Given a string S, an integer K and set of characters Q[], the task is to find the longest substring in string S which contains atmost K characters from the given character set Q[].
Examples:
Input: S = “normal”, Q = {“a”, “o”, “n”, “b”, “r”, “l”}, K = 1
Output: 1
Explanation:
All the characters in the given string S are present in array.
Therefore, we can select any substring of length 1.
Input: S = “giraffe”, Q = {“a”, “f”, “g”, “r”}, K = 2
Output : 3
Explanation:
Possible substrings with atmost 2 characters
From the given set are {“gir”, “ira”, “ffe”}
The maximum length of all the substrings is 3.
Approach: The idea is to use the concept of two pointers to consider the substrings of maximum length, such that it contains at most K character from the given set. Below is the illustration of the approach:
- Maintain two pointers left and right as 0, to consider the string in between these pointers.
- Increment the right pointer until the characters from the given set is at most K.
- Update the longest length substring to be difference between the right pointer and left pointer.
cur_max = max(cur_max, right - left)
- Increment the left pointer and if the character which moved out from the two pointers is the character of the given set, then decrement the count of the characters from the set by 1.
- Similarly, repeat the steps above until the right pointer is not equal to the length of the string.
Below is the implementation of the above approach:
C++
// C++ implementation to find the // longest substring in the string // which contains atmost K characters // from the given set of characters #include <bits/stdc++.h> using namespace std; // Function to find the longest // substring in the string // which contains atmost K characters // from the given set of characters int maxNormalSubstring(string& P, set<char> Q, int K, int N) { // Base Condition if (K == 0) return 0; // Count for Characters // from set in substring int count = 0; // Two pointers int left = 0, right = 0; int ans = 0; // Loop to iterate until // right pointer is not // equal to N while (right < N) { // Loop to increase the substring // length until the characters from // set are at most K while (right < N && count <= K) { // Check if current pointer // points a character from set if (Q.find(P[right]) != Q.end()){ // If the count of the // char is exceeding the limit if (count + 1 > K) break; else count++; } right++; // update answer with // substring length if (count <= K) ans = max(ans, right - left); } // Increment the left pointer until // the count is less than or equal to K while (left < right) { left++; // If the character which comes out is normal character // then decrement the count by 1 if (Q.find(P[left-1]) != Q.end()) count--; if (count < K) break; } } return ans; } // Driver Code int main() { string P = "giraffe"; set<char> Q; // Construction of set Q.insert('a'); Q.insert('f'); Q.insert('g'); Q.insert('r'); int K = 2; int N = P.length(); // output result cout << maxNormalSubstring(P, Q, K, N); return 0; } |
Java
// Java implementation to find the// longest substring in the string// which contains atmost K characters// from the given set of charactersimport java.util.*;class GFG{ // Function to find the longest// substring in the string// which contains atmost K characters// from the given set of charactersstatic int maxNormalSubstring(String P, Set<Character> Q, int K, int N){ // Base Condition if (K == 0) return 0; // Count for Characters // from set in substring int count = 0; // Two pointers int left = 0, right = 0; int ans = 0; // Loop to iterate until // right pointer is not // equal to N while (right < N) { // Loop to increase the substring // length until the characters from // set are at most K while (right < N && count <= K) { // Check if current pointer // points a character from set if (Q.contains(P.charAt(right))) { // If the count of the // char is exceeding the limit if (count + 1 > K) break; else count++; } right++; // update answer with // substring length if (count <= K) ans = Math.max(ans, right - left); } // Increment the left pointer until // the count is less than or equal to K while (left < right) { left++; // If the character which comes out // then decrement the count by 1 if (Q.contains(P.charAt(left-1))) count--; if (count < K) break; } } return ans;}// Driver codepublic static void main(String[] args){ String P = "giraffe"; Set<Character> Q = new HashSet<>(); // Construction of set Q.add('a'); Q.add('f'); Q.add('g'); Q.add('r'); int K = 2; int N = P.length(); // Output result System.out.println(maxNormalSubstring(P, Q, K, N));}}// This code is contributed by offbeat |
