Check if an array can be sorted by swapping pairs from indices consisting of unequal elements in another array
Given an array A[] of size N and a binary array B[] of size N, the task is to check if the array A[] can be converted into a sorted array by swapping pairs (A[i], A[j]) if B[i] is not equal to B[j]. If the array A[] can be sorted, then print “Yes“. Otherwise, print “No“.
Examples:
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Input: A[] = {3, 1, 2}, B[] = {0, 1, 1}
Output: Yes
Explanation:
Swap element at position 1 and position 2 of A[] since B[1]!=B[2]. So, A[] = {1, 3, 2}, B[] = {1, 0, 1}
Now, swap element at position 2 and position 3 of A[] since B[2]!=B[3]. So, A[] = {1, 2, 3}. Hence, it is sorted.Input: A[] = {5, 15, 4}, B[] = {0, 0, 0}
Output: No
Approach: The problem can be solved based on the following observations:
If at least two elements of the array, B[] are different, then it is possible to swap any two elements of the array A[].
Follow the steps below to solve the problem:
- Check if the given array A[] is already sorted in ascending order or not. If found to be true, then print “Yes”.
- Otherwise, count the number of 1s and 0s present in the array B[].
- If the array B[] contains at least one 0 and one 1, then print “Yes”.
- Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ Program for above approach#include <bits/stdc++.h>using namespace std;// Function to check if array, A[] can be converted// into sorted array by swapping (A[i], A[j]) if B[i]// not equal to B[j]bool checkifSorted(int A[], int B[], int N){ // Stores if array A[] is sorted // in descending order or not bool flag = false; // Traverse the array A[] for (int i = 0; i < N - 1; i++) { // If A[i] is greater than A[i + 1] if (A[i] > A[i + 1]) { // Update flag flag = true; break; } } // If array is sorted // in ascending order if (!flag) { return true; } // count = 2: Check if 0s and 1s // both present in the B[] int count = 0; // Traverse the array for (int i = 0; i < N; i++) { // If current element is 0 if (B[i] == 0) { // Update count count++; break; } } // Traverse the array B[] for (int i = 0; i < N; i++) { // If current element is 1 if (B[i] == 1) { count++; break; } } // If both 0s and 1s are present // in the array if (count == 2) { return true; } return false;}// Driver Codeint main(){ // Input array A[] int A[] = { 3, 1, 2 }; // Input array B[] int B[] = { 0, 1, 1 }; int N = sizeof(A) / sizeof(A[0]); // Function call bool check = checkifSorted(A, B, N); // If true,print YES if (check) { cout << "YES" <<endl; } // Else print NO else { cout << "NO" <<endl; } return 0;} |
Java
// Java program of the above approachimport java.io.*;class GFG { // Function to check if array, A[] can be converted // into sorted array by swapping (A[i], A[j]) if B[i] // not equal to B[j] static boolean checkifSorted(int A[], int B[], int N) { // Stores if array A[] is sorted // in descending order or not boolean flag = false; // Traverse the array A[] for (int i = 0; i < N - 1; i++) { // If A[i] is greater than A[i + 1] if (A[i] > A[i + 1]) { // Update flag flag = true; break; } } // If array is sorted // in ascending order if (!flag) { return true; } // count = 2: Check if 0s and 1s // both present in the B[] int count = 0; // Traverse the array for (int i = 0; i < N; i++) { // If current element is 0 if (B[i] == 0) { // Update count count++; break; } } // Traverse the array B[] for (int i = 0; i < N; i++) { // If current element is 1 if (B[i] == 1) { count++; break; } } // If both 0s and 1s are present // in the array if (count == 2) { return true; } return false; } // Driver Code public static void main(String[] args) { // Input array A[] int A[] = { 3, 1, 2 }; // Input array B[] int B[] = { 0, 1, 1 }; int N = A.length; // Function call boolean check = checkifSorted(A, B, N); // If true,print YES if (check) { System.out.println("YES"); } // Else print NO else { System.out.println("NO"); } }} |
