Lexicographically smallest K-length substring containing maximum number of vowels
Last Updated :
13 Apr, 2024
Given string str containing only the lowercase English alphabet and an integer K, the task is to find a K length substring that contains the maximum number of vowels (i.e. ‘a’, ‘e’, ‘i’, ‘o’, ‘u’). If there are multiple such substrings, return the substring which is lexicographically smallest.
Examples:
Input: str = “geeksforgeeks”, K = 4
Output: eeks
Explanation:
The substrings with the maximum count of vowels are “geek”, “eeks” which includes 2 vowels. But “eeks” is lexicographically smallest.
Input: str = “ceebbaceeffo”, K = 3
Output: ace
Explanation:
Lexicographically, substrings with the maximum count of vowels are “ace”.
Naive Approach:
To solve the problem mentioned above, we have to generate all the substrings of length K and store the lexicographically smallest of all such substrings which contain the maximum number of vowels.
Time Complexity: O(N2)
Efficient Approach:
The above-mentioned procedure can be optimized by creating a prefix sum array pref[] of vowels where the ith index contains the count of vowels from 0 to the ith index. The count of vowels for any substring str[l : r] can be given by pref[r]-pref[l-1]. Then, find the lexicographically smallest substring with the maximum count of vowels.
Below is the implementation of the above approach:
C++
// C++ program to find lexicographically smallest K-length
// substring containing maximum number of vowels
#include <bits/stdc++.h>
using namespace std;
// Function to check if reultant string contain any vowel or
// not
bool vowelContain(string s)
{
int i = 0;
while (s[i] != '\0') {
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
|| s[i] == 'o' || s[i] == 'u')
return true;
i++;
}
return false;
}
// Function that prints the lexicographically smallest
// K-length substring containing maximum number of vowels
string maxVowelSubString(string str, int K)
{
// Store the length of the string
int N = str.length();
// Initialize a prefix sum array
int pref[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++) {
// Store 1 at the index
// if it is a vowel
if (str[i] == 'a' or str[i] == 'e' or str[i] == 'i'
or str[i] == 'o' or str[i] == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable to store substring with
// maximum count of vowels
string res = str.substr(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++) {
// Store the current
// count of vowels
int currCount = pref[i] - pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount) {
maxCount = currCount;
res = str.substr(i - K + 1, K);
}
// Update lexicographically smallest substring if
// the current count is equal to the maximum count
else if (currCount == maxCount) {
string temp = str.substr(i - K + 1, K);
if (temp < res)
res = temp;
}
}
// Check if any vowel is present in the whole string or
// not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
// Driver Program
int main()
{
string str = "ceebbaceeffo";
int K = 3;
cout << maxVowelSubString(str, K);
return 0;
}
// This code is modified by Susobhan Akhuli
// Java program to find lexicographically smallest K-length
// substring containing maximum number of vowels
class GFG {
// Function to check if reultant string contain any
// vowel or not
static boolean vowelContain(String s)
{
int n = s.length(), i = 0;
while (i < n) {
if (s.charAt(i) == 'a' || s.charAt(i) == 'e'
|| s.charAt(i) == 'i' || s.charAt(i) == 'o'
|| s.charAt(i) == 'u')
return true;
i++;
}
return false;
}
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
static String maxVowelSubString(String str, int K)
{
// Store the length of the string
int N = str.length();
// Initialize a prefix sum array
int[] pref = new int[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++) {
// Store 1 at the index
// if it is a vowel
if (str.charAt(i) == 'a' || str.charAt(i) == 'e'
|| str.charAt(i) == 'i'
|| str.charAt(i) == 'o'
