Find length of longest substring with at most K normal characters - GeeksforGeeks

03 Jun, 2021
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 : 1 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 : 3 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++
Java
Python3
C#
Javascript
// 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 substringsint 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 codeint 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 implementation to Find// length of longest subString// with at most K normal charactersclass GFG{  // Function to find maximum// length of normal subStringsstatic 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 codepublic 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
# Function to find maximum# length of normal substringsdef 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 codeif(__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# implementation to Find// length of longest subString// with at most K normal charactersusing System; public class GFG{ // Function to find maximum// length of normal subStringsstatic 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 codepublic 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
<script> // Javascript implementation to Find// length of longest substring// with at most K normal character // Function to find maximum// length of normal substringsfunction 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 stringvar P = "giraffe", Q = "01111001111111111011111111";var K = 2;var N = P.length;document.write( maxNormalSubstring(P, Q, K, N));  </script>
3
 
Time Complexity: The above method takes O(N) time. 
skylags
princi singh
rrrtnx
substring
Greedy
Pattern Searching
Strings
Strings
Greedy
Pattern Searching
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