Longest substring consisting of vowels using Binary Search
Given string str of length N, the task is to find the longest substring which contains only vowels using the Binary Search technique.
Examples:
Input: str = “baeicba”
Output: 3
Explanation:
Longest substring which contains vowels only is “aei”.
Input: str = “aeiou”
Output: 5
Approach: Refer to the Longest substring of vowels for an approach in O(N) complexity.
Binary Search Approach: In this article, we are using a Binary Search based approach:
Follow the steps below to solve the problem:
- Apply binary search on the lengths ranging from 1 to N.
- For each mid-value check if there exists a substring of length mid consisting only of vowels in that substring.
- If there exists a substring of length mid, then update the value of max and update l as mid+1 to check if a substring of length greater than mid exists or not which consists only of vowels.
- If no such substring of length mid exists, update r as mid-1 to check if a substring of length smaller than mid exists or not which consists only of vowels.
- Repeat the above three steps until l is less than or equal to r.
- Return the max length obtained finally.
Below is the implementation of the above approach:
C++
// C++ implementation of // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if a character // is vowel or not bool vowel( int vo) { // 0-a 1-b 2-c and so on 25-z if (vo == 0 || vo == 4 || vo == 8 || vo == 14 || vo == 20) return true ; else return false ; } // Function to check if any // substring of length k exists // which contains only vowels bool check(string s, int k) { vector< int > cnt(26, 0); for ( int i = 0; i < k - 1; i++) { cnt[s[i] - 'a' ]++; } // Applying sliding window to get // all substrings of length K for ( int i = k - 1; i < s.size(); i++) { cnt[s[i] - 'a' ]++; int flag1 = 0; for ( int j = 0; j < 26; j++) { if (vowel(j) == false && cnt[j] > 0) { flag1 = 1; break ; } } if (flag1 == 0) return true ; // Remove the occurrence of // (i-k+1)th character cnt[s[i - k + 1] - 'a' ]--; } return false ; } // Function to perform Binary Search int longestSubstring(string s) { int l = 1, r = s.size(); int maxi = 0; // Doing binary search on the lengths while (l <= r) { int mid = (l + r) / 2; if (check(s, mid)) { l = mid + 1; maxi = max(maxi, mid); } else r = mid - 1; } return maxi; } // Driver Code int main() { string s = "sedrewaefhoiu" ; cout << longestSubstring(s); return 0; } |
Java
// Java implementation of // the above approach import java.util.*; class GFG{ // Function to check if a character // is vowel or not static boolean vowel( int vo) { // 0-a 1-b 2-c and so on 25-z if (vo == 0 || vo == 4 || vo == 8 || vo == 14 || vo == 20 ) return true ; else return false ; } // Function to check if any // subString of length k exists // which contains only vowels static boolean check(String s, int k) { int []cnt = new int [ 26 ]; for ( int i = 0 ; i < k - 1 ; i++) { cnt[s.charAt(i) - 'a' ]++; } // Applying sliding window to get // all subStrings of length K for ( int i = k - 1 ; i < s.length(); i++) { cnt[s.charAt(i) - 'a' ]++; int flag1 = 0 ; for ( int j = 0 ; j < 26 ; j++) { if (vowel(j) == false && cnt[j] > 0 ) { flag1 = 1 ; break ; } } if (flag1 == 0 ) return true ; // Remove the occurrence of // (i-k+1)th character cnt[s.charAt(i - k + 1 ) - 'a' ]--; } return false ; } // Function to perform Binary Search static int longestSubString(String s) { int l = 1 , r = s.length(); int maxi = 0 ; // Doing binary search on the lengths while (l <= r) { int mid = (l + r) / 2 ; if (check(s, mid)) { l = mid + 1 ; maxi = Math.max(maxi, mid); } else r = mid - 1 ; } return maxi; } // Driver Code public static void main(String[] args) { String s = "sedrewaefhoiu" ; System.out.print(longestSubString(s)); } } // This code is contributed by sapnasingh4991 |
Python3
