Check whether second string can be formed from characters of first string used any number of times

Given two strings str1 of size N and str2 of size M.The task is to find if it is possible to compose str2 by using only the characters of str1 such that every character of str1 can be used any number of time. 
Note: Lower case letters and upper case letters should be considered different. 

Examples:

Input: str1 = “The quick brown fox jumps”, str2 = “the fox was quicker than the dog” 
Output: No 
Explanation: 
In str2, there is d character, which is not present in str1. So, it is impossible to make the string str2 from str1

Input: str1 = “we all love geeksforgeeks”, str2 = “we all love geeks” 
Output: Yes  

Naive Approach: The simplest approach is to search for every character of str2 in str1. If all the characters are found then print “Yes”. Otherwise, print “No”. 

Implementation of the above approach

C++

#include <iostream>
#include <unordered_map>
using namespace std;

string can_compose_str(string str1, string str2) {
    unordered_map<char, int> freq_map;
    for (char ch : str1) {
        freq_map[ch]++;
    }
    for (char ch : str2) {
        if (freq_map.find(ch) == freq_map.end() || freq_map[ch] < 1) {
            return "No";
        }
        freq_map[ch]--;
    }
    return "Yes";
}

int main() {
    string str1 = "abcdef";
    string str2 = "abcde";
    cout << can_compose_str(str1, str2) << endl; // Output: Yes
    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;

public class Main {
    public static String canComposeStr(String str1, String str2) {
        Map<Character, Integer> freqMap = new HashMap<>();
        for (char ch : str1.toCharArray()) {
            freqMap.put(ch, freqMap.getOrDefault(ch, 0) + 1);
        }
        for (char ch : str2.toCharArray()) {
            if (!freqMap.containsKey(ch) || freqMap.get(ch) < 1) {
                return "No";
            }
            freqMap.put(ch, freqMap.get(ch) - 1);
        }
        return "Yes";
    }

    public static void main(String[] args) {
        String str1 = "abcdef";
        String str2 = "abcde";
        System.out.println(canComposeStr(str1, str2)); // Output: Yes
    }
}

Python

def can_compose_str(str1, str2):
    freq_map = {}
    for ch in str1:
        freq_map[ch] = freq_map.get(ch, 0) + 1
    for ch in str2:
        if ch not in freq_map or freq_map[ch] < 1:
            return "No"
        freq_map[ch] -= 1
    return "Yes"

str1 = "abcdef"
str2 = "abcde"
print(can_compose_str(str1, str2)) # Output: Yes

C#

using System;
using System.Collections.Generic;

public class Program {
    public static string CanComposeStr(string str1, string str2) {
        Dictionary<char, int> freqMap = new Dictionary<char, int>();
        foreach (char ch in str1) {
            if (freqMap.ContainsKey(ch)) {
                freqMap[ch]++;
            } else {
                freqMap[ch] = 1;
            }
        }
        foreach (char ch in str2) {
            if (!freqMap.ContainsKey(ch) || freqMap[ch] < 1) {
                return "No";
            }
            freqMap[ch]--;
        }
        return "Yes";
    }

    public static void Main() {
        string str1 = "abcdef";
        string str2 = "abcde";
        Console.WriteLine(CanComposeStr(str1, str2)); // Output: Yes
    }
}

Javascript

function canComposeStr(str1, str2) {
  let freqMap = new Map();
  for (let ch of str1) {
    freqMap.set(ch, freqMap.get(ch) + 1 || 1);
  }
  for (let ch of str2) {
    if (!freqMap.has(ch) || freqMap.get(ch) < 1) {
      return "No";
    }
    freqMap.set(ch, freqMap.get(ch) - 1);
  }
  return "Yes";
}

let str1 = "abcdef";
let str2 = "abcde";
console.log(canComposeStr(str1, str2)); // Output: Yes

Output

Yes

Time Complexity: O(N*M) 
Auxiliary Space: O(1) 

Efficient Approach: The idea is to mark the presence of all characters of str1 in a count[] array. Then traverse str2 and check if the characters of str2 is present in str1 or not. 

1. Make an array count[] of size 256 (number of total ASCII characters ) and set it to zero.
2. Then iterate over str1 and make count[str[i]] = 1 to mark the occurrence of every character.
3. Iterate over str2 and check if that character is present in str1 or not using count[] array.

Below is the implementation of the above approach: 

CPP

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if
// str2 can be made by characters
// of str1 or not
void isPossible(string str1, string str2)
{
    // To store the
    // occurrence of every
    // character
    int arr[256] = { 0 };

    // Length of the two
    // strings
    int l1 = str1.size();
    int l2 = str2.size();

    int i, j;

    // Assume that it is
    // possible to compose the
    // string str2 from str1
    bool possible = true;

    // Iterate over str1
    for (i = 0; i < l1; i++) {
        // Store the presence of every
        // character
        arr[str1[i]] = 1;
    }

    // Iterate over str2
    for (i = 0; i < l2; i++) {

        // Ignore the spaces
        if (str2[i] != ' ') {

            // Check for the presence
            // of character in str1
            if (arr[str2[i]] == 1)
                continue;

            else {
                possible = false;
                break;
            }
        }
    }

