Sub-sequence of Vowels - algorithm

Question

I was practicing for an interview and came across this question on a website:
A magical sub-sequence of a string S is a sub-sequence of S that
contains all five vowels in order. Find the length of largest magical sub-sequence of a string S.
For example, if S = aeeiooua, then aeiou and aeeioou are magical sub-sequences
but aeio and aeeioua are not.
I am a beginner in dynamic programming and am finding it hard to come up with a recursive formula for this.

Answer 1

I did it with an iterative approach rather than recursive one. I started building solution similar to LIS (Longest Increasing Subsequence) and then optimised it upto O(n).
#include<iostream>
#include<string>
#include<vector>
using namespace std;
string vowel = "aeiou";
int vpos(char c)
{
for (int i = 0; i < 5; ++i)
if (c == vowel[i])
return i;
return -1;
}
int magical(string s)
{
int l = s.length();
int previndex[5] = {-1, -1, -1, -1, -1}; // for each vowel
vector<int> len (l, 0);
int i = 0, maxlen = 0;
// finding first 'a'
while (s[i] != 'a')
{
++i;
if (i == l)
return 0;
}
previndex[0] = i; //prev index of 'a'
len[i] = 1;
for ( ++i; i < l; ++i)
{
if (vpos(s[i]) >= 0) // a vowel
{
/* Need to append to longest subsequence on its left, only for this vowel (for any vowels) and
* its previous vowel (if it is not 'a')
This important observation makes it O(n) -- differnet from typical LIS
*/
if (previndex[vpos(s[i])] >= 0)
len[i] = 1+len[previndex[vpos(s[i])]];
previndex[vpos(s[i])] = i;
if (s[i] != 'a')
{
if (previndex[vpos(s[i])-1] >= 0)
len[i] = max(len[i], 1+len[previndex[vpos(s[i])-1]]);
}
maxlen = max(maxlen, len[i]);
}
}
return maxlen;
}
int main()
{
string s = "aaejkioou";
cout << magical(s);
return 0;
}

Answer 2

O(input string length) runtime
import java.util.*;
public class Main {
/*
algo:
keep map of runningLongestSubsequence that ends in each letter. loop through String s. for each char, try appending
to runningLongestSubsequence for that char, as well as to runningLongestSubsequence for preceding char.
update map with whichever results in longer subsequence.
for String s = "ieaeiouiaooeeeaaeiou", final map is:
terminal letter in longest running subsequence-> longest running subsequence
a -> aaaa
e -> aeeeee
i -> aeeeeei
o -> aeeeeeio
u -> aeeeeeiou
naming:
precCharMap - precedingCharMap
runningLongestSubMap - runningLongestSubsequenceMap
*/
public static int longestSubsequence(String s) {
if (s.length() <= 0) throw new IllegalArgumentException();
Map<Character, Character> precCharMap = new HashMap<>();
precCharMap.put('u', 'o');
precCharMap.put('o', 'i');
precCharMap.put('i', 'e');
precCharMap.put('e', 'a');
Map<Character, String> runningLongestSubMap = new HashMap<>();
for (char currChar : s.toCharArray()) {
//get longest subs
String currCharLongestSub;
String precCharLongestSub = null;
if (currChar == 'a') {
currCharLongestSub = runningLongestSubMap.getOrDefault(currChar, "");
} else {
currCharLongestSub = runningLongestSubMap.get(currChar);
char precChar = precCharMap.get(currChar);
precCharLongestSub = runningLongestSubMap.get(precChar);
}
//update running longest subsequence map
if (precCharLongestSub == null && currCharLongestSub != null) {
updateRunningLongestSubMap(currCharLongestSub, currChar, runningLongestSubMap);
} else if (currCharLongestSub == null && precCharLongestSub != null) {
updateRunningLongestSubMap(precCharLongestSub, currChar, runningLongestSubMap);
} else if (currCharLongestSub != null && precCharLongestSub != null) {
//pick longer
if (currCharLongestSub.length() < precCharLongestSub.length()) {
updateRunningLongestSubMap(precCharLongestSub, currChar, runningLongestSubMap);
} else {
updateRunningLongestSubMap(currCharLongestSub, currChar, runningLongestSubMap);
}
}
}
if (runningLongestSubMap.get('u') == null) {
return 0;
}
return runningLongestSubMap.get('u').length();
}
private static void updateRunningLongestSubMap(String longestSub, char currChar,
Map<Character, String> runningLongestSubMap) {
String currCharLongestSub = longestSub + currChar;
runningLongestSubMap.put(currChar, currCharLongestSub);
}
public static void main(String[] args) {
//String s = "aeeiooua"; //7
//String s = "aeiaaioooaauuaeiou"; //10
String s = "ieaeiouiaooeeeaaeiou"; //9
//String s = "ieaeou"; //0
//String s = "ieaeoooo"; //0
//String s = "aeiou"; //5
//if u have String s beginning in "ao", it'll do nothing with o and
//continue on to index 2.
System.out.println(longestSubsequence(s));
}
}

