Given a string s we have to find the length of the longest substring of s which contain exactly K distinct vowels.
Note: Consider uppercase and lowercase characters as two different characters.
Examples:
Input : s = “tHeracEBetwEEntheTwo”, k = 1
Output : 14
Explanation : Longest substring with only 1 vowel is “cEBetwEEntheTw” and its length is 14.Input : s = “artyebui”, k = 2
Output : 6
Explanation : Longest substring with only 2 vowel is “rtyebu”
Brute-Force Approach: For each substring, we check for the criteria for K distinct vowel and check the length. Finally, the largest length will be the result.
Time Complexity: O(N^2)
Space Complexity: O(1)
Efficient Approach : Here we maintain the count of vowels occurring in the substring. Till K is not zero, we count the distinct vowel occurring in the substring. As K becomes negative, we start deleting the first vowel of the substring we have found till that time, so that it may be possible that new substring( larger length ) is possible afterward. As we delete the vowel we decrease its count so that new substring may contain that vowel occurring in the later part of the string. And as K is 0 we get the length of the substring.
Below is the implementation of the above approach
C++
// CPP program to find the longest substring// with k distinct vowels.#include <bits/stdc++.h>using namespace std;#define MAX 128// Function to check whether a character is// vowel or notbool isVowel(char x){ return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U');}int KDistinctVowel(char s[], int k){ // length of string int n = strlen(s); // array for count of characters int c[MAX]; memset(c, 0, sizeof(c)); // Initialize result to be // negative int result = -1; for (int i = 0, j = -1; i < n; ++i) { int x = s[i]; // If letter is vowel then we // increment its count value // and decrease the k value so // that if we again encounter the // same vowel character then we // don't consider it for our result if (isVowel(x)) { if (++c[x] == 1) { // Decrementing the K value --k; } } // Till k is 0 above if condition run // after that this while loop condition // also become active. Here what we have // done actually is that, if K is less // than 0 then we eliminate the first // vowel we have encountered till that // time . Now K is incremented and k // becomes 0. So, now we calculate the // length of substring from the present // index to the deleted index of vowel // which result in our results. while (k < 0) { x = s[++j]; if (isVowel(x)) { // decreasing the count // so that it may appear // in another substring // appearing after this // present substring if (--c[x] == 0) { // incrementing the K value ++k; } } } // Checking the maximum value // of the result by comparing // the length of substring // whenever K value is 0 means // K distinct vowel is present // in substring if (k == 0) result = max(result, i - j); } return result;}// Driver codeint main(void){ char s[] = "tHeracEBetwEEntheTwo"; int k = 1; cout << KDistinctVowel(s, k); return 0;} |
Java
// Java program to find the longest substring // with k distinct vowels. class GFG { static int MAX = 128; // Function to check whether a character is // vowel or not static boolean isVowel(char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U'); } static int KDistinctVowel(String s, int k) { // length of string int n = s.length(); // array for count of characters int[] c = new int[MAX]; //Array.Clear(c, 0, c.Length); // Initialize result to be // negative int result = -1; for (int i = 0, j = -1; i < n; ++i) { char x = s.charAt(i); // If letter is vowel then we // increment its count value // and decrease the k value so // that if we again encounter the // same vowel character then we // don't consider it for our result if (isVowel(x)) { if (++c[x] == 1) { // Decrementing the K value --k; } } // Till k is 0 above if condition run // after that this while loop condition // also become active. Here what we have // done actually is that, if K is less // than 0 then we eliminate the first // vowel we have encountered till that // time . Now K is incremented and k // becomes 0. So, now we calculate the // length of substring from the present // index to the deleted index of vowel // which result in our results. while (k < 0) { x = s.charAt(++j); if (isVowel(x)) { // decreasing the count // so that it may appear // in another substring // appearing after this // present substring if (--c[x] == 0) { // incrementing the K value ++k; } } } // Checking the maximum value // of the result by comparing // the length of substring // whenever K value is 0 means // K distinct vowel is present // in substring if (k == 0) { result = Math.max(result, i - j); } } return result; } // Driver code public static void main(String[] args) { String s = "tHeracEBetwEEntheTwo"; int k = 1; System.out.println(KDistinctVowel(s, k)); }}/* This Java code is contributed by Rajput-Ji*/ |
Python3
# Python3 program to find the longest substring# with k distinct vowels.MAX = 128# Function to check whether a character is# vowel or notdef isVowel(x): return (x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u' or x == 'A' or x == 'E' or x == 'I' or x == 'O' or x == 'U')def KDistinctVowel(c,k): n = len(s) c = [0 for i in range(MAX)] result = -1 j = -1 for i in range(n): x=s[i] # If letter is vowel then we # increment its count value # and decrease the k value so # that if we again encounter the # same vowel character then we # don't consider it for our result if isVowel(x): c[ord(x)] += 1 if c[ord(x)] == 1: k -= 1 # Till k is 0 above if condition run # after that this while loop condition # also become active. Here what we have # done actually is that, if K is less # than 0 then we eliminate the first # vowel we have encountered till that # time . Now K is incremented and k # becomes 0. So, now we calculate the # length of substring from the present # index to the deleted index of vowel # which result in our results. while k < 0: j += 1 x = s[j] if isVowel(x): # decreasing the count # so that it may appear # in another substring # appearing after this # present substring c[ord(x)] -= 1 k += 1 # Checking the maximum value # of the result by comparing # the length of substring # whenever K value is 0 means # K distinct vowel is present # in substring if k == 0: result = max(result, i - j) return results = "tHeracEBetwEEntheTwo"k = 1print(KDistinctVowel(s, k))# This code is contributed by mohit kumar 29 |
C#
