检查第二个字符串是否可以由使用任意次数的第一个字符串的字符组成

鉴于两个字符串的大小N str1和M尺寸.The任务的STR2是寻找是否有可能撰写STR2使用STR1这样的只有字符str1中的每一个字符，可以使用任意数量的时间。
注意：小写字母和大写字母应该被认为是不同的。

例子：

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 
 

天真的方法：最简单的方法是在str1 中搜索str2 的每个字符。如果找到所有字符，则打印“是”。否则打印“否”。

时间复杂度： O(N*M)
辅助空间： O(1)

有效的方法：这个想法是在count[]数组中标记str1的所有字符的存在。然后遍历str2并检查str2的字符是否存在于str1 中。

1. 创建一个大小为 256（ASCII字符总数）的数组count[]并将其设置为零。
2. 然后迭代str1并使count[str[i]] = 1来标记每个字符的出现。
3. 迭代str2并使用count[]数组检查该字符是否存在于str1 中。

下面是上述方法的实现：

CPP

// C++ implementation of
// the above approach
#include 
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

 
// 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);
 

输出：

Yes

时间复杂度： O(N+M)
辅助空间： O(1)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。