# Minimum number of subsequences required to convert one string to another

Given two strings A and B consisting of only lowercase letters, the task is to find the minimum number of subsequences required from A to form B.

Examples:

Input: A = “abbace” B = “acebbaae”
Output: 3
Explanation:
Sub-sequences “ace”, “bba”, “ae” from string A used to form string B

Input: A = “abc” B = “cbacbacba”
Output: 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Maintain an array for each character of A which will store its indexes in increasing order.
• Traverse through each element of B and increase the counter whenever there is a need for new subsequence.
• Maintain a variable minIndex which will show that elements greater than this index can be taken in current subsequence otherwise increase the counter and update the minIndex to -1.

Below is the implementation of the above approach.

## C++

 `// C++ program to find the Minimum number ` `// of subsequences required to convert ` `// one string to another ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the no of subsequences ` `int` `minSubsequnces(string A, string B) ` `{ ` `    ``vector<``int``> v; ` `    ``int` `minIndex = -1, cnt = 1, j = 0; ` `    ``int` `flag = 0; ` ` `  `    ``for` `(``int` `i = 0; i < A.length(); i++) { ` ` `  `        ``// Push the values of indexes of each character ` `        ``int` `p = (``int``)A[i] - 97; ` `        ``v[p].push_back(i); ` `    ``} ` ` `  `    ``while` `(j < B.length()) { ` `        ``int` `p = (``int``)B[j] - 97; ` ` `  `        ``// Find the next index available in the array ` `        ``int` `k = upper_bound(v[p].begin(), ` `                            ``v[p].end(), minIndex) ` `                ``- v[p].begin(); ` ` `  `        ``// If Character is not in string A ` `        ``if` `(v[p].size() == 0) { ` `            ``flag = 1; ` `            ``break``; ` `        ``} ` ` `  `        ``// Check if the next index is not equal to the ` `        ``// size of array which means there is no index ` `        ``// greater than minIndex in the array ` `        ``if` `(k != v[p].size()) { ` ` `  `            ``// Update value of minIndex with this index ` `            ``minIndex = v[p][k]; ` `            ``j = j + 1; ` `        ``} ` `        ``else` `{ ` ` `  `            ``// Update the value of counter ` `            ``// and minIndex for next operation ` `            ``cnt = cnt + 1; ` `            ``minIndex = -1; ` `        ``} ` `    ``} ` `    ``if` `(flag == 1) { ` `        ``return` `-1; ` `    ``} ` `    ``return` `cnt; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string A1 = ``"abbace"``; ` `    ``string B1 = ``"acebbaae"``; ` `    ``cout << minSubsequnces(A1, B1) << endl; ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program to find the Minimum number ` `# of subsequences required to convert ` `# one to another ` `from` `bisect ``import` `bisect as upper_bound ` ` `  `# Function to find the no of subsequences ` `def` `minSubsequnces(A, B): ` `    ``v ``=` `[[] ``for` `i ``in` `range``(``26``)] ` `    ``minIndex ``=` `-``1` `    ``cnt ``=` `1` `    ``j ``=` `0` `    ``flag ``=` `0` ` `  `    ``for` `i ``in` `range``(``len``(A)): ` ` `  `        ``# Push the values of indexes of each character ` `        ``p ``=` `ord``(A[i]) ``-` `97` `        ``v[p].append(i) ` ` `  `    ``while` `(j < ``len``(B)): ` `        ``p ``=` `ord``(B[j]) ``-` `97` ` `  `        ``# Find the next index available in the array ` `        ``k ``=` `upper_bound(v[p], minIndex) ` ` `  `        ``# If Character is not in A ` `        ``if` `(``len``(v[p]) ``=``=` `0``): ` `            ``flag ``=` `1` `            ``break` ` `  `        ``# Check if the next index is not equal to the ` `        ``# size of array which means there is no index ` `        ``# greater than minIndex in the array ` `        ``if` `(k !``=` `len``(v[p])): ` ` `  `            ``# Update value of minIndex with this index ` `            ``minIndex ``=` `v[p][k] ` `            ``j ``=` `j ``+` `1` `        ``else``: ` ` `  `            ``# Update the value of counter ` `            ``# and minIndex for next operation ` `            ``cnt ``=` `cnt ``+` `1` `            ``minIndex ``=` `-``1` `    ``if` `(flag ``=``=` `1``): ` `        ``return` `-``1` `    ``return` `cnt ` ` `  `# Driver Code ` `A1 ``=` `"abbace"` `B1 ``=` `"acebbaae"` `print``(minSubsequnces(A1, B1)) ` ` `  `# This code is contriuted by mohit kumar 29 `

Output:

```3
```

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