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Longest substring consisting of vowels using Binary Search

Last Updated : 03 Mar, 2023
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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:  

  1. Apply binary search on the lengths ranging from 1 to N.
  2. For each mid-value check if there exists a substring of length mid consisting only of vowels in that substring.
  3. 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.
  4. 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.
  5. Repeat the above three steps until l is less than or equal to r.
  6. 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. 


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