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Least count of words required to construct a target String

  • Last Updated : 20 Jul, 2021

Given an array of strings, words[] of size M and a string target of size N. The task is to find the minimum number of words required to spell out the string target by cutting individual letters from the collection of words and rearranging them, provided that there are infinite supplies of each word. If it is not possible, print -1.

Examples:

Input: words[] = {“with”, “example”, “science”}, target = “thehat”
Output: 3
Explanation: The target string, “thehat” consists of {‘h’ = 2, ‘t’ = 2, ‘e’ = 1, ‘a’ = 1}.
The two “with” strings are required to get two ‘h’ and two ‘t’ and one “example” string is required to get one ‘e’ and one ‘a’ to form the target string “thehat”.
So, total word required to construct the given target is 3.

Input: words[] = {“notice”, “possible”}, target = “basicbasic”
Output: -1
Explanation: It is not possible to form the given target string.

Approach: The idea is to use backtracking. Follow the steps to solve the problem:



  • Declare a 2D array, say countMap[][], where countMap[i][j] stores the count of jth character in ith string.
  • Declare an array charAvavilable[26], denoting the current count of available characters.
  • Define a recursive function that takes in the current index, i (initially 0) of the string, target as the input.
    • If the current count is greater than the overall count, then return. Also, if the current index, i is equal to N, update the overall count and return.
    • If there is an occurrence of target[i] in charAvailable[],
      • Decrement the occurrence of the character target[i] in charAvailable[].
      • Recursively call for index i+1, simultaneously updating count by 1.
      • Increment the occurrence of the character target[i] by 1 after the function call for backtracking.
    • Otherwise, iterate the countMap[][] to find the occurrence of target[i]. If found then, recursively call the function by updating count by 1 and also perform the backtracking step after the function call.
  • Print the minimum number of words required.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// countMap[][] to store
// count of characters
vector<vector<int> > countMap;
 
int cnt = INT_MAX;
 
// Function to get minimum number of
// stickers for a particular state
void count(int curCnt, int pos, vector<int>charAvailable,
           string target, vector<string> stickers)
{
     
    // If an optimal solution is
    // already there, return
    if (curCnt >= cnt)
        return;
 
    int m = stickers.size();
    int n = target.size();
 
    // If Target has been constructed
    // update cnt and return
    if (pos == n)
    {
        cnt = min(cnt, curCnt);
        return;
    }
 
    char c = target[pos];
 
    if (charAvailable > 0)
    {
         
        // Update charAvailable[]
        charAvailable--;
 
        // Recursizevely function call
        // for (pos + 1)
        count(curCnt, pos + 1, charAvailable,
              target, stickers);
 
        // Update charAvailable[]
        charAvailable++;
    }
    else
    {
        for(int i = 0; i < m; i++)
        {
            if (countMap[i] == 0)
                continue;
 
            // Update charAvailable[]
            for(int j = 0; j < 26; j++)
            {
                charAvailable[j] += countMap[i][j];
            }
 
            // Recursizeve Call
            count(curCnt + 1, pos, charAvailable,
                  target, stickers);
 
            // Update charAvailable[]
            for(int j = 0; j < 26; j++)
            {
                charAvailable[j] -= countMap[i][j];
            }
        }
    }
}
 
// Function to find the minimum
// number of stickers
int minStickers(vector<string> stickers,
                string target)
{
     
    // Base Case
    if (target == "")
        return -1;
 
    if (target.size() == 0)
        return 0;
 
    if (stickers.size() == 0)
        return -1;
 
    int m = stickers.size();
    countMap.resize(m, vector<int>(26, 0));
     
    // Fill the countMap Array
    for(int i = 0; i < stickers.size(); i++)
    {
        string s = stickers[i];
        for(char c : s)
        {
            countMap[i]++;
        }
    }
 
    // Recusizeve function call to get
    // minimum number of stickers
    vector<int> temp(26);
    count(0, 0, temp, target, stickers);
 
    return cnt == INT_MAX ? -1 : cnt;
}
 
// Driver Code
int main()
{
     
    // Given Input
    vector<string> str = {"with", "example", "science"};
    string target = "thehat";
 
    // Function Call
    int Result = minStickers(str, target);
 
    // Print the result
    cout << Result;
}
 
// This code is contributed by mohit kumar 29

Java




// Java program of the above approach
import java.io.*;
import java.util.*;
 
class Sol {
 
    // countMap[][] to store
    // count of characters
    int[][] countMap;
 
    int cnt = Integer.MAX_VALUE;
 
    // Function to find the minimum
    // number of stickers
    public int minStickers(String[] stickers
                           , String target)
    {
        // Base Case
        if (target == null)
            return -1;
 
        if (target.length() == 0)
            return 0;
 
        if (stickers == null || stickers.length == 0)
            return -1;
 
        int m = stickers.length;
        countMap = new int[m][26];
 
        // Fill the countMap Array
        for (int i = 0; i < stickers.length; i++) {
            String s = stickers[i];
            for (char c : s.toCharArray()) {
                countMap[i]++;
            }
        }
 
        // Recursive function call to get
        // minimum number of stickers
        count(0, 0, new int[26], target, stickers);
 
        return cnt == Integer.MAX_VALUE ? -1 : cnt;
    }
 
    // Function to get minimum number of
    // stickers for a particular state
    private void count(int curCnt, int pos,
                       int[] charAvailable, String target,
                       String[] stickers)
    {
        // If an optimal solution is
        // already there, return
        if (curCnt >= cnt)
            return;
 
        int m = stickers.length;
        int n = target.length();
 
        // If Target has been constructed
        // update cnt and return
        if (pos == n) {
            cnt = Math.min(cnt, curCnt);
            return;
        }
 
        char c = target.charAt(pos);
 
        if (charAvailable > 0) {
 
            // Update charAvailable[]
            charAvailable--;
 
            // Recursively function call
            // for (pos + 1)
            count(curCnt, pos + 1, charAvailable, target,
                  stickers);
 
            // Update charAvailable[]
            charAvailable++;
        }
        else {
            for (int i = 0; i < m; i++) {
 
                if (countMap[i] == 0)
                    continue;
 
                // Update charAvailable[]
                for (int j = 0; j < 26; j++) {
                    charAvailable[j] += countMap[i][j];
                }
 
                // Recursive Call
                count(curCnt + 1, pos, charAvailable,
                      target, stickers);
 
                // Update charAvailable[]
                for (int j = 0; j < 26; j++) {
                    charAvailable[j] -= countMap[i][j];
                }
            }
        }
    }
}
 
class GFG {
 
    // Driver Code
    public static void main(String[] args)
    {
        Sol st = new Sol();
 
        // Given Input
        String[] str = { "with", "example", "science" };
        String target = "thehat";
 
        // Function Call
        int Result = st.minStickers(str, target);
 
        // Print the result
        System.out.println(Result);
    }
}
Output: 
3

 

Time Complexity: O(M*26*2N)
Auxiliary Space: O(M*26)

 

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