|| str.charAt(i) == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i != 0)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
String res = str.substring(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++) {
// Store the current
// count of vowels
int currCount = pref[i] - pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount) {
maxCount = currCount;
res = str.substring(i - K + 1, i + 1);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount == maxCount) {
String temp
= str.substring(i - K + 1, i + 1);
if (temp.compareTo(res) < 0)
res = temp;
}
}
// Check if any vowel is present in the whole string
// or not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
// Driver Code
public static void main(String[] args)
{
String str = "ceebbaceeffo";
int K = 3;
System.out.print(maxVowelSubString(str, K));
}
}
// This code is contributed by Chitranayal
// This code is modified by Susobhan Akhuli
# Python3 program to find lexicographically smallest
# K-length substring containing maximum number of vowels
# Function to check if reultant string contain any vowel or
# not
def vowelContain(s):
i = 0
N = len(s)
while (i < N):
if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i'
or s[i] == 'o' or s[i] == 'u'):
return True
i += 1
return False
# Function that prints the
# lexicographically smallest
# K-length substring containing
# maximum number of vowels
def maxVowelSubString(str1, K):
# Store the length of the string
N = len(str1)
# Initialize a prefix sum array
pref = [0 for i in range(N)]
# Loop through the string to
# create the prefix sum array
for i in range(N):
# Store 1 at the index
# if it is a vowel
if (str1[i] == 'a' or
str1[i] == 'e' or
str1[i] == 'i' or
str1[i] == 'o' or
str1[i] == 'u'):
pref[i] = 1
# Otherwise, store 0
else:
pref[i] = 0
# Process the prefix array
if (i):
pref[i] += pref[i - 1]
# Initialize the variable to
# store maximum count of vowels
maxCount = pref[K - 1]
# Initialize the variable
# to store substring with
# maximum count of vowels
res = str1[0:K]
# Loop through the prefix array
for i in range(K, N):
# Store the current
# count of vowels
currCount = pref[i] - pref[i - K]
# Update the result if current count
# is greater than maximum count
if (currCount > maxCount):
maxCount = currCount
res = str1[i - K + 1 : i + 1]
# Update lexicographically smallest
# substring if the current count
# is equal to the maximum count
elif (currCount == maxCount):
temp = str1[i - K + 1 : i + 1]
if (temp < res):
res = temp
# Check if any vowel is present in the whole string or
# not
if vowelContain(res) == False:
return "No vowel is present in the whole string"
# Return the result
return res
# Driver code
if __name__ == '__main__':
str1 = "ceebbaceeffo"
K = 3
print(maxVowelSubString(str1, K))
# This code is contributed by Surendra_Gangwar
# This code is modified by Susobhan Akhuli
// C# program to find lexicographically smallest K-length
// substring containing maximum number of vowels
using System;
class GFG {
// Function to check if reultant string contain any
// vowel or not
static bool vowelContain(String s)
{
int n = s.Length, i = 0;
while (i < n) {
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
|| s[i] == 'o' || s[i] == 'u')
return true;
i += 1;
}
return false;
}
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
static string maxVowelSubString(string str, int K)
{
// Store the length of the string
int N = str.Length;
// Initialize a prefix sum array
int[] pref = new int[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++) {
// Store 1 at the index
// if it is a vowel
if (str[i] == 'a' || str[i] == 'e'
|| str[i] == 'i' || str[i] == 'o'
|| str[i] == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i != 0)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
string res = str.Substring(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++) {
// Store the current
// count of vowels
int currCount = pref[i] - pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount) {
maxCount = currCount;
