Special Subsequence

  • 0

    Consider a string, s = "abc". An alphabetically-ordered sequence of substrings of s would be {"a", "ab", "abc", "b", "bc", "c"}. If we reduce this sequence to only those substrings that start with a vowel and end with a consonant, we're left with {"ab", "abc"}. The alphabetically first element in this reduced list is "ab", and the alphabetically last element is "abc". As a reminder:
    Vowels: a, e, i, o, and u.
    Consonants: b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, and z.

    Complete the findSubstrings function in your editor. It has 1 parameter: a string, s, consisting of lowercase English letters (a − z). The function must find the substrings of s that start with a vowel and end with a consonant, then print the alphabetically first and alphabetically last of these substrings.

    Input Format
    The locked stub code in your editor reads a single string, s, from stdin and passes it to your function.

    3 ≤ length of s ≤ 5 × 10^5

    Output Format
    Your function must print two lines of output denoting the alphabetically first and last substrings of s that start with a vowel and end with a consonant. Print the alphabetically first qualifying substring on the first line, and the alphabetically last qualifying substring on the second line.

    Sample Input 1

    Sample Output 1

    Explanation 1
    "ab" is the only possible substring which starts with a vowel (a) and ends with a consonant (b). Because we only have 1 qualifying substring, "ab" is both the alphabetically first and last qualifying substring and we print it as our first and second lines of output.

    Sample Input 2

    Sample Output 2

    Explanation 2
    There are 2 possible substrings which start with a vowel and end with a consonant: "aab" and "ab". When ordered alphabetically, "aab" comes before "ab". This means that we print "aab" (the alphabetically first qualifying substring) as our first line of output, and we print "ab" (the alphabetically last qualifying substring) as our second line of output.

    Sample Input 3

    Sample Output 3

    Explanation 3
    There are 4830 substrings of s, but only 676 of them start with a vowel and end with a consonant. When ordered alphabetically, the first substring is "aaop" and the last substring is "utyrulqaeuouiecodjlmjeaummaoqkexylwaaopnfvlbiiiidyckzfh".

  • 0

    I ponder over the following idea .
    First cut all vowels at the end of the string. for example if s = aabaaaae, we can get s = aab, because there is no substring which starts with vowel and end with consonant after character 'b'.
    Then we can generate suffix array of string s.
    The last element which starts with vowel in the suffix array is the last substrings of s that start with a vowel and end with a consonant.
    The first element from the suffix array which starts with vowel need to be trimmed till to the first consonant and this is the first substrings of s that start with a vowel and end with a consonant.
    Time complexity depends on how suffix array was created

  • 1

    My implementation of the idea above O(nlog^2(n)) but could be reduced to O(n) or at least O(nlogn) if we use radix sort not quicksort

    int[] createSuffixArray(String str) {     
          class Triplet implements Comparable<Triplet> {  
              int originalIndex;   
              int firstHalf;      
              int secondHalf;
            public int compareTo(Triplet arg) {
                if (arg.firstHalf != firstHalf)
                    return firstHalf - arg.firstHalf;
                return secondHalf - arg.secondHalf;
            Triplet(int id, int fh, int sh) {
                originalIndex = id;
                firstHalf = fh;
                secondHalf = sh;
          int n = str.length();
          int suffixRank[] = new int[n];
          Triplet sa[] = new Triplet[n];   
          for (int i = 0; i < n;i++) {
              suffixRank[i] = str.charAt(i) -'a';
              sa[i] = new Triplet(i,suffixRank[i],0);
          for (int cnt = 1 ; cnt < n; cnt *= 2) {
               for(int i = 0; i < n; i++) {
                   sa[i].firstHalf = suffixRank[i];
                   sa[i].secondHalf = i + cnt < n ? suffixRank[i + cnt] : -1;
                   sa[i].originalIndex = i;
               suffixRank[sa[0].originalIndex] = 0;
               for(int i = 1, currRank = 0; i < n; i++) {            
                   if(sa[i - 1].firstHalf != sa[i].firstHalf || sa[i - 1].secondHalf != sa[i].secondHalf)
                   suffixRank[sa[i].originalIndex] = currRank;
           int[] res = new int[n];
           for(int i = 0; i < n; i++) 
               res[i] = sa[i].originalIndex;
           return res;
      void findSubstrings(String s) {    
          if (s.length() >= 2) {
              Set<Character> volSet = new HashSet<>((Arrays.asList(new Character[] {'a', 'e', 'u', 'o', 'i'})));
              int last = s.length();
              for (int i = s.length() - 1; i >= 0; i--) {
                  if (volSet.contains(s.charAt(i))) {
                      last --;
                  } else 
              s = s.substring(0, last);
              int[] sa = createSuffixArray(s);
              int i = 0;          
              while (i < sa.length) {
                  if (volSet.contains(s.charAt(sa[i]))) {
              if (i < sa.length) {
                  String first = s.substring(sa[i]);
                  for (i = 0; i < first.length(); i++) {            
                      if (!volSet.contains(first.charAt(i))) 
                  System.out.println(first.substring(0, i + 1));
                  for (i = sa.length - 1; i >= 0; i--) {            
                      if (volSet.contains(s.substring(sa[i]).charAt(0))) {

  • 0

    @elmirap your approach is wrong, in case of "eab", your code will give output as

    where as the answer should be

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