检查一个数组是否可以放入另一个数组中，以重新排列该数组中的元素

给定两个具有相同大小N的数组A和B。检查数组A是否可以适合于数组B。如果通过排列两个数组的元素都存在一个解决方案，使得第i^个元素适合，则称该数组适合另一个数组第一个数组的^第一个元素的^个数小于或等于第二个数组的^第i^个元素。

例子：

Input : A[] = { 7, 5, 3, 2 }, B[] = { 5, 4, 8, 7 }
Output : YES 
Rearrange the first array to {3, 2, 7, 5} 
Do not rearrange the second array's element. 
After rearranging, all Ai<=Bi. 

Input : A[] = { 7, 5, 3, 2, 5, 105, 45, 10 }, B[] = { 2, 4, 0, 5, 6, 9, 75, 84 }
Output : NO

方法：对两个数组都进行排序，并检查所有0≤i≤N的A _i是否小于或等于B _i 。如果在^第i^个位置A _i大于B _i ，则返回false，否则返回true。
下面是上述方法的实现：

C++

// C++ Program to check whether an array
// can be fit into another array with given
// condition.
#include 
using namespace std;
  
// Returns true if the array A can be fit into
// array B, otherwise false
bool checkFittingArrays(int A[], int B[], int N)
{
    // Sort both the arrays
    sort(A, A + N);
    sort(B, B + N);
  
    // Iterate over the loop and check whether every
    // array element of A is less than or equal to
    // its corresponding array element of B
    for (int i = 0; i < N; i++)
        if (A[i] > B[i])
            return false;
  
    return true;
}
  
// Driver Code
int main()
{
    int A[] = { 7, 5, 3, 2 };
    int B[] = { 5, 4, 8, 7 };
    int N = sizeof(A) / sizeof(A[0]);
  
    if (checkFittingArrays(A, B, N))
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java Program to check
// whether an array can
// be fit into another
// array with given
// condition.
import java.io.*;
import java.util.*;
import java.lang.*;
  
class GFG 
{
  
// Returns true if the
// array A can be fit 
// into array B, 
// otherwise false
static boolean checkFittingArrays(int []A,
                                  int []B, 
                                  int N)
{
      
    // Sort both the arrays
    Arrays.sort(A);
    Arrays.sort(B);
  
    // Iterate over the loop
    // and check whether every
    // array element of A is 
    // less than or equal to
    // its corresponding array 
    // element of B
    for (int i = 0; i < N; i++)
        if (A[i] > B[i])
            return false;
  
    return true;
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = {7, 5, 3, 2};
    int B[] = {5, 4, 8, 7};
    int N = A.length;
      
    if (checkFittingArrays(A, B, N))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}

Python3

# Python3 Program to check whether an array
# can be fit into another array with given
# condition.
  
# Returns true if the array A can be fit into
# array B, otherwise false
def checkFittingArrays(A, B, N):
      
    # Sort both the arrays
    A = sorted(A)
    B = sorted(B)
  
    # Iterate over the loop and check whether 
    # every array element of A is less than 
    # or equal to its corresponding array 
    # element of B
    for i in range(N):
        if (A[i] > B[i]):
            return False
  
    return True
  
# Driver Code
A = [7, 5, 3, 2]
B = [5, 4, 8, 7]
N = len(A)
  
if (checkFittingArrays(A, B, N)):
    print("YES")
else:
    print("NO")
  
# This code is contributed 
# by mohit kumar

C#

// C# Program to check
// whether an array can
// be fit into another
// array with given
// condition.
using System;
  
class GFG 
{
  
// Returns true if the
// array A can be fit 
// into array B, 
// otherwise false
static bool checkFittingArrays(int []A,
                               int []B, 
                               int N)
{
      
    // Sort both the arrays
    Array.Sort(A);
    Array.Sort(B);
  
    // Iterate over the loop
    // and check whether every
    // array element of A is 
    // less than or equal to
    // its corresponding array 
    // element of B
    for (int i = 0; i < N; i++)
        if (A[i] > B[i])
            return false;
  
    return true;
}
  
// Driver Code
public static void Main () 
{
    int []A = {7, 5, 3, 2};
    int []B = {5, 4, 8, 7};
    int N = A.Length;
      
    if (checkFittingArrays(A, B, N))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
  
// This code is contributed 
// by anuj_67.

PHP

 $B[$i])
            return false;
  
    return true;
}
  
// Driver Code
$A = array( 7, 5, 3, 2 );
$B = array( 5, 4, 8, 7 );
$N = count($A);
  
if (checkFittingArrays($A, $B, $N))
    echo "YES";
else
    echo "NO";
      
// This code is contributed by shs
?>

输出：

YES

时间复杂度：O(N * logN)，其中N是数组的大小。