Python3
# Python3 implementation to find the # longest substring in the string # which contains atmost K characters # from the given set of characters # Function to find the longest # substring in the string # which contains atmost K characters # from the given set of characters def maxNormalSubstring(P, Q, K, N): # Base Condition if (K == 0): return 0 # Count for Characters # from set in substring count = 0 # Two pointers left = 0 right = 0 ans = 0 # Loop to iterate until # right pointer is not # equal to N while (right < N): # Loop to increase the substring # length until the characters from # set are at most K while (right < N and count <= K): # Check if current pointer # points a character from set if (P[right] in Q): # If the count of the # char is exceeding the limit if (count + 1 > K): break else: count += 1 right += 1 # update answer with # substring length if (count <= K): ans = max(ans, right - left) # Increment the left pointer until # the count is less than or equal to K while (left < right): left += 1 # If the character which comes out # then decrement the count by 1 if (P[left-1] in Q): count -= 1 if (count < K): break return ans# Driver CodeP = "giraffe"Q = {chr}# Construction of set Q.add('a')Q.add('f')Q.add('g')Q.add('r')K = 2N = len(P)# Output result print(maxNormalSubstring(P, Q, K, N))# This code is contributed by Sanjit_Prasad |
C#
// C# implementation to find the// longest substring in the string// which contains atmost K characters// from the given set of charactersusing System;using System.Collections.Generic;// Function to find the longest// substring in the string// which contains atmost K characters// from the given set of charactersclass Program { static int MaxNormalSubstring(string P, HashSet<char> Q, int K, int N) { // Base Condition if (K == 0) { return 0; } // Count for Characters // from set in substring int count = 0; // Two pointers int left = 0; int right = 0; int ans = 0; // Loop to iterate until // right pointer is not // equal to N while (right < N) { // Loop to increase the substring // length until the characters from // set are at most K while (right < N && count <= K) { // Check if current pointer // points a character from set if (Q.Contains(P[right])) { // If the count of the // char is exceeding the limit if (count + 1 > K) { break; } else { count++; } } right++; // update answer with // substring length if (count <= K) { ans = Math.Max(ans, right - left); } } // Increment the left pointer until // the count is less than or equal to K while (left < right) { left++; // If the character which comes out // then decrement the count by 1 if (Q.Contains(P[left - 1])) { count--; } if (count < K) { break; } } } return ans; }// Driver code static void Main(string[] args) { string P = "giraffe"; // Construction of set HashSet<char> Q = new HashSet<char> { 'a', 'f', 'g', 'r' }; int K = 2; int N = P.Length; // Output result Console.WriteLine(MaxNormalSubstring(P, Q, K, N)); }}// This code is contributed by Aman Kumar |
Javascript
<script>// JavaScript implementation of above approach// Function to find the longest // substring in the string // which contains atmost K characters // from the given set of characters function maxNormalSubstring(P, Q, K, N) { // Base Condition if (K == 0) return 0; // Count for Characters // from set in substring let count = 0; // Two pointers let left = 0, right = 0; let ans = 0; // Loop to iterate until // right pointer is not // equal to N while (right < N) { // Loop to increase the substring // length until the characters from // set are at most K while (right < N && count <= K) { // Check if current pointer // points a character from set if (Q.has(P[right])){ // If the count of the // char is exceeding the limit if (count + 1 > K) break; else count++; } right++; // update answer with // substring length if (count <= K) ans = Math.max(ans, right - left); } // Increment the left pointer until // the count is less than or equal to K while (left < right) { left++; // If the character which comes out is normal character // then decrement the count by 1 if (Q.has(P[left-1])) count--; if (count < K) break; } } return ans; } // Driver Code let P = "giraffe"let Q = new Set() // Construction of set Q.add('a') Q.add('f') Q.add('g') Q.add('r') let K = 2; let N = P.length; // output result document.write(maxNormalSubstring(P, Q, K, N),"</br>"); // This code is contributed by shinjanpatra</script> |
Output:
3
Performance Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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