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Count the number of ways to construct the target string
  • Difficulty Level : Hard
  • Last Updated : 16 Nov, 2020

Given an array of strings A, each of the same length and a target string S, find the number of ways to construct the target string using characters from strings in the given array such that the indices of the characters used in string construction form a strictly increasing sequence. Multiple characters can also be used from the same string. Since the answer can be very large, take modulo with (10^9 + 7)
 

Examples:  

Input :  A = ["adc", "aec", "erg"], S = "ac"
Output : 4
Target string can be formed in following ways :
1) 1st character of "adc" and the 3rd character of "adc".
2) 1st character of "adc" and the 3rd character of "aec".
3) 1st character of "aec" and the 3rd character of "adc".
4) 1st character of "aec" and the 3rd character of "aec".

Input : A = ["afsdc", "aeeeedc", "ddegerg"], S = "ae"
Output : 12

 

Approach :

  • We will use Dynamic Programming to solve the problem. 
  • For each string in the array, store the positions in which characters occurred in the string in a common list (L). Our aim is to use the characters whose indices form a strictly increasing sequence, so it doesn’t matter which string they come from. 
  • Traverse the target string, and keep the information of the previously picked index (prev). For the current position of the target, string check whether this character has occurred at any index greater than prev by searching in L. This can be done naively using recursion but we can memorize it using dp table. 
     

 

Following are the states of dp :
dp[pos][prev], 
where pos represents the position we are at in the target string, 
and prev represents the previously picked index.

 



C++

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// C++ Program to Count the number of ways to
// construct the target string
#include <bits/stdc++.h>
 
using namespace std;
 
int mod = 1000000007;
 
int dp[1000][1000];
 
int calculate(int pos, int prev, string s, vector<int>* index)
{
 
    // base case
    if (pos == s.length())
        return 1;
 
    // If current subproblem has been solved, use the value
    if (dp[pos][prev] != -1)
        return dp[pos][prev];
 
    // current character
    int c = s[pos] - 'a';
 
    // search through all the indiced at which the current
    // character occurs. For each index greater than prev,
    // take the index and move
    // to the next position, and add to the answer.
    int answer = 0;
    for (int i = 0; i < index.size(); i++) {
        if (index[i] > prev) {
            answer = (answer % mod + calculate(pos + 1,
                         index[i], s, index) % mod) % mod;
        }
    }
 
    // Store and return the solution for this subproblem
    return dp[pos][prev] = answer;
}
 
int countWays(vector<string>& a, string s)
{
    int n = a.size();
 
    // preprocess the strings by storing for each
    // character of every string, the index of their
    // occurrence we will use a common list for all
    // because of only the index matter
    // in the string from which the character was picked
    vector<int> index[26];
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < a[i].length(); j++) {
 
            // we are storing j+1 because the initial picked index
            // in the recursive step will ne 0.
            // This is just for ease of implementation
            index[a[i][j] - 'a'].push_back(j + 1);
        }
    }
 
    // initialise dp table. -1 represents that
    // the subproblem hasn't been solved
    memset(dp, -1, sizeof(dp));
 
    return calculate(0, 0, s, index);
}
 
// Driver Code
int main()
{
    vector<string> A;
    A.push_back("adc");
    A.push_back("aec");
    A.push_back("erg");
 
    string S = "ac";
 
    cout << countWays(A, S);
 
    return 0;
}

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Java

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// Java Program to Count the number of ways to
// conthe target String
import java.util.*;
 
class GFG{
  
static int mod = 1000000007;
  
static int [][]dp = new int[1000][1000];
  
static int calculate(int pos, int prev, String s, Vector<Integer> index)
{
  
    // base case
    if (pos == s.length())
        return 1;
  
    // If current subproblem has been solved, use the value
    if (dp[pos][prev] != -1)
        return dp[pos][prev];
 
