Given a string P consisting of small English letters and a 26-digit bit string Q, where 1 represent special character and 0 represent normal character for the 26 English alphabets. The task is to find the length of 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 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 is it 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 |
3
Time Complexity: The above method takes O(N) time.
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