Given an array arr[] of size N consisting of distinct integers from range [0, N – 1] arranged in a random order. Also given a few pairs where each pair denotes the indices where the elements of the array can be swapped. There is no limit on the number of swaps allowed. The task is to find if it is possible to arrange the array in ascending order using these swaps. If possible then print Yes else print No.
Examples:
Input: arr[] = {0, 4, 3, 2, 1, 5}, pairs[][] = {{1, 4}, {2, 3}}
Output: Yes
swap(arr[1], arr[4]) -> arr[] = {0, 1, 3, 2, 4, 5}
swap(arr[2], arr[3]) -> arr[] = {0, 1, 2, 3, 4, 5}
Input: arr[] = {1, 2, 3, 0, 4}, pairs[][] = {{2, 3}}
Output: No
Approach: The given problem can be considered as a graph problem where N denotes the total number of nodes in the graph and each swapping pair denotes an undirected edge in the graph. We have to find out if it is possible to convert the input array in the form of {0, 1, 2, 3, …, N – 1}.
Let us call the above array as B. Now find out all the connected components of both the arrays and if the elements differ for at least one component then the answer is No else the answer is Yes.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if the array elements// can be sorted with the given operationbool canBeSorted(int N, vector<int> a, int P, vector<pair<int, int> > vp){ // To create the adjacency list of the graph vector<int> v[N]; // Boolean array to mark the visited nodes bool vis[N] = { false }; // Creating adjacency list for undirected graph for (int i = 0; i < P; i++) { v[vp[i].first].push_back(vp[i].second); v[vp[i].second].push_back(vp[i].first); } for (int i = 0; i < N; i++) { // If not already visited // then apply BFS if (!vis[i]) { queue<int> q; vector<int> v1; vector<int> v2; // Set visited to true vis[i] = true; // Push the node to the queue q.push(i); // While queue is not empty while (!q.empty()) { int u = q.front(); v1.push_back(u); v2.push_back(a[u]); q.pop(); // Check all the adjacent nodes for (auto s : v[u]) { // If not visited if (!vis[s]) { // Set visited to true vis[s] = true; q.push(s); } } } sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); // If the connected component does not // contain same elements then return false if (v1 != v2) return false; } } return true;}// Driver codeint main(){ vector<int> a = { 0, 4, 3, 2, 1, 5 }; int n = a.size(); vector<pair<int, int> > vp = { { 1, 4 }, { 2, 3 } }; int p = vp.size(); if (canBeSorted(n, a, p, vp)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approach import java.io.*;import java.util.*;class GFG { // Function that returns true if the array elements // can be sorted with the given operation static boolean canBeSorted(int N, ArrayList<Integer> a, int p, ArrayList<ArrayList<Integer>> vp) { // To create the adjacency list of the graph ArrayList<ArrayList<Integer>> v = new ArrayList<ArrayList<Integer>>(); for(int i = 0; i < N; i++) { v.add(new ArrayList<Integer>()); } // Boolean array to mark the visited nodes boolean[] vis = new boolean[N]; // Creating adjacency list for undirected graph for (int i = 0; i < p; i++) { v.get(vp.get(i).get(0)).add(vp.get(i).get(1)); v.get(vp.get(i).get(1)).add(vp.get(i).get(0)); } for (int i = 0; i < N; i++) { // If not already visited // then apply BFS if (!vis[i]) { Queue<Integer> q = new LinkedList<>(); ArrayList<Integer> v1 = new ArrayList<Integer>(); ArrayList<Integer> v2 = new ArrayList<Integer>(); // Set visited to true vis[i] = true; // Push the node to the queue q.add(i); // While queue is not empty while (q.size() > 0) { int u = q.poll(); v1.add(u); v2.add(a.get(u)); // Check all the adjacent nodes for(int s: v.get(u)) { // If not visited if (!vis[s]) { // Set visited to true vis[s] = true; q.add(s); } } } Collections.sort(v1); Collections.sort(v2); // If the connected component does not // contain same elements then return false if(!v1.equals(v2)) { return false; } } } return true; } // Driver code public static void main (String[] args) { ArrayList<Integer> a = new ArrayList<Integer>(Arrays.asList(0, 4, 3, 2, 1, 5)); int n = a.size(); ArrayList<ArrayList<Integer>> vp = new ArrayList<ArrayList<Integer>>(); vp.add(new ArrayList<Integer>(Arrays.asList(1, 4))); vp.add(new ArrayList<Integer>(Arrays.asList(2, 3))); int p = vp.size(); if (canBeSorted(n, a, p, vp)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 implementation of the approachfrom collections import deque as queue# Function that returns true if the array elements# can be sorted with the given operationdef canBeSorted(N, a, P, vp): # To create the adjacency list of the graph v = [[] for i in range(N)] # Boolean array to mark the visited nodes vis = [False]*N # Creating adjacency list for undirected graph for i in range(P): v[vp[i][0]].append(vp[i][1]) v[vp[i][1]].append(vp[i][0]) for i in range(N): # If not already visited # then apply BFS if (not vis[i]): q = queue() v1 = [] v2 = [] # Set visited to true vis[i] = True # Push the node to the queue q.append(i) # While queue is not empty while (len(q) > 0): u = q.popleft() v1.append(u) v2.append(a[u]) # Check all the adjacent nodes for s in v[u]: # If not visited if (not vis[s]): # Set visited to true vis[s] = True q.append(s) v1 = sorted(v1) v2 = sorted(v2) # If the connected component does not # contain same elements then return false if (v1 != v2): return False return True# Driver codeif __name__ == '__main__': a = [0, 4, 3, 2, 1, 5] n = len(a) vp = [ [ 1, 4 ], [ 2, 3 ] ] p = len(vp) if (canBeSorted(n, a, p, vp)): print("Yes") else: print("No")# This code is contributed by mohit kumar 29 |
Yes
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.