  
    // search through all the indiced at which the current
    // character occurs. For each index greater than prev,
    // take the index and move
    // to the next position, and add to the answer.
    int answer = 0;
    for (int i = 0; i < index.size(); i++) {
        if (index.get(i).compareTo(prev) >= 0) {
            answer = (answer % mod + calculate(pos + 1,
                         index.get(i), s, index) % mod) % mod;
        }
    }
  
    // Store and return the solution for this subproblem
    return dp[pos][prev] = answer;
}
  
static int countWays(Vector<String> a, String s)
{
    int n = a.size();
  
    // preprocess the Strings by storing for each
    // character of every String, the index of their
    // occurrence we will use a common list for all
    // because of only the index matter
    // in the String from which the character was picked
    Vector<Integer> []index = new Vector[26];
    for (int i = 0; i < 26; i++)
        index[i] = new Vector<Integer>();
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < a.get(i).length(); j++) {
  
            // we are storing j+1 because the initial picked index
            // in the recursive step will ne 0.
            // This is just for ease of implementation
            index[a.get(i).charAt(j) - 'a'].add(j + 1);
        }
    }
  
    // initialise dp table. -1 represents that
    // the subproblem hasn't been solved
    for(int i = 0;i< 1000;i++)
    {
        for (int j = 0; j < 1000; j++) {
            dp[i][j] = -1;
        }
    }
  
    return calculate(0, 0, s, index[0]);
}
  
// Driver Code
public static void main(String[] args)
{
    Vector<String> A = new Vector<String>();
    A.add("adc");
    A.add("aec");
    A.add("erg");
  
    String S = "ac";
  
    System.out.print(countWays(A, S));
}
}
 
// This code is contributed by Princi Singh

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Python 3

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# Python 3 Program to Count the number of ways to
# construct the target string
 
mod = 1000000007
 
dp = [[-1 for i in range(1000)] for j in range(1000)];
 
def calculate(pos, prev, s,index):
    # base case
    if (pos == len(s)):
        return 1
 
    # If current subproblem has been solved, use the value
    if (dp[pos][prev] != -1):
        return dp[pos][prev]
 
    # current character
    c = ord(s[pos]) - ord('a');
 
    # search through all the indiced at which the current
    # character occurs. For each index greater than prev,
    # take the index and move
    # to the next position, and add to the answer.
    answer = 0
    for i in range(len(index)):
        if (index[i] > prev):
            answer = (answer % mod + calculate(pos + 1,index[i], s, index) % mod) % mod
             
    dp[pos][prev] = 4
 
    # Store and return the solution for this subproblem
    return dp[pos][prev]
 
def countWays(a, s):
    n = len(a)
 
    # preprocess the strings by storing for each
    # character of every string, the index of their
    # occurrence we will use a common list for all
    # because of only the index matter
    # in the string from which the character was picked
    index = [[] for i in range(26)]
 
    for i in range(n):
        for j in range(len(a[i])):
            # we are storing j+1 because the initial picked index
            # in the recursive step will ne 0.
            # This is just for ease of implementation
            index[ord(a[i][j]) - ord('a')].append(j + 1);
 
    # initialise dp table. -1 represents that
    # the subproblem hasn't been solve
 
    return calculate(0, 0, s, index[0])
 
# Driver Code
if __name__ == '__main__':
    A = []
    A.append("adc")
    A.append("aec")
    A.append("erg")
 
    S = "ac"
 
    print(countWays(A, S))
 
# This code is contributed by Surendra_Gangwar

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C#

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// C# Program to Count the number of ways to
// conthe target String
using System;
using System.Collections.Generic;
 
class GFG{
   
static int mod = 1000000007;
   
static int [,]dp = new int[1000,1000];
   
static int calculate(int pos, int prev, String s, List<int> index)
{
   
    // base case
    if (pos == s.Length)
        return 1;
   
    // If current subproblem has been solved, use the value
    if (dp[pos,prev] != -1)
        return dp[pos,prev];
  
