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:3Explanation:

Sub-sequences “ace”, “bba”, “ae” from string A used to form string B

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

**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 <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to find the no of subsequences` `int` `minSubsequnces(string A, string B)` `{` ` ` `vector<` `int` `> v[26];` ` ` `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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