Number of Concatenations to Create Subsequence- Google Top Interview Questions


Problem Statement :


You are given two lowercase alphabet strings s and t. 

Return the minimum number of times we must concatenate s such that t is a subsequence of s. 

For example, if we concatenate "abc" three times, we'd get "abcabcabc". If it's not possible, return -1.

Constraints

1 ≤ n ≤ 100,000 where n is the length of s

1 ≤ m ≤ 100,000 where m is the length of t

Example 1

Input

s = "dab"

t = "abbd"

Output

3

Explanation

If we concatenate a = "dab" three times, we can get "dabdabdab". And "abbd" is a subsequence of 
"dabdabdab".



Example 2

Input

s = "abc"

t = "def"

Output

-1

Explanation

It's impossible to make t a subsequence of s.


Solution :



title-img 




                        Solution in C++ :

int solve(string s, string t) {
    vector<vector<int>> v(128);
    for (int i = 0; i < s.length(); i++) {
        v[s[i]].push_back(i);
    }
    int ans = 1, j = 0, currind = -1;
    while (j < t.length()) {
        vector<int> &arr = v[t[j]];
        if (arr.size() == 0) return -1;
        int nextind = upper_bound(arr.begin(), arr.end(), currind) - arr.begin();
        if (nextind == arr.size()) {
            currind = -1;
            ans += 1;
        } else {
            currind = arr[nextind];
            j += 1;
        }
    }
    return j == t.length() ? ans : -1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    int v(char c) {
        return (int) (c - 'a');
    }
    public int solve(String s, String t) {
        int m = s.length(), n = t.length();
        Integer[][] indices = new Integer[m][26];
        indices[m - 1][v(s.charAt(m - 1))] = m - 1;
        for (int i = m - 2; i >= 0; i--) {
            int c = v(s.charAt(i));
            for (int j = 0; j < 26; j++) indices[i][j] = indices[i + 1][j];
            indices[i][c] = i;
        }
        int i = 0, j = 0;

        int use = 1;
        while (j < n) {
            int c = v(t.charAt(j));
            if (indices[0][c] == null)
                return -1;
            if (indices[i % m][c] == null) {
                i = 0;
                use++;
            } else {
                i = indices[i % m][c] + 1;
                if (i >= m) {
                    i = 0;
                    use++;
                }
                j++;
            }
        }
        if (i == 0)
            return use - 1;
        return use;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s, t):
        if not set(t).issubset(set(s)):
            return -1

        pos = defaultdict(list)
        for i, c in enumerate(s):
            pos[c].append(i)

        def match(t_i):
            cache = t_i
            prev_pos = -1
            while t_i < len(t):
                s_i = bisect_right(pos[t[t_i]], prev_pos)
                if s_i == len(pos[t[t_i]]):
                    break

                prev_pos = pos[t[t_i]][s_i]
                t_i += 1

            return t_i

        ans = 0
        t_i = 0
        while t_i < len(t):
            t_i = match(t_i)
            ans += 1
        return ans


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