26) :

        if visited[i] != True :
            count += 1

    # Comparison of count 
    # with given value K 
    if count <= k :
        return True
    return False
        
# Driver Code
if __name__ == "__main__" :
    
    S = "thequickquickfoxmumpsoverthelazydog"
    k = 15
    
    # function calling 
    print(isPangram(S,k))
        
# This code is contributed by ANKITRAI1

// C# Program to check if a 
// String can be converted to 
// Pangram by atmost k modifications 
using System;

class GFG 
{ 

// Function to find if string 
// can be converted to Pangram 
// by atmost k modifications 
static bool isPangram(String S, int k) 
{ 
    if (S.Length < 26) 
        return false; 

    // visit array to keep track 
    // of all the characters present 
    // in the array 
    bool[] visited = new bool[26]; 

    for (int i = 0; i < S.Length; i++) 
    { 
        visited[S[i] - 'a'] = true; 
    } 

    // A variable to keep count 
    // of characters missing in
    // the string 
    int count = 0; 

    for (int i = 0; i < 26; i++)
    { 
        if (!visited[i]) 
            count++; 
    } 
    
    // Comparison of count with 
    // given value K 
    if (count <= k) 
        return true; 
    return false; 
} 

// Driver code 
public static void Main() 
{ 
    string S = "thequickquickfoxmumpsoverthelazydog"; 
    
    int k = 15; 
    
    Console.WriteLine(isPangram(S, k)); 
} 
} 

// This code is contributed
// by inder_verma.

<?php 
// PHP program to check if a 
// String can be converted 
// to Pangram by atmost k modifications 

// Function to find if string 
// can be converted to Pangram
// by atmost k modifications 
function isPangram($S, $k) 
{ 
    if (strlen($S) < 26)
        return false;

    // visit array to keep track 
    // of all the characters 
    // present in the array 
    $visited = array_fill(0, 26, NULL);

    for($i = 0; $i < strlen($S); $i++) 
        $visited[ord($S[$i]) - 
                 ord('a')] = true;

    // A variable to keep count 
    // of characters missing
    // in the string 
    $count = 0;

    for($i = 0; $i < 26; $i++) 
    {
        if ($visited[$i] != true) 
            $count += 1;
    }

    // Comparison of count 
    // with given value K 
    if ($count <= $k )
        return true;
    return false;
}
        
// Driver Code
$S = "thequickquickfoxmumpsoverthelazydog";
$k = 15;
    
// function calling 
echo isPangram($S, $k)? "true" : "false";
    
// This code is contributed by ChitraNayal
?>

<script>
      // JavaScript Program to check if a
      // String can be converted to
      // Pangram by atmost k modifications
      // Function to find if string
      // can be converted to Pangram
      // by atmost k modifications
      function isPangram(S, k) {
        if (S.length < 26) return false;

        // visit array to keep track
        // of all the characters present
        // in the array
        var visited = new Array(26);

        for (var i = 0; i < S.length; i++) {
          visited[S[i].charCodeAt(0) - "a".charCodeAt(0)] = true;
        }

        // A variable to keep count
        // of characters missing in
        // the string
        var count = 0;

        for (var i = 0; i < 26; i++) {
          if (!visited[i]) count++;
        }

        // Comparison of count with
        // given value K
        if (count <= k) return true;
        return false;
      }

      // Driver code
      var S = "thequickquickfoxmumpsoverthelazydog";
      var k = 15;

      document.write(isPangram(S, k));
    </script>

true

Complexity Analysis:

• Time Complexity: O(|S|) ,where S is the given string
• Space Complexity : O(26) ,to store characters.