Skip to content
Related Articles

Related Articles

Improve Article
Check if an array of pairs can be sorted by swapping pairs with different first elements
  • Last Updated : 20 May, 2021

Given an array arr[] consisting of N pairs, where each pair represents the value and ID respectively, the task is to check if it is possible to sort the array by the first element by swapping only pairs having different IDs. If it is possible to sort, then print “Yes”. Otherwise, print “No”.

Examples:

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

Approach: The given problem can be solved based on the observation that the array can be sorted if there exist any two array elements with different IDs. Follow the steps below to solve the problem:

  • Initialize a variable, say X that stores the ID of the pair at index 0.
  • Traverse the array arr[] and if there exists any pair whose ID is different from X, then print “Yes” and break out of the loop.
  • After completing the above steps, if all the elements have the same IDs and if the array is already sorted then print “Yes”. Otherwise, Print “No”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if an
// array is sorted or not
bool isSorted(pair<int, int>* 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<int, int>* 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<int, int> 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




<script>
        // 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
    </script>
Output: 
Yes

 

Time Complexity: O(N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :