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Find length of longest substring with at most K normal characters

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Given a string P consisting of small English letters and a 26-digit bit string Q, where 1 represents the special character and 0 represents a normal character for the 26 English alphabets. The task is to find the length of the longest substring with at most K normal characters.


Examples: 

Input : P = “normal”, Q = “00000000000000000000000000”, K=1 
Output :
Explanation : In string Q all characters are normal. 
Hence, we can select any substring of length 1.


Input : P = “giraffe”, Q = “01111001111111111011111111”, K=2 
Output :
Explanation : Normal characters in P from Q are {a, f, g, r}. 
Therefore, possible substrings with at most 2 normal characters are {gir, ira, ffe}. 
The maximum length of all substring is 3. 

Approach: 
To solve the problem mentioned above we will be using the concept of two pointers. Hence, maintain left and right pointers of the substring, and a count of normal characters. Increment the right index till the count of normal characters is at most K. Then update the answer with a maximum length of substring encountered till now. Increment left index and decrement count till it is greater than K.
Below is the implementation of the above approach: 

C++

// C++ implementation to Find
// length of longest substring
// with at most K normal characters
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum
// length of normal substrings
int maxNormalSubstring(string& P, string& Q,
                       int K, int N)
{
 
    if (K == 0)
        return 0;
 
    // keeps count of normal characters
    int count = 0;
 
    // indexes of substring
    int left = 0, right = 0;
 
    // maintain length of longest substring
    // with at most K normal characters
    int ans = 0;
 
    while (right < N) {
 
        while (right < N && count <= K) {
 
            // get position of character
            int pos = P[right] - 'a';
 
            // check if current character is normal
            if (Q[pos] == '0') {
 
                // check if normal characters
                // count exceeds K
                if (count + 1 > K)
 
                    break;
 
                else
                    count++;
            }
 
            right++;
 
            // update answer with substring length
            if (count <= K)
                ans = max(ans, right - left);
        }
 
        while (left < right) {
 
            // get position of character
            int pos = P[left] - 'a';
 
            left++;
 
            // check if character is
            // normal then decrement count
            if (Q[pos] == '0')
 
                count--;
 
            if (count < K)
                break;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    // initialise the string
    string P = "giraffe", Q = "01111001111111111011111111";
 
    int K = 2;
 
    int N = P.length();
 
    cout << maxNormalSubstring(P, Q, K, N);
 
    return 0;
}

                    

Java

// Java implementation to Find
// length of longest subString
// with at most K normal characters
class GFG{
  
// Function to find maximum
// length of normal subStrings
static int maxNormalSubString(char []P, char []Q,
                       int K, int N)
{
  
    if (K == 0)
        return 0;
  
    // keeps count of normal characters
    int count = 0;
  
    // indexes of subString
    int left = 0, right = 0;
  
    // maintain length of longest subString
    // with at most K normal characters
    int ans = 0;
  
    while (right < N) {
  
        while (right < N && count <= K) {
  
            // get position of character
            int pos = P[right] - 'a';
  
            // check if current character is normal
            if (Q[pos] == '0') {
  
                // check if normal characters
                // count exceeds K
                if (count + 1 > K)
  
                    break;
  
                else
                    count++;
            }
  
            right++;
  
            // update answer with subString length
            if (count <= K)
                ans = Math.max(ans, right - left);
        }
  
        while (left < right) {
  
            // get position of character
            int pos = P[left] - 'a';
  
            left++;
  
            // check if character is
            // normal then decrement count
            if (Q[pos] == '0')
  
                count--;
  
            if (count < K)
                break;
        }
    }
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    // initialise the String
    String P = "giraffe", Q = "01111001111111111011111111";
  
    int K = 2;
  
    int N = P.length();
  
    System.out.print(maxNormalSubString(P.toCharArray(), Q.toCharArray(), K, N));
}
}
 
// This code is contributed by Princi Singh

                    

Python3

# Function to find maximum
# length of normal substrings
def maxNormalSubstring(P, Q, K, N):
     
    if (K == 0):
        return 0
   
    # keeps count of normal characters
    count = 0
   
    # indexes of substring
    left, right = 0, 0
     
    # maintain length of longest substring
    # with at most K normal characters
    ans = 0
   
    while (right < N):
   
        while (right < N and count <= K):
   
            # get position of character
            pos = ord(P[right]) - ord('a')
   
            # check if current character is normal
            if (Q[pos] == '0'):
   
                # check if normal characters
                # count exceeds K
                if (count + 1 > K):
                    break
                else:
                    count += 1
   
            right += 1
   
            # update answer with substring length
            if (count <= K):
                ans = max(ans, right - left)
   
        while (left < right):
   
            # get position of character
            pos = ord(P[left]) - ord('a')
   
            left += 1
   
            # check if character is
            # normal then decrement count
            if (Q[pos] == '0'):
                count -= 1
   
            if (count < K):
                break
   
    return ans
   
# Driver code
if(__name__ == "__main__"):
    # initialise the string
    P = "giraffe"
    Q = "01111001111111111011111111"
   
    K = 2
   
    N = len(P)
   
    print(maxNormalSubstring(P, Q, K, N))
 
# This code is contributed by skylags

                    

C#

// C# implementation to Find
// length of longest subString
// with at most K normal characters
using System;
 
public class GFG{
 
// Function to find maximum
// length of normal subStrings
static int maxNormalSubString(char []P, char []Q,
                    int K, int N)
{
 
    if (K == 0)
        return 0;
 
    // keeps count of normal characters
    int count = 0;
 
    // indexes of subString
    int left = 0, right = 0;
 
    // maintain length of longest subString
    // with at most K normal characters
    int ans = 0;
 
    while (right < N) {
 
        while (right < N && count <= K) {
 
            // get position of character
            int pos = P[right] - 'a';
 
            // check if current character is normal
            if (Q[pos] == '0') {
 
                // check if normal characters
                // count exceeds K
                if (count + 1 > K)
 
                    break;
 
                else
                    count++;
            }
 
            right++;
 
            // update answer with subString length
            if (count <= K)
                ans = Math.Max(ans, right - left);
        }
 
        while (left < right) {
 
            // get position of character
            int pos = P[left] - 'a';
 
            left++;
 
            // check if character is
            // normal then decrement count
            if (Q[pos] == '0')
 
                count--;
 
            if (count < K)
                break;
        }
    }
 
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    // initialise the String
    String P = "giraffe", Q = "01111001111111111011111111";
 
    int K = 2;
 
    int N = P.Length;
 
    Console.Write(maxNormalSubString(P.ToCharArray(),
                     Q.ToCharArray(), K, N));
}
}
 
// This code contributed by Princi Singh

                    

Javascript

<script>
 
// Javascript implementation to Find
// length of longest substring
// with at most K normal character
 
// Function to find maximum
// length of normal substrings
function maxNormalSubstring(P, Q, K, N)
{
 
    if (K == 0)
        return 0;
 
    // keeps count of normal characters
    var count = 0;
 
    // indexes of substring
    var left = 0, right = 0;
 
    // maintain length of longest substring
    // with at most K normal characters
    var ans = 0;
 
    while (right < N) {
 
        while (right < N && count <= K) {
 
            // get position of character
            var pos = P[right].charCodeAt(0) - 'a'.charCodeAt(0);
 
            // check if current character is normal
            if (Q[pos] == '0') {
 
                // check if normal characters
                // count exceeds K
                if (count + 1 > K)
 
                    break;
 
                else
                    count++;
            }
 
            right++;
 
            // update answer with substring length
            if (count <= K)
                ans = Math.max(ans, right - left);
        }
 
        while (left < right) {
 
            // get position of character
            var pos = P[left].charCodeAt(0) - 'a'.charCodeAt(0);
 
            left++;
 
            // check if character is
            // normal then decrement count
            if (Q[pos] == '0')
                count--;
 
            if (count < K)
                break;
        }
    }
 
    return ans;
}
 
// Driver code
// initialise the string
var P = "giraffe", Q = "01111001111111111011111111";
var K = 2;
var N = P.length;
document.write( maxNormalSubstring(P, Q, K, N));
 
 
</script>

                    

Output
3

Time Complexity: O(N*N)
Auxiliary Space: O(1)

Another Approach:

Initialize two variables, start and end, to zero, a variable max_len to zero, and an array freq of size 26 to zero.

Traverse the string from left to right:

a. Increment the frequency of the current character in the freq array.

b. If the frequency of the current character exceeds k, move the start pointer to the right and decrement the frequency of the character at the previous start position.

c. Update the max_len if the current substring length is greater than max_len.

Return max_len.

