从给定数组中找到 N-1 对 (X, Y)，使得 X 和 Y 不同，并且 X 模 Y 未在数组中预设

给定一个大小为N的数组Arr[] ，由N对不同的正整数组成。任务是找到满足以下条件的N – 1 个不同的正整数X、Y对：

X≠Y
X、Y 都属于数组。
X mod Y 不属于数组。

Note: It is proved that N-1 such pairs always exists.

例子：

Input: N = 4 , Arr [ ] = { 2 , 3 ,4 ,5 }
Output: (5 , 2) , (4 , 2) , (3 , 2)
Explanation: 4 – 1 = 3, hence 3 such pairs printed.
In the first pair 5 and 2 both belongs to the array, 5 not equals to 2 , 5 mod 2 does not belongs to the array.
Same is applicable for all the printed pairs.

Input: N = 2, Arr [ ] = { 1 , 3 }
Output: 3 , 1
Explanation: 2 – 1 = 1. Hence only one such pair exists. That is 3 , 1.

编程需要懂一点英语

方法：上述问题可以使用贪心法来解决。在这种方法中，永远不要寻找所有可能的情况。相反，寻找我们需要的部分或数量。请按照以下步骤解决此问题：

将变量min_element初始化为Arr[0]。
使用变量i遍历范围[0, N)并执行以下任务：
- 计算min_element的值作为数组Arr[] 的最小元素。
使用变量i遍历范围[0, N)并执行以下任务：
- 如果Arr[i]不等于min_element，则打印Arr[i], min_element作为对。

这种方法基于以下观察：

Say X is the minimum element in the given array.
The value Arr[i] % X will always be less than X and no element in array is less than X

编程需要懂一点英语

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the possible pairs
void find(int N, int Arr[])
{
    int min_element = Arr[0];
 
    // Get the minimum element
    for (int i = 0; i < N; i++) {
        if (Arr[i] < min_element) {
            min_element = Arr[i];
        }
    }
 
    // Pair rest N - 1 elements with
    // the minimum element and print
    for (int i = 0; i < N; i++) {
        if (Arr[i] != min_element) {
            cout << Arr[i] << " " <<
                min_element << endl;
        }
    }
}
 
// Driver Code
int main()
{
 
    // Initialize N and the Array
    int N = 4;
    int Arr[4] = { 2, 3, 4, 5 };
 
    find(N, Arr);
 
    return 0;
}

Java

// Java  program for the above approach
import java.util.*;
public class GFG
{
   
// Function to find the possible pairs
static void find(int N, int Arr[])
{
    int min_element = Arr[0];
 
    // Get the minimum element
    for (int i = 0; i < N; i++) {
        if (Arr[i] < min_element) {
            min_element = Arr[i];
        }
    }
 
    // Pair rest N - 1 elements with
    // the minimum element and print
    for (int i = 0; i < N; i++) {
        if (Arr[i] != min_element) {
            System.out.println(Arr[i] + " " +
                min_element);
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int N = 4;
    int Arr[] = { 2, 3, 4, 5 };
 
    find(N, Arr);
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program for the above approach
 
# Function to find the possible pairs
def find(N, Arr):
 
    min_element = Arr[0]
 
    # Get the minimum element
    for i in range(0, N):
        if (Arr[i] < min_element):
            min_element = Arr[i]
 
    # Pair rest N - 1 elements with
    # the minimum element and print
    for i in range(0, N):
        if (Arr[i] != min_element):
            print(f"{Arr[i]} {min_element}")
 
# Driver Code
if __name__ == "__main__":
 
    # Initialize N and the Array
    N = 4
    Arr = [2, 3, 4, 5]
 
    find(N, Arr)
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG
{
   
// Function to find the possible pairs
static void find(int N, int []Arr)
{
    int min_element = Arr[0];
 
    // Get the minimum element
    for (int i = 0; i < N; i++) {
        if (Arr[i] < min_element) {
            min_element = Arr[i];
        }
    }
 
    // Pair rest N - 1 elements with
    // the minimum element and print
    for (int i = 0; i < N; i++) {
        if (Arr[i] != min_element) {
            Console.WriteLine(Arr[i] + " " +
                min_element);
        }
    }
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    int []Arr = { 2, 3, 4, 5 };
 
    find(N, Arr);
 
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

3 2
4 2
5 2

时间复杂度： O(N)
辅助空间： O(1)