Answer 3

#include <iostream>
#include<string>
#include<cstring>
using namespace std;
unsigned int getcount(string a, unsigned int l,unsigned int r );
int main()
{
std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoo"
"oooeeeeiiioooouuuu");
//std::string a("aaaaaeeeeaaaaiiioooeeeeuuuuuuiiiiiaaaaaaoooooeeeeiiioooo");
//std::string a("aaaaaeeeeaaaaiiioooeeeeiiiiiaaaaaaoooooeeeeiiioooo"); //sol0
//std::string a{"aeiou"};
unsigned int len = a.length();
unsigned int i=0,cnt =0,countmax =0;
bool newstring = true;
while(i<len)
{
if(a.at(i) == 'a' && newstring == true)
{
newstring = false;
cnt = getcount(a,i,len);
if(cnt > countmax)
{
countmax = cnt;
cnt = 0;
}
}
else if(a.at(i)!='a')
{
newstring = true;
}
i++;
}
cout<<countmax;
return 0;
}
unsigned int getcount(string a, unsigned int l,unsigned int r )
{
std::string b("aeiou");
unsigned int seq=0,cnt =0;
unsigned int current =l;
bool compstr = false;
while(current<r)
{
if(a.at(current) == b.at(seq))
{
cnt++;
}
else if((seq <= (b.size()-2)) && (a.at(current) == b.at(seq+1)))
{
seq++;
cnt++;
if (seq == 4)
compstr =true;
}
current++;
}
if (compstr == true)
return cnt;
return 0;
}

Answer 4

you can use recursive approach here (this should work for string length upto max int (easily memorization can be used)
public class LMV {
static final int NOT_POSSIBLE = -1000000000;
// if out put is this i.e soln not possible
static int longestSubsequence(String s, char[] c) {
//exit conditions
if(s.length() ==0 || c.length ==0){
return 0;
}
if(s.length() < c.length){
return NOT_POSSIBLE;
}
if(s.length() == c.length){
for(int i=0; i<s.length(); i++){
if(s.charAt(i) !=c [i]){
return NOT_POSSIBLE;
}
}
return s.length();
}
if(s.charAt(0) < c[0]){
// ignore, go ahead with next item
return longestSubsequence(s.substring(1), c);
} else if (s.charAt(0) == c[0]){
// <case 1> include item and start search for next item in chars
// <case 2> include but search for same item again in chars
// <case 3> don't include item
return Math.max(
Math.max( ( 1+longestSubsequence(s.substring(1), Arrays.copyOfRange(c, 1, c.length) ) ),
( 1+longestSubsequence(s.substring(1), c ) ) ),
( longestSubsequence(s.substring(1), c )) );
} else {
//ignore
return longestSubsequence(s.substring(1), c);
}
}
public static void main(String[] args) {
char[] chars = {'a', 'e', 'i', 'o', 'u'};
String s1 = "aeio";
String s2 = "aaeeieou";
String s3 = "aaeeeieiioiiouu";
System.out.println(longestSubsequence(s1, chars));
System.out.println(longestSubsequence(s2, chars));
System.out.println(longestSubsequence(s3, chars));
}
}