C++

#include <iostream>
#include <string.h>
 
using namespace std;
 
int main() {
char str[] = "aabbccddd";
int k = 2;
int n = strlen(str);
  int start = 0;
int end = 0;
int max_len = 0;
int freq[26] = {0};
 
// Traverse the string from left to right and
  // find the length of the longest substring with at most K normal characters
while (end < n) {
    freq[str[end] - 'a']++;
    if (freq[str[end] - 'a'] > k) {
        freq[str[start] - 'a']--;
        start++;
    }
    max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;
    end++;
}
 
cout << "Length of longest substring with at most " << k << " normal characters is: " << max_len << endl;
 
return 0;
}

                    

C

#include <stdio.h>
#include <string.h>
 
int main() {
    char str[] = "aabbccddd";
    int k = 2;
    int n = strlen(str);
     
    int start = 0;
    int end = 0;
    int max_len = 0;
    int freq[26] = {0};
     
    // Traverse the string from left to right and find the length of the longest substring with at most K normal characters
    while (end < n) {
        freq[str[end] - 'a']++;
        if (freq[str[end] - 'a'] > k) {
            freq[str[start] - 'a']--;
            start++;
        }
        max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;
        end++;
    }
     
    printf("Length of longest substring with at most %d normal characters is: %d\n", k, max_len);
     
    return 0;
}

                    

Java

// Java program for the above approach
 
import java.util.*;
 
public class Main {
    public static void main(String[] args) {
        String str = "aabbccddd";
        int k = 2;
        int n = str.length();
 
        int start = 0;
        int end = 0;
        int max_len = 0;
        int[] freq = new int[26];
 
        // Traverse the string from left to right and
       // find the length of the longest substring with
      // at most K normal characters
        while (end < n) {
            freq[str.charAt(end) - 'a']++;
            if (freq[str.charAt(end) - 'a'] > k) {
                freq[str.charAt(start) - 'a']--;
                start++;
            }
            max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;
            end++;
        }
 
        System.out.printf("Length of longest substring with at most %d normal characters is: %d\n", k, max_len);
    }
}
 
// This code is contributed by rishabmalhdijo

                    

Python3

# Python implementation of longest substring with at most k normal characters
def longest_substring(str, k):
    n = len(str)
    start = 0
    end = 0
    max_len = 0
    freq = [0] * 26
     
    # Traverse the string from left to right and find the length of the longest substring with at most K normal characters
    while end < n:
        freq[ord(str[end]) - ord('a')] += 1
        if freq[ord(str[end]) - ord('a')] > k:
            freq[ord(str[start]) - ord('a')] -= 1
            start += 1
        max_len = max(max_len, end - start + 1)
        end += 1
     
    print("Length of longest substring with at most", k, "normal characters is:", max_len)
 
# Driver code
str = "aabbccddd"
k = 2
longest_substring(str, k)

                    

C#

using System;
 
class MainClass {
public static void Main (string[] args) {
char[] str = "aabbccddd".ToCharArray();
int k = 2;
int n = str.Length;
int start = 0;
int end = 0;
int max_len = 0;
int[] freq = new int[26];
  // Traverse the string from left to right and
// find the length of the longest substring with at most K normal characters
while (end < n) {
  freq[str[end] - 'a']++;
  if (freq[str[end] - 'a'] > k) {
    freq[str[start] - 'a']--;
    start++;
  }
  max_len = (end - start + 1 > max_len) ? end - start + 1 : max_len;
  end++;
}
 
Console.WriteLine("Length of longest substring with at most {0} normal characters is: {1}", k, max_len);
 
}
}

                    

Javascript

function findLongestSubstring(str, k) {
    const n = str.length;
    let start = 0;
    let end = 0;
    let max_len = 0;
    let freq = new Array(26).fill(0);
 
    // Traverse the string from left to right and
    // find the length of the longest substring with at most K
    // normal characters
    while (end < n) {
        freq[str.charCodeAt(end) - 'a'.charCodeAt()]++;
        if (freq[str.charCodeAt(end) - 'a'.charCodeAt()] > k) {
            freq[str.charCodeAt(start) - 'a'.charCodeAt()]--;
            start++;
        }
        max_len = Math.max(end - start + 1, max_len);
        end++;
    }
 
    return max_len;
}
 
// Driver program
const str = "aabbccddd";
const k = 2;
const longestSubstringLength = findLongestSubstring(str, k);
 
console.log(`Length of longest substring with at most ${k} normal characters is: ${longestSubstringLength}`);

                    

Output
Length of longest substring with at most 2 normal characters is: 8

time complexity of O(n), where n is the string length

 space complexity of O(1)


 



Last Updated : 02 Aug, 2023
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