**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 :

` ````
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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