    // If it is possible to make
    // str2 from str1
    if (possible) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}

// Driver Code
int main()
{
    // Given strings
    string str1 = "we all love geeksforgeeks";
    string str2 = "we all love geeks";

    // Function Call
    isPossible(str1, str2);
    return 0;
}

Java

// Java implementation of
// the above approach

import java.util.Arrays;

class GFG {

    // Function to check if
    // str2 can be made by characters
    // of str1 or not
    public static void isPossible(String str1, String str2) {
        // To store the
        // occurrence of every
        // character
        int arr[] = new int[256];

        Arrays.fill(arr, 0);
        // Length of the two
        // strings
        int l1 = str1.length();
        int l2 = str2.length();

        int i, j;

        // Assume that it is
        // possible to compose the
        // string str2 from str1
        boolean possible = true;

        // Iterate over str1
        for (i = 0; i < l1; i++) {
            // Store the presence of every
            // character
            arr[str1.charAt(i)] = 1;
        }

        // Iterate over str2
        for (i = 0; i < l2; i++) {

            // Ignore the spaces
            if (str2.charAt(i) != ' ') {

                // Check for the presence
                // of character in str1
                if (arr[str2.charAt(i)] == 1)
                    continue;

                else {
                    possible = false;
                    break;
                }
            }
        }

        // If it is possible to make
        // str2 from str1
        if (possible) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

    // Driver Code
    public static void main(String args[])
    {
        // Given strings
        String str1 = "we all love geeksforgeeks";
        String str2 = "we all love geeks";

        // Function Call
        isPossible(str1, str2);

    }

}

// This code is contributed by saurabh_jaiswal.

C#

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to check if
// str2 can be made by characters
// of str1 or not
static void isPossible(string str1, string str2)
{

    // To store the
    // occurrence of every
    // character
    int []arr = new int[256];
    Array.Clear(arr,0,256);

    // Length of the two
    // strings
    int l1 = str1.Length;
    int l2 = str2.Length;

    int i;

    // Assume that it is
    // possible to compose the
    // string str2 from str1
    bool possible = true;

    // Iterate over str1
    for (i = 0; i < l1; i++) {
        // Store the presence of every
        // character
        arr[str1[i]] = 1;
    }

    // Iterate over str2
    for (i = 0; i < l2; i++) {

        // Ignore the spaces
        if (str2[i] != ' ') {

            // Check for the presence
            // of character in str1
            if (arr[str2[i]] == 1)
                continue;

            else {
                possible = false;
                break;
            }
        }
    }

    // If it is possible to make
    // str2 from str1
    if (possible) {
        Console.Write("Yes");
    }
    else {
        Console.Write("No");
    }
}

// Driver Code
public static void Main()
{
    // Given strings
    string str1 = "we all love geeksforgeeks";
    string str2 = "we all love geeks";

    // Function Call
    isPossible(str1, str2);
}
}

// This code is contributed by ipg2016107.

Python3

# Python3 implementation of
# the above approach

# Function to check if
# str2 can be made by characters
# of str1 or not
def isPossible(str1, str2):

      # To store the
    # occurrence of every
    # character
    arr = {}

    # Length of the two
    # strings
    l1 = len(str1)
    l2 = len(str2)

        # Assume that it is
    # possible to compose the
    # string str2 from str1
    possible = True

        # Iterate over str1
    for i in range(l1):

      # Store the presence or every element
        arr[str1[i]] = 1

        # Iterate over str2
    for i in range(l2):

      # Ignore the spaces
        if str2[i] != ' ':

              # Check for the presence
           # of character in str1
            if arr[str2[i]] == 1:
                continue
            else:
                possible = False
                break

     # If it is possible to make
    # str2 from str1
    if possible:
        print("Yes")
    else:
        print("No")


# Driver code
str1 = "we all love geeksforgeeks"
str2 = "we all love geeks"

# Function call.
isPossible(str1, str2)

# This code is contributed by Parth Manchanda

Javascript

<script>

// JavaScript implementation of
// the above approach

// Function to check if
// str2 can be made by characters
// of str1 or not
function isPossible(str1, str2) {
  // To store the
  // occurrence of every
  // character
  let arr = new Array(256).fill(0);

  // Length of the two
  // strings
  let l1 = str1.length;
  let l2 = str2.length;

  let i, j;

  // Assume that it is
  // possible to compose the
  // string str2 from str1
  let possible = true;

  // Iterate over str1
  for (i = 0; i < l1; i++) {
    // Store the presence of every
    // character
    arr[str1[i]] = 1;
  }

  // Iterate over str2
  for (i = 0; i < l2; i++) {
    // Ignore the spaces
    if (str2[i] != " ") {
      // Check for the presence
      // of character in str1
      if (arr[str2[i]] == 1) continue;
      else {
        possible = false;
        break;
      }
    }
  }

  // If it is possible to make
  // str2 from str1
  if (possible) {
    document.write("Yes<br>");
  } else {
    document.write("No<br>");
  }
}

// Driver Code

// Given strings
let str1 = "we all love geeksforgeeks";
let str2 = "we all love geeks";

// Function Call
isPossible(str1, str2);

</script>

Output

Yes

Time Complexity: O(N+M) 
Auxiliary Space: O(1)