Open In App
Related Articles

Minimum number of subsequences required to convert one string to another

Improve Article
Improve
Save Article
Save
Like Article
Like

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

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

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;
}


Java




// Java program to find the Minimum number
// of subsequences required to convert
// one to another
import java.util.*;
 
class GFG {
 
  // This function finds the upper bound of a value
  // in an array
  static int upper_bound(int[] arr, int value)
  {
    int left = 0;
    int right = arr.length;
 
    // Using the binary search method
    while (left < right) {
      int mid = (left + right) / 2;
      if (arr[mid] <= value) {
        left = mid + 1;
      }
      else {
        right = mid;
      }
    }
    return left;
  }
 
  // Function to find the no of subsequences
  static int minSubsequnces(String A, String B)
  {
    int[][] v = new int[26][];
    for (int i = 0; i < v.length; i++) {
      v[i] = new int[0];
    }
    int minIndex = -1;
    int cnt = 1;
    int 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.charAt(i) - 97;
      v[p] = Arrays.copyOf(v[p], v[p].length + 1);
      v[p][v[p].length - 1] = i;
    }
 
    while (j < B.length()) {
      int p = (int)B.charAt(j) - 97;
 
      // Find the next index available in the array
      int k = upper_bound(v[p], minIndex);
 
      // If Character is not in A
      if (v[p].length == 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].length) {
 
        // 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
  public static void main(String[] args)
  {
    String A1 = "abbace";
    String B1 = "acebbaae";
    System.out.println(minSubsequnces(A1, B1));
  }
}
 
// This code is contributed by phasing17


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 contributed by mohit kumar 29


C#




// C# program to find the Minimum number
// of subsequences required to convert
// one to another
using System;
 
class GFG {
 
  // This function finds the upper bound of a value
  // in an array
  static int upper_bound(int[] arr, int value)
  {
    int left = 0;
    int right = arr.Length;
 
    // Using the binary search method
    while (left < right) {
      int mid = (left + right) / 2;
      if (arr[mid] <= value) {
        left = mid + 1;
      }
      else {
        right = mid;
      }
    }
    return left;
  }
 
  // Function to find the no of subsequences
  static int minSubsequnces(string A, string B)
  {
    int[][] v = new int[26][];
    for (int i = 0; i < v.Length; i++) {
      v[i] = new int[0];
    }
    int minIndex = -1;
    int cnt = 1;
    int 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;
      Array.Resize(ref v[p], v[p].Length + 1);
      v[p][v[p].Length - 1] = 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], minIndex);
 
      // If Character is not in A
      if (v[p].Length == 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].Length) {
 
        // 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
  static void Main(string[] args)
  {
    string A1 = "abbace";
    string B1 = "acebbaae";
    Console.WriteLine(minSubsequnces(A1, B1));
  }
}
 
// This code is contributed by phasing17


Javascript




// Javascript program to find the Minimum number
// of subsequences required to convert
// one to another
 
// This function finds the upper bound of a value
// in an array
function upper_bound(arr, value) {
  let left = 0;
  let right = arr.length;
   
  // Using the binary search method
  while (left < right) {
    const mid = Math.floor((left + right) / 2);
    if (arr[mid] <= value) {
      left = mid + 1;
    } else {
      right = mid;
    }
  }
  return left;
}
 
 
// Function to find the no of subsequences
function minSubsequnces(A, B) {
let v = Array.from({length: 26}, () => []);
let minIndex = -1;
let cnt = 1;
let j = 0;
let flag = 0;
 
for (let i = 0; i < A.length; i++) {
 
    // Push the values of indexes of each character
    let p = A.charCodeAt(i) - 97;
    v[p].push(i);
}
 
while (j < B.length) {
    let p = B.charCodeAt(j) - 97;
 
    // Find the next index available in the array
    let k = upper_bound(v[p], minIndex);
 
    // If Character is not in A
    if (v[p].length == 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].length) {
 
        // 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
let A1 = "abbace";
let B1 = "acebbaae";
console.log(minSubsequnces(A1, B1));
 
// This code is contributed by phasing17


Output: 

3

 

Time Complexity: O(N1+N2) // N1 is the length of string A and N2 is the length of string B

Auxiliary Space: O(26)


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 24 Feb, 2023
Like Article
Save Article
Similar Reads
Related Tutorials