Check if the array can be sorted only if the elements on given positions can be swapped

Given an array arr[] of length N and another array P[] containing {a₁, a₂, … a_k} which represents the positions of the given array arr[], the task is to check if the array can be sorted by only swapping the elements’ arr[a_i], arr[a_i+1] where ‘i’ is some element in the array P[].
Examples:

Input: arr[] = {3, 2, 1}, P[] = {1, 2} 
Output: Yes 
Explanation:
Initially, i = 1 (i.e.) first element and second element are swapped. Therefore, arr[0] <=> arr[1]. arr[] = {2, 3, 1}. 
Similarly, i = 2 (i.e.) second element and third element are swapped. arr[] = {2, 1, 3}. 
Finally, i = 1 (i.e.) first element and second element are swapped. arr[] = {1, 2, 3}. 
Since this array is sorted, therefore, the given array can be sorted. 
Input: arr[] = {5, 3, -4, 1, 12}, P[] = {2, 4, 3} 
Output: No 

Approach: The idea is to use two pointer approach to check if the array can be sorted or not.  

• Initially, we create a position array pos[] of size N. This array will be used to mark the given positions in the array P[]. That is:

if j = ai (1 ≤ i ≤ K)
then the element pos[ai-1] will be 1
else 0

• Now, iterate through the array and check if pos[i] = 1 or not.
• If we encounter the pos[i]=1, we store the iterator in a temporary variable, and then we increment the iterator with value 1, till we have pos[i]=1 continuously, i.e.,

j = i
while (j < N and pos[j]) 
    j=j+1

• After this increment, we sort this segment that we obtained from i to j+1 and finally, check after the position j, in the vector that we have to check, because we have sorted till this segment.

Sort(arr[i] to arr[j+1])
i=j

• Finally, after this loop completion, we have to check if the array has been sorted or not.

Below is the implementation of the above approach: 

C++

// C++ program to check if the array
// can be sorted only if the elements
// on the given positions can be swapped

#include <bits/stdc++.h>
using namespace std;

// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
void check_vector(vector<int> A, int n,
                vector<int> p)
{

    // Creating an array for marking
    // the positions
    vector<int> pos(A.size());

    // Iterating through the array and
    // mark the positions
    for (int i = 0; i < p.size(); i++) {
        pos[p[i] - 1] = 1;
    }

    int flag = 1;

    // Iterating through the given array
    for (int i = 0; i < n; i++) {
        if (pos[i] == 0)
            continue;
        int j = i;

        // If pos[i] is 1, then incrementing
        // till 1 is continuously present in pos
        while (j < n && pos[j])
            ++j;

        // Sorting the required segment
        sort(A.begin() + i, A.begin() + j + 1);
        i = j;
    }

    // Checking if the vector is sorted or not
    for (int i = 0; i < n - 1; i++) {
        if (A[i] > A[i + 1]) {
            flag = 0;
            break;
        }
    }

    // Print yes if it is sorted
    if (flag == 1)
        cout << "Yes";
    else
        cout << "No";
}

// Driver code
int main()
{
    vector<int> A{ 3, 2, 1 };
    vector<int> p{ 1, 2 };

    check_vector(A, A.size(), p);
    return 0;
}

Java

// Java program to check if the array
// can be sorted only if the elements
// on the given positions can be swapped
import java.util.Arrays;

class GFG{

// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
public static void check_vector(int[] A,
                                int n,
                                int[] p)
{

    // Creating an array for marking
    // the positions
    int[] pos = new int[A.length];

    // Iterating through the array and
    // mark the positions
    for(int i = 0; i < p.length; i++)
    {
        pos[p[i] - 1] = 1;
    }

    int flag = 1;

    // Iterating through the given array
    for(int i = 0; i < n; i++)
    {
        if (pos[i] == 0)
            continue;

        int j = i;

        // If pos[i] is 1, then incrementing
        // till 1 is continuously present in pos
        while (j < n && pos[j] != 0)
            ++j;

        // Sorting the required segment
        Arrays.sort(A, i, j + 1);
        i = j;
    }

    // Checking if the vector is sorted or not
    for(int i = 0; i < n - 1; i++)
    {
        if (A[i] > A[i + 1])
        {
            flag = 0;
            break;
        }
    }

    // Print yes if it is sorted
    if (flag == 1)
        System.out.print("Yes");
    else
        System.out.print("No");
}

