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📜  通过重新排列第二个数组,将两个数组的相同索引元素的总和减少到小于 K

📅  最后修改于: 2021-09-07 02:13:26             🧑  作者: Mango

给定两个数组arr1[]arr2[] ,大小均为N和整数X ,任务是检查在重新排列第二个数组后,对应索引处的两个数组的相同索引元素的总和是否最多为 K大批。如果可能,则打印“是”,否则打印“否”

例子:

朴素方法:最简单的方法是生成数组B[] 的所有可能排列,如果任何排列满足给定条件,则打印Yes 。否则,打印No

时间复杂度: O(N!)
辅助空间: O(1)

高效的方法:为了优化上述方法,想法是使用排序。请按照以下步骤解决问题:

  • 按升序对数组arr1[]进行排序,按降序对arr2[]进行排序。
  • 现在,同时遍历两个数组,如果存在任何对使得arr1[i]arr2[]的总和超过X ,则打印“No” 。否则,打印“是”

下面是上述方法的实现:

C++
// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if elements
// of B[] can be rearranged
// such that A[i] + B[i] <= X
void rearrange(int A[], int B[],
               int N, int X)
{
    // Checks the given condition
    bool flag = true;
 
    // Sort A[] in ascending order
    sort(A, A + N);
 
    // Sort B[] in descending order
    sort(B, B + N, greater());
 
    // Traverse the arrays A[] and B[]
    for (int i = 0; i < N; i++) {
 
        // If A[i] + B[i] exceeds X
        if (A[i] + B[i] > X) {
 
            // Rearrangement not possible,
            // set flag to false
            flag = false;
            break;
        }
    }
 
    // If flag is true
    if (flag)
        cout << "Yes";
 
    // Otherwise
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3 };
    int B[] = { 1, 1, 2 };
    int X = 4;
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    rearrange(A, B, N, X);
 
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to check if elements
  // of B[] can be rearranged
  // such that A[i] + B[i] <= X
  static void rearrange(int A[], int B[],
                        int N, int X)
  {
 
    // Checks the given condition
    boolean flag = true;
 
    // Sort A[] in ascending order
    Arrays.sort(A);
 
    // Sort B[] in descending order
    Arrays.sort(B);
 
    // Traverse the arrays A[] and B[]
    for (int i = 0; i < N; i++)
    {
 
      // If A[i] + B[i] exceeds X
      if (A[i] + B[N - 1 - i] > X)
      {
 
        // Rearrangement not possible,
        // set flag to false
        flag = false;
        break;
      }
    }
 
    // If flag is true
    if (flag == true)
      System.out.print("Yes");
 
    // Otherwise
    else
      System.out.print("No");
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int A[] = { 1, 2, 3 };
    int B[] = { 1, 1, 2 };
    int X = 4;
    int N = A.length;
 
    // Function Call
    rearrange(A, B, N, X);
  }
}
 
// This code is contributed by AnkThon


Python3
# Python3 program for the above approach
 
# Function to check if elements
# of B can be rearranged
# such that A[i] + B[i] <= X
def rearrange(A, B, N, X):
   
    # Checks the given condition
    flag = True
 
    # Sort A in ascending order
    A = sorted(A)
 
    # Sort B in descending order
    B = sorted(B)[::-1]
 
    # Traverse the arrays A and B
    for i in range(N):
 
        # If A[i] + B[i] exceeds X
        if (A[i] + B[i] > X):
 
            # Rearrangement not possible,
            # set flag to false
            flag = False
            break
 
    # If flag is true
    if (flag):
        print("Yes")
 
    # Otherwise
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    A = [ 1, 2, 3 ]
    B = [ 1, 1, 2 ]
    X = 4
    N = len(A)
 
    # Function Call
    rearrange(A, B, N, X)
 
# This code is contributed by mohit kumar 29


C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to check if elements
  // of B[] can be rearranged
  // such that A[i] + B[i] <= X
  static void rearrange(int[] A, int[] B,
                        int N, int X)
  {
 
    // Checks the given condition
    bool flag = true;
 
    // Sort A[] in ascending order
    Array.Sort(A);
 
    // Sort B[] in descending order
    Array.Sort(B);
 
    // Traverse the arrays A[] and B[]
    for (int i = 0; i < N; i++)
    {
 
      // If A[i] + B[i] exceeds X
      if (A[i] + B[N - 1 - i] > X)
      {
 
        // Rearrangement not possible,
        // set flag to false
        flag = false;
        break;
      }
    }
 
    // If flag is true
    if (flag == true)
      Console.WriteLine("Yes");
 
    // Otherwise
    else
      Console.WriteLine("No");
  }
 
  // Driver Code
  public static void Main()
  {
    int []A = { 1, 2, 3 };
    int []B = { 1, 1, 2 };
    int X = 4;
    int N = A.Length;
 
    // Function Call
    rearrange(A, B, N, X);
  }
}
 
// This code is contributed by AnkThon


Javascript


输出:
Yes

时间复杂度: O(N*log N)
辅助空间: O(1)

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