Given a string S consisting of N lowercase characters, the task is to find the length of the longest substring that does not contain any vowel.
Examples:
Input: S = “neveropen”
Output: 3
The substring “ksf” is the longest substring that does not contain any vowel. The length of this substring is 3.Input: S = “ceebbaceeffo”
Output: 2
Naive Approach: The simplest approach to solve the given problem is to generate all substrings of the given string S and print the length of the substring of maximum length that does not contain any vowel.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized using Sliding Window Technique.
Follow the steps below to solve the problem:
- Initialize two variables, say count and res as 0, to store the length of string without any vowel, and the maximum length of resultant substring found respectively.
- Traverse the given string S using the variable i and perform the following steps:
- If the current character S[i] is a vowel, then update the value of count to 0. Otherwise, increment the value of the count by 1.
- Update the value of res to the maximum of res and count.
- After completing the above steps, print the value of res as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the// character is a vowel or notbool vowel(char ch){ if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') { return true; } return false;}// Function to find the length of// the longest substring that// doesn't contain any vowelint maxLengthString(string s){ // Stores the length of // the longest substring int maximum = 0; int count = 0; // Traverse the string, S for (int i = 0; i < s.length(); i++) { // If the current character // is vowel, set count as 0 if (vowel(s[i])) { count = 0; } // If the current // character is a consonant else { // Increment count by 1 count++; } // Update the maximum length maximum = max(maximum, count); } // Return the result return maximum;}// Driver Codeint main(){ string S = "neveropen"; cout << maxLengthString(S); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { public static boolean vowel(char ch) { if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') { return true; } return false; } // Function to find the length of // the longest substring that // doesn't contain any vowel public static int maxLengthString(String s) { // Stores the length of // the longest substring int maximum = 0; int count = 0; // Traverse the string, S for (int i = 0; i < s.length(); i++) { // If the current character // is vowel, set count as 0 if (vowel(s.charAt(i))) { count = 0; } // If the current // character is a consonant else { // Increment count by 1 count++; } // Update the maximum length maximum = Math.max(maximum, count); } // Return the result return maximum; } public static void main(String[] args) { String S = "neveropen"; System.out.println(maxLengthString(S)); // This code is contributed by Potta Lokesh } |
Python
# Python program for the above approach# Function to check if the# character is a vowel or notdef vowel(ch): if (ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u' or ch == 'A' or ch == 'E' or ch == 'I' or ch == 'O' or ch == 'U'): return True return False# Function to find the length of# the longest substring that# doesn't contain any voweldef maxLengthString(s): # Stores the length of # the longest substring maximum = 0 count = 0; # Traverse the string, S for i in range(len(s)): # If the current character # is vowel, set count as 0 if (vowel(s[i])): count = 0; # If the current # character is a consonant else: # Increment count by 1 count += 1 # Update the maximum length maximum = max(maximum, count) # Return the result return maximum# Driver CodeS = 'neveropen'print(maxLengthString(S))# This code is contributed by shivanisinghss2110 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to check if the// character is a vowel or notstatic bool vowel(char ch){ if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') { return true; } return false;}// Function to find the length of// the longest substring that// doesn't contain any vowelstatic int maxLengthString(string s){ // Stores the length of // the longest substring int maximum = 0; int count = 0; // Traverse the string, S for(int i = 0; i < s.Length; i++) { // If the current character // is vowel, set count as 0 if (vowel(s[i]) == true) { count = 0; } // If the current // character is a consonant else { // Increment count by 1 count++; } // Update the maximum length maximum = Math.Max(maximum, count); } // Return the result return maximum;}// Driver Codepublic static void Main(){ string S = "neveropen"; Console.Write(maxLengthString(S));}}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script> // JavaScript program for the above approach // Function to check if the // character is a vowel or not function vowel(ch) { if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' || ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U') { return true; } return false; } // Function to find the length of // the longest substring that // doesn't contain any vowel function maxLengthString(s) { // Stores the length of // the longest substring let maximum = 0; let count = 0; // Traverse the string, S for (let i = 0; i < s.length; i++) { // If the current character // is vowel, set count as 0 if (vowel(s[i])) { count = 0; } // If the current // character is a consonant else { // Increment count by 1 count++; } // Update the maximum length maximum = Math.max(maximum, count); } // Return the result return maximum; } // Driver Code var S = "neveropen"; document.write(maxLengthString(S));// This code is contributed by Potta Lokesh</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