   
    // search through all the indiced at which the current
    // character occurs. For each index greater than prev,
    // take the index and move
    // to the next position, and add to the answer.
    int answer = 0;
    for (int i = 0; i < index.Count; i++) {
        if (index[i].CompareTo(prev) >= 0) {
            answer = (answer % mod + calculate(pos + 1,
                         index[i], s, index) % mod) % mod;
        }
    }
   
    // Store and return the solution for this subproblem
    return dp[pos,prev] = answer;
}
   
static int countWays(List<String> a, String s)
{
    int n = a.Count;
   
    // preprocess the Strings by storing for each
    // character of every String, the index of their
    // occurrence we will use a common list for all
    // because of only the index matter
    // in the String from which the character was picked
    List<int> []index = new List<int>[26];
    for (int i = 0; i < 26; i++)
        index[i] = new List<int>();
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < a[i].Length; j++) {
   
            // we are storing j+1 because the initial picked index
            // in the recursive step will ne 0.
            // This is just for ease of implementation
            index[a[i][j] - 'a'].Add(j + 1);
        }
    }
   
    // initialise dp table. -1 represents that
    // the subproblem hasn't been solved
    for(int i = 0;i< 1000;i++)
    {
        for (int j = 0; j < 1000; j++) {
            dp[i,j] = -1;
        }
    }
   
    return calculate(0, 0, s, index[0]);
}
   
// Driver Code
public static void Main(String[] args)
{
    List<String> A = new List<String>();
    A.Add("adc");
    A.Add("aec");
    A.Add("erg");
   
    String S = "ac";
   
    Console.Write(countWays(A, S));
}
}
 
// This code is contributed by sapnasingh4991

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Output: 

4



 

Time complexity : O(M * N2), where M is the length of the target string, and N is the length of each of the array strings.

Approach: (Dynamic programming + counter+twopointers)

  1. The idea is to First store all the occurrences of the string  by traversing all the possible positions in a default dictionary
  2. Now define another function and start from the 0th position by placing the two pointers over the 0th index  and start recursion
  3. For the recursion first, we should define the  base cases in order to terminate the recursion at some point

Base cases are mentioned below:

1.if i==n: # if i is equal to n

 return 1 #return

2. if start==m: # if start is equal to m

return 0 #

1.In the first base case if the given i ==  n that means  only one character exists  we simple return 1



2. if the start == length of the first character of the string we return 0 since we just terminate our recursion

Main condition:

  • Now we will perform  the pair  recursion and check whether the target can be formed from the given string if yes we will simply increment the counter:

ans = 0 # initilaize the ans with 0

      ans += back(i, start + 1) # recursive call and stroe the result in ans variable

  • Likewise, we will multiply the recursion call with the given string and check whether the target can be generated from the string:

 if positions[start][target[i]]: # check the condition

          ans += positions[start][target[i]] * back(i + 1, start + 1) # multiply with the each charecter and the recursive call

  • Finally, we return the answer
Time complexity:  O(2^n)
space complexity: O(N)

Implementation:

Python3

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from collections import defaultdict,Counter
 
# total ways calculator function
def numWays(words,target):
   
    # length of first word
    m = len(words[0])
     
    # length of the target
    n = len(target)
     
    # stroing in default dict
    positions = defaultdict(Counter)
     
     # traversing array
    for i in range(len(words)):
       
        # traversing like 2d row and col
        for j in range(m):
           
            # store the occurences in dict
            positions[j][words[i][j]] += 1
                 
    # define the back function   
    def back(i, start):
       
        # if i is equal to n
        if i==n:
            return 1
         
        # if start is equal to m
        if start==m:
            return 0
           
        # initilaize the ans with 0 
        ans = 0
         
        # recursive call and store
        # the result in ans variable
        ans += back(i, start + 1)
         
        # check the condition
        if positions[start][target[i]]:
           
            # multiply with the each charecter and the recursive call
            ans += positions[start][target[i]] * back(i + 1, start + 1)
        return ans
  
    return back(0,0)
 
# Function Call
words = ["abba","baab"]
target = "bab"
print(numWays(words,target))
 
# This code is contributed by saikumar kudikala

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Output

4


 

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