检查是否可以通过交换具有不同第一个元素的对来对一组对进行排序

给定一个由N对组成的数组arr[] ，其中每对分别代表值和ID ，任务是检查是否可以通过仅交换具有不同ID 的对来按第一个元素对数组进行排序。如果可以排序，则打印“是” 。否则，打印“否” 。

例子：

Input: arr[] = {{340000, 2}, {45000, 1}, {30000, 2}, {50000, 4}}
Output: Yes
Explanation:
One of the possible way to sort the array is to swap the array elements in the following order:

1. Swap, arr[0] and arr[3], which modifies the array to arr[] = {{50000, 4}, {45000, 1}, {30000, 2}, {340000, 2}}.
2. Swap, arr[0] and arr[2], which modifies the array to arr[] = {{30000, 2}, {45000, 1}, {50000, 4}, {340000, 2}}.

Therefore, after the above steps the given array is sorted by the first element..

Input: arr[] = {{15000, 2}, {34000, 2}, {10000, 2}}
Output: No

方法：如果存在任意两个具有不同ID 的数组元素，则可以基于对数组进行排序的观察来解决给定的问题。请按照以下步骤解决问题：

• 初始化一个变量，比如X ，它将对的ID存储在索引0 处。
• 遍历数组arr[] ，如果存在任何ID与X不同的对，则打印“Yes”并跳出循环。
• 完成上述步骤后，如果所有元素都具有相同的ID，并且数组已经排序，则打印“Yes” 。否则，打印“否” 。

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 an
// array is sorted or not
bool isSorted(pair* arr,
              int N)
{
    // Traverse the array arr[]
    for (int i = 1; i < N; i++) {
 
        if (arr[i].first
            > arr[i - 1].first) {
            return false;
        }
    }
 
    // Return true
    return true;
}
 
// Function to check if it is possible
// to sort the array w.r.t. first element
string isPossibleToSort(
    pair* arr, int N)
{
    // Stores the ID of the first element
    int group = arr[0].second;
 
    // Traverse the array arr[]
    for (int i = 1; i < N; i++) {
 
        // If arr[i].second is not
        // equal to that of the group
        if (arr[i].second != group) {
            return "Yes";
        }
    }
 
    // If array is sorted
    if (isSorted(arr, N)) {
        return "Yes";
    }
    else {
        return "No";
    }
}
 
// Driver Code
int main()
{
    pair arr[]
        = { { 340000, 2 }, { 45000, 1 },
            { 30000, 2 }, { 50000, 4 } };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << isPossibleToSort(arr, N);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to check if an
    // array is sorted or not
    static boolean isSorted(int[][] arr, int N)
    {
        // Traverse the array arr[]
        for (int i = 1; i < N; i++) {
 
            if (arr[i][0] > arr[i - 1][0]) {
                return false;
            }
        }
 
        // Return true
        return true;
    }
 
    // Function to check if it is possible
    // to sort the array w.r.t. first element
    static String isPossibleToSort(int[][] arr, int N)
    {
        // Stores the ID of the first element
        int group = arr[0][1];
 
        // Traverse the array arr[]
        for (int i = 1; i < N; i++) {
 
            // If arr[i].second is not
            // equal to that of the group
            if (arr[i][1] != group) {
                return "Yes";
            }
        }
 
        // If array is sorted
        if (isSorted(arr, N)) {
            return "Yes";
        }
        else {
            return "No";
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int arr[][] = { { 340000, 2 },
                        { 45000, 1 },
                        { 30000, 2 },
                        { 50000, 4 } };
 
        int N = arr.length;
        System.out.print(isPossibleToSort(arr, N));
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to check if an
# array is sorted or not
def isSorted(arr, N):
   
    # Traverse the array arr[]
    for i in range(1, N):
        if (arr[i][0] > arr[i - 1][0]):
            return False
 
    # Return true
    return True
 
# Function to check if it is possible
# to sort the array w.r.t. first element
def isPossibleToSort(arr, N):
   
    # Stores the ID of the first element
    group = arr[0][1]
 
    # Traverse the array arr[]
    for i in range(1, N):
       
        # If arr[i][1] is not
        # equal to that of the group
        if (arr[i][1] != group):
            return "Yes"
 
    # If array is sorted
    if (isSorted(arr, N)):
        return "Yes"
    else:
        return "No"
 
# Driver Code
if __name__ == '__main__':
    arr = [ [ 340000, 2 ], [ 45000, 1 ],[ 30000, 2 ], [ 50000, 4 ] ]
 
    N = len(arr)
 
    print (isPossibleToSort(arr, N))
 
# This code is contributted by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG {
    // Function to check if an
    // array is sorted or not
    static bool isSorted(int[, ] arr, int N)
    {
        // Traverse the array arr[]
        for (int i = 1; i < N; i++) {
 
            if (arr[i, 0] > arr[i - 1, 0]) {
                return false;
            }
        }
 
        // Return true
        return true;
    }
 
    // Function to check if it is possible
    // to sort the array w.r.t. first element
    static string isPossibleToSort(int[, ] arr, int N)
    {
        // Stores the ID of the first element
        int group = arr[0, 1];
 
        // Traverse the array arr[]
        for (int i = 1; i < N; i++) {
 
            // If arr[i].second is not
            // equal to that of the group
            if (arr[i, 1] != group) {
                return "Yes";
            }
        }
 
        // If array is sorted
        if (isSorted(arr, N)) {
            return "Yes";
        }
        else {
            return "No";
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int[, ] arr = { { 340000, 2 },
                        { 45000, 1 },
                        { 30000, 2 },
                        { 50000, 4 } };
 
        int N = arr.GetLength(0);
 
        Console.WriteLine(isPossibleToSort(arr, N));
    }
}
 
// This code is contributed by ukasp.

Javascript

        // Javascript program for the above approach
 
        // Function to check if an
        // array is sorted or not
        function isSorted(arr, N) {
            // Traverse the array arr[]
            for (let i = 1; i < N; i++) {
 
                if (arr[i][0] > arr[i - 1][0]) {
                    return false;
                }
            }
 
            // Return true
            return true;
        }
 
        // Function to check if it is possible
        // to sort the array w.r.t. first element
        function isPossibleToSort(arr, N) {
            // Stores the ID of the first element
            let group = arr[0][1];
 
            // Traverse the array arr[]
            for (let i = 1; i < N; i++) {
 
                // If arr[i].second is not
                // equal to that of the group
                if (arr[i][1] != group) {
                    return "Yes";
                }
            }
 
            // If array is sorted
            if (isSorted(arr, N)) {
                return "Yes";
            }
            else {
                return "No";
            }
        }
 
        // Driver Code
 
        let arr = [[340000, 2],
        [15000, 2],
        [34000, 2],
        [10000, 2]];
 
        let N = arr.length;
        document.write(isPossibleToSort(arr, N));
 
        // This code is contributed by Hritik
    

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)

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