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++ 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 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 |
# 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# 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 |
<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> |
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.
#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;
} |
#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 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 |
# 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)
|
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);
} } |
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}`);
|
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)