res = str.Substring(i - K + 1, K);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount == maxCount) {
string temp = str.Substring(i - K + 1, K);
if (string.Compare(temp, res) == -1)
res = temp;
}
}
// Check if any vowel is present in the whole string
// or not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
// Driver Code
public static void Main()
{
string str = "ceebbaceeffo";
int K = 3;
Console.Write(maxVowelSubString(str, K));
}
}
// This code is contributed by Code_Mech
// This code is modified by Susobhan Akhuli
<script>
// Javascript program to find lexicographically smallest
// K-length substring containing maximum number of vowels
// Function to check if resultant string contains any vowel or not
function vowelContain(s) {
var N = s.length;
for (var i = 0; i < N; i++) {
if (s[i] === 'a' || s[i] === 'e' || s[i] === 'i'
|| s[i] === 'o' || s[i] === 'u')
return true;
}
return false;
}
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
function maxVowelSubString(str, K) {
// Store the length of the string
var N = str.length;
// Initialize a prefix sum array
var pref = Array(N);
// Loop through the string to
// create the prefix sum array
for (var i = 0; i < N; i++) {
// Store 1 at the index
// if it is a vowel
if (str[i] === 'a'
|| str[i] === 'e'
|| str[i] === 'i'
|| str[i] === 'o'
|| str[i] === 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
var maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
var res = str.substring(0, K);
// Loop through the prefix array
for (var i = K; i < N; i++) {
// Store the current
// count of vowels
var currCount
= pref[i]
- pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount) {
maxCount = currCount;
res = str.substring(i - K + 1, i + 1);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount === maxCount) {
var temp
= str.substring(
i - K + 1, i + 1);
if (temp < res)
res = temp;
}
}
// Check if any vowel is present in the whole string or not
if (vowelContain(res) === false) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
// Driver Program
var str = "ceebbaceeffo";
var K = 3;
document.write(maxVowelSubString(str, K));
// This code is modified by Susobhan Akhuli
</script>
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the length of the given string.
Time Optimized Approach :
we are using sliding window concept by means of mapping (map stl cpp).
let’s see approach/algorithm:
- first declare a map.
- go on traversing and mapping each character.
- if i>=k then remove i-kth element in map then.
- count all vowels in substr or map.
- take max of count if count is greater then max then traverse over string (i-k)+1 to i then store substr.
C++
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
bool isVowel(char ch)
{
return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
|| ch == 'u');
}
int main()
{
string s = "geeksforgeeks";
int k = 4;
unordered_map<char, int> mp;
int maxelement = 0;
string res;
for (int i = 0; i < s.size(); i++) {
mp[s[i]]++;
int cnt = 0;
string temp = "";
if (i >= k) {
--mp[s[i - k]];
}
for (auto& it : mp) {
if (isVowel(it.first)) {
cnt += it.second;
}
}
if (maxelement < cnt && i >= k) {
maxelement = cnt;
temp = s.substr((i - k) + 1, i);
// for(int
// j=(i-k)+1;j<=i;j++)temp.push_back(s[j]);
res = temp;
}
}
// If no vowel is present in the whole string
if (maxelement == 0) {
cout << "No vowel is present in the whole string"
<< endl;
}
// Print the result and the maximum vowel count
else {
cout << res << endl;
cout << maxelement << endl;
}
return 0;
// code and approach contributed by Sanket Gode.
// follow https://www.linkedin.com/in/sanketgode/
}
// This code is modified by Susobhan Akhuli
import java.util.HashMap;
import java.util.Map;
public class Main {
public static boolean isVowel(char ch)
{
return (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o' || ch == 'u');
}
public static void main(String[] args)
{
String s = "geeksforgeeks";
int k = 4;
Map<Character, Integer> mp = new HashMap<>();
int maxElement = 0;
String res = "";