// C# program to find the longest substring// with k distinct vowels.using System;class GFG { static int MAX = 128; // Function to check whether a character is // vowel or not static bool isVowel(char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U'); } static int KDistinctVowel(string s, int k) { // length of string int n = s.Length; // array for count of characters int []c = new int[MAX]; Array.Clear(c, 0, c.Length); // Initialize result to be // negative int result = -1; for (int i = 0, j = -1; i < n; ++i) { char x = s[i]; // If letter is vowel then we // increment its count value // and decrease the k value so // that if we again encounter the // same vowel character then we // don't consider it for our result if (isVowel(x)) { if (++c[x] == 1) { // Decrementing the K value --k; } } // Till k is 0 above if condition run // after that this while loop condition // also become active. Here what we have // done actually is that, if K is less // than 0 then we eliminate the first // vowel we have encountered till that // time . Now K is incremented and k // becomes 0. So, now we calculate the // length of substring from the present // index to the deleted index of vowel // which result in our results. while (k < 0) { x = s[++j]; if (isVowel(x)) { // decreasing the count // so that it may appear // in another substring // appearing after this // present substring if (--c[x] == 0) { // incrementing the K value ++k; } } } // Checking the maximum value // of the result by comparing // the length of substring // whenever K value is 0 means // K distinct vowel is present // in substring if (k == 0) { result = Math.Max(result, i - j); } } return result; } // Driver code static void Main() { string s = "tHeracEBetwEEntheTwo"; int k = 1; Console.Write(KDistinctVowel(s, k)); }}// This code is contributed Manish Shaw// (manishshaw1) |
PHP
<?php // PHP program to find the longest // substring with k distinct vowels.$MAX = 128;// Function to check whether a // character is vowel or notfunction isVowel($x){ return ($x == 'a' || $x == 'e' || $x == 'i' || $x == 'o' || $x == 'u' || $x == 'A' || $x == 'E' || $x == 'I' || $x == 'O' || $x == 'U');}function KDistinctVowel($s, $k){ global $MAX; // length of string $n = strlen($s); // array for count of characters $c = array_fill(0, $MAX, NULL); // Initialize result to be negative $result = -1; for ($i = 0, $j = -1; $i < $n; ++$i) { $x = $s[$i]; // If letter is vowel then we increment // its count value and decrease the k // value so that if we again encounter // the same vowel character then we // don't consider it for our result if (isVowel($x)) { if (++$c[ord($x)] == 1) { // Decrementing the K value --$k; } } // Till k is 0 above if condition run // after that this while loop condition // also become active. Here what we have // done actually is that, if K is less // than 0 then we eliminate the first // vowel we have encountered till that // time . Now K is incremented and k // becomes 0. So, now we calculate the // length of substring from the present // index to the deleted index of vowel // which result in our results. while ($k < 0) { $x = $s[++$j]; if (isVowel($x)) { // decreasing the count so that it // may appear in another substring // appearing after this present // substring if (--$c[ord($x)] == 0) { // incrementing the K value ++$k; } } } // Checking the maximum value of the // result by comparing the length of // substring whenever K value is 0 // means K distinct vowel is present // in substring if ($k == 0) $result = max($result, $i - $j); } return $result;}// Driver code$s = "tHeracEBetwEEntheTwo";$k = 1;echo KDistinctVowel($s, $k);// This code is contributed by ita_c?> |
Javascript
<script>// Javascript program to find the longest substring // with k distinct vowels. let MAX = 128; // Function to check whether a character is // vowel or not function isVowel(x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U'); } function KDistinctVowel(s,k) { // length of string let n = s.length; // array for count of characters let c = new Array(MAX); for(let i=0;i<c.length;i++) { c[i]=0; } //Array.Clear(c, 0, c.Length); // Initialize result to be // negative let result = -1; for (let i = 0, j = -1; i < n; ++i) { let x = s[i]; // If letter is vowel then we // increment its count value // and decrease the k value so // that if we again encounter the // same vowel character then we // don't consider it for our result if (isVowel(x)) { if (++c[x.charCodeAt(0)] == 1) { // Decrementing the K value --k; } } // Till k is 0 above if condition run // after that this while loop condition // also become active. Here what we have // done actually is that, if K is less // than 0 then we eliminate the first // vowel we have encountered till that // time . Now K is incremented and k // becomes 0. So, now we calculate the // length of substring from the present // index to the deleted index of vowel // which result in our results. while (k < 0) { x = s[++j]; if (isVowel(x)) { // decreasing the count // so that it may appear // in another substring // appearing after this // present substring if (--c[x.charCodeAt(0)] == 0) { // incrementing the K value ++k; } } } // Checking the maximum value // of the result by comparing // the length of substring // whenever K value is 0 means // K distinct vowel is present // in substring if (k == 0) { result = Math.max(result, i - j); } } return result; } // Driver code let s = "tHeracEBetwEEntheTwo"; let k = 1; document.write(KDistinctVowel(s, k)); // This code is contributed by rag2127</script> |
Output
7
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
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