Python3
# Python program of the above approach # Function to check if array, A[] can be converted# into sorted array by swapping (A[i], A[j]) if B[i]# not equal to B[j]def checkifSorted(A, B, N): # Stores if array A[] is sorted # in descending order or not flag = False # Traverse the array A[] for i in range( N - 1): # If A[i] is greater than A[i + 1] if (A[i] > A[i + 1]): # Update flag flag = True break # If array is sorted # in ascending order if (not flag): return True # count = 2: Check if 0s and 1s # both present in the B[] count = 0 # Traverse the array for i in range(N): # If current element is 0 if (B[i] == 0): # Update count count += 1 break # Traverse the array B[] for i in range(N): # If current element is 1 if B[i]: count += 1 break # If both 0s and 1s are present # in the array if (count == 2): return True return False# Driver Code# Input array A[]A = [ 3, 1, 2 ]# Input array B[]B = [ 0, 1, 1 ]N = len(A)# Function callcheck = checkifSorted(A, B, N)# If true,print YESif (check): print("YES") # Else print NOelse: print("NO") # This code is contributed by rohitsingh07052 |
C#
// C# program of the above approachusing System;public class GFG{ // Function to check if array, A[] can be converted // into sorted array by swapping (A[i], A[j]) if B[i] // not equal to B[j] static bool checkifSorted(int []A, int []B, int N) { // Stores if array A[] is sorted // in descending order or not bool flag = false; // Traverse the array A[] for (int i = 0; i < N - 1; i++) { // If A[i] is greater than A[i + 1] if (A[i] > A[i + 1]) { // Update flag flag = true; break; } } // If array is sorted // in ascending order if (!flag) { return true; } // count = 2: Check if 0s and 1s // both present in the B[] int count = 0; // Traverse the array for (int i = 0; i < N; i++) { // If current element is 0 if (B[i] == 0) { // Update count count++; break; } } // Traverse the array B[] for (int i = 0; i < N; i++) { // If current element is 1 if (B[i] == 1) { count++; break; } } // If both 0s and 1s are present // in the array if (count == 2) { return true; } return false; } // Driver Code public static void Main(string[] args) { // Input array A[] int []A = { 3, 1, 2 }; // Input array B[] int []B = { 0, 1, 1 }; int N = A.Length; // Function call bool check = checkifSorted(A, B, N); // If true,print YES if (check) { Console.WriteLine("YES"); } // Else print NO else { Console.WriteLine("NO"); } }}// This code is contributed by AnkThon |
Javascript
<script>// javascript program of the above approach // Function to check if array, A can be converted // into sorted array by swapping (A[i], A[j]) if B[i] // not equal to B[j] function checkifSorted(A , B , N) { // Stores if array A is sorted // in descending order or not var flag = false; // Traverse the array A for (i = 0; i < N - 1; i++) { // If A[i] is greater than A[i + 1] if (A[i] > A[i + 1]) { // Update flag flag = true; break; } } // If array is sorted // in ascending order if (!flag) { return true; } // count = 2: Check if 0s and 1s // both present in the B var count = 0; // Traverse the array for (i = 0; i < N; i++) { // If current element is 0 if (B[i] == 0) { // Update count count++; break; } } // Traverse the array B for (i = 0; i < N; i++) { // If current element is 1 if (B[i] == 1) { count++; break; } } // If both 0s and 1s are present // in the array if (count == 2) { return true; } return false; } // Driver Code // Input array A var A = [ 3, 1, 2 ]; // Input array B var B = [ 0, 1, 1 ]; var N = A.length; // Function call var check = checkifSorted(A, B, N); // If true,print YES if (check) { document.write("YES"); } // Else print NO else { document.write("NO"); }// This code contributed by Rajput-Ji</script> |
Output:
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