for (int i = 0; i < s.length(); i++) {
mp.put(s.charAt(i),
mp.getOrDefault(s.charAt(i), 0) + 1);
int cnt = 0;
String temp = "";
if (i >= k) {
char removedChar = s.charAt(i - k);
mp.put(removedChar,
mp.get(removedChar) - 1);
if (mp.get(removedChar) == 0) {
mp.remove(removedChar);
}
}
for (Map.Entry<Character, Integer> entry :
mp.entrySet()) {
if (isVowel(entry.getKey())) {
cnt += entry.getValue();
}
}
if (maxElement < cnt && i >= k) {
maxElement = cnt;
temp = s.substring(i - k + 1, i + 1);
res = temp;
}
}
// If no vowel is present in the whole string
if (maxElement == 0) {
System.out.println(
"No vowel is present in the whole string");
}
// Print the result and the maximum vowel count
else {
System.out.println(res);
System.out.println(maxElement);
}
}
}
// This code is modified by Susobhan Akhuli
def isVowel(ch):
return ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u'
s = "geeksforgeeks"
k = 4
mp = {}
maxelement = 0
res = ""
for i in range(len(s)):
if s[i] not in mp:
mp[s[i]] = 0
mp[s[i]] += 1
cnt = 0
temp = ""
if i >= k:
mp[s[i-k]] -= 1
if mp[s[i-k]] == 0:
del mp[s[i-k]]
for key in mp:
if isVowel(key):
cnt += mp[key]
if maxelement < cnt and i >= k:
maxelement = cnt
temp = s[(i-k+1):(i+1)]
res = temp
# If no vowel is present in the whole string
if (maxelement == 0):
print("No vowel is present in the whole string");
# Print the result and the maximum vowel count
else:
print(res)
print(maxelement)
# This code is modified by Susobhan Akhuli
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
// Function to check if a character is a vowel
static bool IsVowel(char ch)
{
return (ch == 'a' || ch == 'e' || ch == 'i'
|| ch == 'o' || ch == 'u');
}
static void Main()
{
string s = "geeksforgeeks";
int k = 4;
Dictionary<char, int> mp = new Dictionary<
char, int>(); // Create a dictionary to store
// character frequencies
int maxelement = 0;
string res = "";
for (int i = 0; i < s.Length; i++) {
// Update character frequency in the dictionary
if (!mp.ContainsKey(s[i]))
mp[s[i]] = 0;
mp[s[i]]++;
int cnt = 0; // Initialize a count for vowel
// frequencies
string temp
= ""; // Initialize a temporary string to
// store the current substring
if (i >= k) {
char toRemove = s[i - k];
mp[toRemove]--; // Reduce the frequency of
// the character being
// removed
if (mp[toRemove] == 0)
mp.Remove(
toRemove); // Remove the character
// from the dictionary if
// its frequency becomes
// zero
}
// Calculate the count of vowels in the current
// substring
foreach(var item in mp)
{
if (IsVowel(item.Key)) {
cnt += item.Value;
}
}
// Update the result if the current substring
// has more vowels
if (maxelement < cnt && i >= k) {
maxelement = cnt;
temp = s.Substring(i - k + 1, k);
res = temp;
}
}
// If no vowel is present in the whole string
if (maxelement == 0) {
Console.WriteLine(
"No vowel is present in the whole string");
}
// Print the result and the maximum vowel count
else {
Console.WriteLine(res);
Console.WriteLine(maxelement);
}
}
}
// This code is modified by Susobhan Akhuli
function isVowel(ch){
return (ch=='a'||ch=='e'||ch=='i'||ch=='o'||ch=='u');
}
let s="geeksforgeeks";
let k=4;
let mp=new Map();
let maxelement = 0;
let res="";
for(let i=0;i<s.length;i++){
if (!mp.has(s[i])) {
mp.set(s[i], 0);
}
mp.set(s[i], mp.get(s[i]) + 1);
let cnt=0;
let temp="";
if(i>=k){
const charToRemove = s[i - k];
mp.set(charToRemove, mp.get(charToRemove) - 1);
if (mp.get(charToRemove) === 0) {
mp.delete(charToRemove);
}
}
for (const [key, value] of mp) {
if (isVowel(key)) {
cnt += value;
}
}
if(maxelement<cnt && i>=k){
maxelement=cnt;
temp=s.substring((i-k)+1,i+1);
// for(int j=(i-k)+1;j<=i;j++)temp.push_back(s[j]);
res=temp;
}
}
// If no vowel is present in the whole string
if (maxelement == 0) {
console.log("No vowel is present in the whole string");
}
// Print the result and the maximum vowel count
else {
console.log(res);
console.log(maxelement);
}
// This code is modified by Susobhan Akhuli
Complexity Analysis:
Time Complexity : O(n*k).
Space Complexity : O(k).