# Python3 implementation of # the above approach # Function to check if a character # is vowel or not def vowel(vo): # 0-a 1-b 2-c and so on 25-z if (vo = = 0 or vo = = 4 or vo = = 8 or vo = = 14 or vo = = 20 ): return True else : return False # Function to check if any # substring of length k exists # which contains only vowels def check(s, k): cnt = [ 0 ] * 26 for i in range (k - 1 ): cnt[ ord (s[i]) - ord ( 'a' )] + = 1 # Applying sliding window to get # all substrings of length K for i in range (k - 1 , len (s)): cnt[ ord (s[i]) - ord ( 'a' )] + = 1 flag1 = 0 for j in range ( 26 ): if (vowel(j) = = False and cnt[j] > 0 ): flag1 = 1 break if (flag1 = = 0 ): return True # Remove the occurrence of # (i-k+1)th character cnt[ ord (s[i - k + 1 ]) - ord ( 'a' )] - = 1 return False # Function to perform Binary Search def longestSubstring(s): l = 1 r = len (s) maxi = 0 # Doing binary search on the lengths while (l < = r): mid = (l + r) / / 2 if (check(s, mid)): l = mid + 1 maxi = max (maxi, mid) else : r = mid - 1 return maxi # Driver Code if __name__ = = "__main__" : s = "sedrewaefhoiu" print (longestSubstring(s)) # This code is contributed by Chitranayal |
C#
// C# implementation of // the above approach using System; class GFG{ // Function to check if a character // is vowel or not static bool vowel( int vo) { // 0-a 1-b 2-c and so on 25-z if (vo == 0 || vo == 4 || vo == 8 || vo == 14 || vo == 20) return true ; else return false ; } // Function to check if any // subString of length k exists // which contains only vowels static bool check(String s, int k) { int []cnt = new int [26]; for ( int i = 0; i < k - 1; i++) { cnt[s[i] - 'a' ]++; } // Applying sliding window to get // all subStrings of length K for ( int i = k - 1; i < s.Length; i++) { cnt[s[i] - 'a' ]++; int flag1 = 0; for ( int j = 0; j < 26; j++) { if (vowel(j) == false && cnt[j] > 0) { flag1 = 1; break ; } } if (flag1 == 0) return true ; // Remove the occurrence of // (i-k+1)th character cnt[s[i - k + 1] - 'a' ]--; } return false ; } // Function to perform Binary Search static int longestSubString(String s) { int l = 1, r = s.Length; int maxi = 0; // Doing binary search on the lengths while (l <= r) { int mid = (l + r) / 2; if (check(s, mid)) { l = mid + 1; maxi = Math.Max(maxi, mid); } else r = mid - 1; } return maxi; } // Driver Code public static void Main(String[] args) { String s = "sedrewaefhoiu" ; Console.Write(longestSubString(s)); } } // This code is contributed by sapnasingh4991 |
Javascript
// JavaScript code to implement above approach. // Function to check if a character // is vowel or not function vowel(vo) { // 0-a 1-b 2-c and so on 25-z if (vo == 0 || vo == 4 || vo == 8 || vo == 14 || vo == 20) { return true ; } else { return false ; } } // Function to check if any // substring of length k exists // which contains only vowels function check(s, k) { let cnt = new Array(26).fill(0); for (let i = 0; i < k - 1; i++) { cnt[s.charCodeAt(i) - 'a' .charCodeAt(0)] += 1; } // Applying sliding window to get // all substrings of length K for (let i = k - 1; i < s.length; i++) { cnt[s.charCodeAt(i) - 'a' .charCodeAt(0)] += 1; let flag1 = 0; for (let j = 0; j < 26; j++) { if (vowel(j) == false && cnt[j] > 0) { flag1 = 1; break ; } } if (flag1 == 0) { return true ; } // Remove the occurrence of // (i-k+1)th character cnt[s.charCodeAt(i - k + 1) - 'a' .charCodeAt(0)] -= 1; } return false ; } // Function to perform Binary Search function longestSubstring(s) { let l = 1; let r = s.length; let maxi = 0; // Doing binary search on the lengths while (l <= r) { let mid = Math.floor((l + r) / 2); if (check(s, mid)) { l = mid + 1; maxi = Math.max(maxi, mid); } else { r = mid - 1; } } return maxi; } // Driver Code let s = "sedrewaefhoiu" ; console.log(longestSubstring(s)); // contributed by adityasha4x71 |
Output:
3
Time Complexity: O(NlogN)
Auxiliary Space: O(26) => O(1), no extra space is required, so it is a constant.