Answer 5

int func( char *p)
{
char *temp = p;
char ae[] = {'a','e','i','o','u'};
int size = strlen(p), i = 0;
int chari = 0, count_aeiou=0;
for (i=0;i<=size; i++){
if (temp[i] == ae[chari]) {
count_aeiou++;
}
else if ( temp[i] == ae[chari+1]) {
count_aeiou++;
chari++;
}
}
if (chari == 4 ) {
printf ("Final count : %d ", count_aeiou);
} else {
count_aeiou = 0;
}
return count_aeiou;
}
The solution to retrun the VOWELS count as per the hackerrank challenge.

Answer 6

int findsubwithcontinuousvowel(string str){
int curr=0;
int start=0,len=0,maxlen=0,i=0;
for(i=0;i<str.size();i++){
if(str[i]=='u' && (current[curr]=='u' || (curr+1<5 && current[curr+1]=='u'))){
//len++;
maxlen=max(len+1,maxlen);
}
if(str[i]==current[curr]){
len++;
}
else if(curr+1<5 && str[i]==current[curr+1]){
len++;
curr++;
}
else{
len=0;
curr=0;
if(str[i]=='a'){
len=1;
}
}
}
return maxlen;
}

Answer 7

Check if vowels are available in sequence in isInSequence and process the result on processor.
public class one {
private char[] chars = {'a','e','i','o','u'};
private int a = 0;
private boolean isInSequence(char c){
// check if char is repeating
if (c == chars[a]){
return true;
}
// if vowels are in sequence and just passed by 'a' and so on...
if (c == 'e' && a == 0){
a++;
return true;
}
if (c == 'i' && a == 1){
a++;
return true;
}
if (c == 'o' && a == 2){
a++;
return true;
}
if (c == 'u' && a == 3){
a++;
return true;
}
return false;
}
private char[] processor(char[] arr){
int length = arr.length-1;
int start = 0;
// In case if all chars are vowels, keeping length == arr
char array[] = new char[length];
for (char a : arr){
if (isInSequence(a)){
array[start] = a;
start++;
}
}
return array;
}
public static void main(String args[]){
char[] arr = {'m','a','e','l','x','o','i','o','u','a'};
one o = new one();
System.out.print(o.processor(arr));
}
}

Answer 8

#include <bits/stdc++.h>
#define ios ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
#define ll unsigned long long
using namespace std;
int main() {
// your code goes here
ios
string s;
cin>>s;
int n=s.length();
int dp[n+1][5]={0};
for(int i=1;i<=n;i++)
{
if(s[i-1]=='a')
{
dp[i][0]=1+dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='e')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][0]>0)
{dp[i][1]=1+max(dp[i-1][1],dp[i-1][0]);}
else
dp[i-1][1]=0;
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='i')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][1]>0)
{dp[i][2]=1+max(dp[i-1][1],dp[i-1][2]);}
else
dp[i-1][2]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='o')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][2]>0)
{dp[i][3]=1+max(dp[i-1][3],dp[i-1][2]);}
else
dp[i-1][3]=0;
dp[i][2]=dp[i-1][2];
dp[i][1]=dp[i-1][1];
dp[i][4]=dp[i-1][4];
}
else if(s[i-1]=='u')
{dp[i][0]=dp[i-1][0];
if(dp[i-1][3]>0)
{dp[i][4]=1+max(dp[i-1][4],dp[i-1][3]);}
else
dp[i-1][4]=0;
dp[i][1]=dp[i-1][1];
dp[i][3]=dp[i-1][3];
dp[i][2]=dp[i-1][2];
}
else
{
dp[i][0]=dp[i-1][0];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2];
dp[i][3]=dp[i-1][3];
dp[i][4]=dp[i-1][4];
}
}
cout<<dp[n][4];
return 0;
}