Space Optimized Approach :
Instead of storing prefix sums, we can use a sliding window and update our result at each maximum.
C++
#include <iostream>
using namespace std;
// Helper function to check if a character is a vowel
bool isVowel(char c)
{
return c == 'a' or c == 'e' or c == 'i' or c == 'o'
or c == 'u';
}
// Function to check if reultant string contain any vowel or
// not
bool vowelContain(string s)
{
int i = 0;
while (s[i] != '\0') {
if (isVowel(s[i]))
return true;
i++;
}
return false;
}
// Function to find the maximum vowel substring
string maxVowelSubstring(string s, int k)
{
int maxCount = 0; // initialize maxCount as 0
string res = s.substr(
0, k); // and result as first substring of size k
for (int i = 0, count = 0; i < s.size();
i++) // iterate through the string
{
if (isVowel(
s[i])) // if current character is a vowel
count++; // then increase count
if (i >= k
and isVowel(
s[i - k])) // if character that is leaving
// the window is a vowel
count--; // then decrease count
if (count > maxCount) // if we get a substring
// having more vowels
{
maxCount = count; // update count
if (i >= k)
res = s.substr(i - k + 1,
k); // and update result
}
if (count == maxCount
and i >= k) // if we get a substring with same
// maximum number of vowels
{
string t = s.substr(i - k + 1, k);
if (t < res) // then check if it is
// lexicographically smaller than
// current result and update it
res = t;
}
}
// Check if any vowel is present in the whole string or
// not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int k = 4;
cout << maxVowelSubstring(str, k);
return 0;
}
// This code is modified by Susobhan Akhuli
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
static boolean isVowel(char c)
{
return c == 'a' || c == 'e' || c == 'i' || c == 'o'
|| c == 'u';
}
// Function to check if reultant string contain any
// vowel or not
static boolean vowelContain(String s)
{
int n = s.length(), i = 0;
while (i < n) {
if (isVowel(s.charAt(i)))
return true;
i++;
}
return false;
}
// Function to find the maximum vowel subString
static String maxVowelSubString(String s, int k)
{
int maxCount = 0; // initialize maxCount as 0
String res = s.substring(
0,
k); // and result as first subString of size k
for (int i = 0, count = 0; i < s.length();
i++) // iterate through the String
{
if (isVowel(s.charAt(
i))) // if current character is a vowel
count++; // then increase count
if (i >= k
&& isVowel(s.charAt(
i - k))) // if character that is leaving
// the window is a vowel
count--; // then decrease count
if (count > maxCount) // if we get a subString
// having more vowels
{
maxCount = count; // update count
if (i >= k)
res = s.substring(
i - k + 1,
i + 1); // and update result
}
if (count == maxCount
&& i >= k) // if we get a subString with
// same maximum number of vowels
{
String t = s.substring(i - k + 1, i + 1);
if (t.compareTo(res)
< 0) { // then check if it is
// lexicographically smaller than
// current result and update it
res = t;
}
}
}
// Check if any vowel is present in the whole string
// or not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
public static void main(String[] args)
{
String str = "geeksforgeeks";
int k = 4;
System.out.println(maxVowelSubString(str, k));
}
}
// This code is contributed by shinjanpatra.