// Driver code
public static void main(String[] args)
{
    int A[] = { 3, 2, 1 };
    int p[] = { 1, 2 };

    check_vector(A, A.length, p);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to check if the array
# can be sorted only if the elements
# on the given positions can be swapped

# Function to check if the array can
# be sorted only if the elements on
# the given positions can be swapped
def check_vector(A, n, p):

    # Creating an array for marking
    # the positions
    pos = [0 for i in range(len(A))]

    # Iterating through the array and
    # mark the positions
    for i in range(len(p)):
        pos[p[i] - 1] = 1

    flag = 1

    # Iterating through the given array
    for i in range(n):
        if (pos[i] == 0):
            continue
        j = i

        # If pos[i] is 1, then incrementing
        # till 1 is continuously present in pos
        while (j < n and pos[j]):
            j += 1

        # Sorting the required segment
        p = A[: i]
        q = A[i : i + j + 1]
        r = A[i + j + 1 : len(A)]

        q.sort(reverse = False)
        A = p + q + r
        i = j

    # Checking if the vector is sorted or not
    for i in range(n - 1):
        if (A[i] > A[i + 1]):
            flag = 0
            break

    # Print yes if it is sorted
    if (flag == 1):
        print("Yes")
    else:
        print("No");

# Driver code
if __name__ == '__main__':

    A = [ 3, 2, 1 ]
    p = [ 1, 2 ]

    check_vector(A,len(A), p)

# This code is contributed by Samarth

C#

// C# program to check
// if the array can be
// sorted only if the
// elements on the given
// positions can be swapped
using System;
class GFG{

// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
public static void check_vector(int[] A,
                                int n,
                                int[] p)
{
  // Creating an array for marking
  // the positions
  int[] pos = new int[A.Length];

  // Iterating through the array and
  // mark the positions
  for(int i = 0; i < p.Length; i++)
  {
    pos[p[i] - 1] = 1;
  }

  int flag = 1;

  // Iterating through the given array
  for(int i = 0; i < n; i++)
  {
    if (pos[i] == 0)
      continue;

    int j = i;

    // If pos[i] is 1, then
    // incrementing till 1
    // is continuously present in pos
    while (j < n && pos[j] != 0)
      ++j;

    // Sorting the required segment
    Array.Sort(A, i, j + 1);
    i = j;
  }

  // Checking if the vector
  // is sorted or not
  for(int i = 0; i < n - 1; i++)
  {
    if (A[i] > A[i + 1])
    {
      flag = 0;
      break;
    }
  }

  // Print yes if it is sorted
  if (flag == 1)
    Console.Write("Yes");
  else
    Console.Write("No");
}

// Driver code
public static void Main()
{
  int[] A = {3, 2, 1};
  int[] p = {1, 2};
  check_vector(A, A.Length, p);
}
}

// This code is contributed by Chitranayal

Javascript

<script>
// Javascript program to check if the array
// can be sorted only if the elements
// on the given positions can be swapped

// Function to check if the array can
// be sorted only if the elements on
// the given positions can be swapped
function check_vector(A, n, p)
{

    // Creating an array for marking
    // the positions
    var pos = A.length;

    // Iterating through the array and
    // mark the positions
    for (var i = 0; i < p.length; i++) {
        pos[p[i] - 1] = 1;
    }

    var flag = 1;

    // Iterating through the given array
    for (var i = 0; i < n; i++) {
        if (pos[i] == 0)
            continue;
        var j = i;

        // If pos[i] is 1, then incrementing
        // till 1 is continuously present in pos
        while (j < n && pos[j])
            ++j;

        // Sorting the required segment
         A = Array.prototype.concat(
         A.slice(0, i),
         A.slice(i+1, j+1).sort((a,b)=>a-b),
         A.slice(j+2).sort((a,b)=>a-b));

        i = j;
    }

    // Checking if the vector is sorted or not
    for (var i = 0; i < n - 1; i++) {
        if (A[i] > A[i + 1]) {
            flag = 0;
            break;
        }
    }

    // Print yes if it is sorted
    if (flag == 1)
        document.write( "Yes");
    else
        document.write( "No");
}

// Driver code
var A = [ 3, 2, 1 ];
var p = [ 1, 2 ];
check_vector(A, A.length, p);

// This code is contributed by noob2000.
</script>

Output:

Yes

Time Complexity: O(N * log(N)), where N is the size of the array.