// This code is modified by Susobhan Akhuli
# Helper function to check if a character is a vowel
def isVowel(c):
return (c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u')
# Function to check if reultant string contain any vowel or
# not
def vowelContain(s):
i = 0
N = len(s)
while (i < N):
if (isVowel(s[i])):
return True
i += 1
return False
# Function to find the maximum vowel substring
def maxVowelSubstring(s, k):
# initialize maxCount as 0
maxCount = 0
# and result as first substring of size k
res = s[0:k]
# iterate through the string
count = 0
for i in range(len(s)):
# if current character is a vowel
if (isVowel(s[i])):
count += 1 # then increase count
# if character that is leaving
if (i >= k and isVowel(s[i - k])): # the window is a vowel
count -= 1 # then decrease count
if (count > maxCount): # if we get a substring
# having more vowels
maxCount = count # update count
if (i >= k):
# and update result
res = s[i - k + 1: i + 1]
# if we get a substring with same
if (count == maxCount and i >= k):
# maximum number of vowels
t = s[i - k + 1: i+1]
if (t < res): # then check if it is
# lexicographically smaller than
# current result and update it
res = t
# Check if any vowel is present in the whole string or
# not
if vowelContain(res) == False:
return "No vowel is present in the whole string"
# Return the result
return res
# driver code
str = "geeksforgeeks"
k = 4
print(maxVowelSubstring(str, k))
# This code is contributed by shinjanpatra
# This code is modified by Susobhan Akhuli
// C# code for above approach
using System;
class GFG {
static bool IsVowel(char c)
{
return c == 'a' || c == 'e' || c == 'i' || c == 'o'
|| c == 'u';
}
// Function to check if reultant string contain any
// vowel or not
static bool vowelContain(String s)
{
int n = s.Length, i = 0;
while (i < n) {
if (IsVowel(s[i]))
return true;
i += 1;
}
return false;
}
// Function to find the maximum vowel subString
static string maxVowelSubString(string s, int k)
{
int maxCount = 0; // initialize maxCount as 0
string res = s.Substring(
0,
k); // and result as first subString of size k
for (int i = 0, count = 0; i < s.Length;
i++) // iterate through the String
{
if (IsVowel(s[i])) // if current character is a
// vowel
count++; // then increase count
if (i >= k
&& IsVowel(s[i - k])) // if character that
// is leaving the
// window is a vowel
count--; // then decrease count
if (count > maxCount) // if we get a subString
// having more vowels
{
maxCount = count; // update count
if (i >= k)
res = s.Substring(
i - k + 1, k); // and update result
}
if (count == maxCount
&& i >= k) // if we get a subString with
// same maximum number of vowels
{
string t = s.Substring(i - k + 1, k);
if (string.Compare(t, res)
< 0) // then check if it is
// lexicographically smaller than
// current result and update it
{
res = t;
}
}
}
// Check if any vowel is present in the whole string
// or not
if (!vowelContain(res)) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
static void Main(string[] args)
{
string str = "geeksforgeeks";
int k = 4;
Console.WriteLine(maxVowelSubString(str, k));
}
}
// This code is contributed by Pushpesh Raj.
// This code is modified by Susobhan Akhuli
<script>
// Helper function to check if a character is a vowel
function isVowel(c)
{
return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
|| c == 'u');
}
// Function to check if resultant string contains any vowel or not
function vowelContain(s) {
var N = s.length;
for (var i = 0; i < N; i++) {
if (isVowel(s[i]))
return true;
}
return false;
}
// Function to find the maximum vowel substring
function maxVowelSubstring(s, k)
{
// initialize maxCount as 0
let maxCount = 0;
// and result as first substring of size k
let res = s.substr(0, k);
// iterate through the string
for (let i = 0, count = 0; i < s.length; i++)
{
// if current character is a vowel
if (isVowel(s[i]))
count++; // then increase count
// if character that is leaving
if (i >= k && isVowel(s[i - k]))
// the window is a vowel
count--; // then decrease count
if (count > maxCount) // if we get a substring
// having more vowels
{
maxCount = count; // update count
if (i >= k)
// and update result
res = s.substr(i - k + 1, k);
}
// if we get a substring with same
if (count == maxCount && i >= k)
// maximum number of vowels
{
let t = s.substr(i - k + 1, k);
if (t < res) // then check if it is
// lexicographically smaller than
// current result and update it
res = t;
}
}
// Check if any vowel is present in the whole string or not
if (vowelContain(res) === false) {
return "No vowel is present in the whole string";
}
// Return the result
return res;
}
let str = "geeksforgeeks";
let k = 4;
document.write(maxVowelSubstring(str, k));
// This code is modified by Susobhan Akhuli
</script>
Time Complexity: O(N)
Space Complexity: O(1)
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