Count Array elements that occur before any of its prefix value of another Array

Given two arrays A[] and B[] of size N each, the task is to find the number of elements in array B[] that occur before any element that was present before it in array A[].

Example:

Input: N = 5, A[] = {3, 5, 1, 2, 4}, B[] = {4, 3, 1, 5, 2}
Output: 2
Explanation: Array A represent that 3 comes first then followed by 5, 1, 2, 4.
In array B, 4 must comes after 1, 2, 3, 5 but 4 comes before them. 
The value 1 is also in invalid order. It comes before 5.

Input: N = 3, A[] = {1, 3, 2}, B[] = {2, 1, 3}
Output: 1

Approach: The problem can be solved based on the following idea:

Find the number of elements that are present after all the elements that were present in its prefix in array A[]. Then the remaining elements are the ones that do not follow the rule.

Follow the steps mentioned below to implement the idea:

• Initialize i = 0 and j = 0 to iterate through array A[] and B[] respectively.
• While i and j are both less than N:
  • Increment j till B[j] is the same as A[i] or j goes out of the bound of the array size.
  • If A[i] and B[j] are the same then B[j] is satisfying the conditions. So increment the count (say C) for this element.
  • After the above iteration, increment the value of i to point to the next elements in array A[].
• When the loop ends, return the value of (N-C) as the required answer.

Below is the implementation of the above approach.

// C++ code to implement the approach

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

// Function to count the number of elements
// that occur before any of the element
// present in its prefix in array A[]
int InvalidOrder(int a[], int b[], int n)
{
    int i = 0, j = 0, temp, validcount = 0;

    // Loop to count the elements
    // that are in valid order
    while (i < n && j < n) {
        temp = j;
        while (temp < n && a[i] != b[temp]) {
            temp++;
        }
        if (temp != n) {

            validcount++;
            j = temp + 1;
        }
        i++;
    }

    // Return the answer
    return (n - validcount);
}

// Driver code
int main()
{
    int A[] = { 3, 5, 1, 2, 4 };
    int B[] = { 4, 3, 1, 5, 2 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function call
    cout << InvalidOrder(A, B, N);

    return 0;
}

// Java code to implement the approach
import java.io.*;

class GFG {

  // Function to count the number of elements
  // that occur before any of the element
  // present in its prefix in array A[]
  static int InvalidOrder(int[] a, int[] b, int n)
  {
    int i = 0, j = 0, temp, validcount = 0;

    // Loop to count the elements
    // that are in valid order
    while (i < n && j < n) {
      temp = j;
      while (temp < n && a[i] != b[temp]) {
        temp++;
      }
      if (temp != n) {
        validcount++;
        j = temp + 1;
      }
      i++;
    }

    // Return the answer
    return (n - validcount);
  }

  public static void main(String[] args)
  {
    int[] A = { 3, 5, 1, 2, 4 };
    int[] B = { 4, 3, 1, 5, 2 };
    int N = A.length;

    // Function call
    System.out.print(InvalidOrder(A, B, N));
  }
}

// This code is contributed by lokeshmvs21.

# python3 code to implement the approach

# Function to count the number of elements
# that occur before any of the element
# present in its prefix in array A[]
def InvalidOrder(a, b, n):

    i, j, temp, validcount = 0, 0, 0, 0

    # Loop to count the elements
    # that are in valid order
    while (i < n and j < n):
        temp = j
        while (temp < n and a[i] != b[temp]):
            temp += 1

        if (temp != n):

            validcount += 1
            j = temp + 1

        i += 1

    # Return the answer
    return (n - validcount)

# Driver code
if __name__ == "__main__":

    A = [3, 5, 1, 2, 4]
    B = [4, 3, 1, 5, 2]
    N = len(A)

    # Function call
    print(InvalidOrder(A, B, N))

    # This code is contributed by rakeshsahni

// C# implementation
using System;

public class GFG{

  static int InvalidOrder(int[] a, int[] b, int n)
  {
    int i = 0, j = 0, temp, validcount = 0;

    // Loop to count the elements
    // that are in valid order
    while (i < n && j < n) {
      temp = j;
      while (temp < n && a[i] != b[temp]) {
        temp++;
      }
      if (temp != n) {

        validcount++;
        j = temp + 1;
      }
      i++;
    }

    // Return the answer
    return (n - validcount);
  }

  static public void Main (){
    int[] A = { 3, 5, 1, 2, 4 };
    int[] B = { 4, 3, 1, 5, 2 };
    int N = A.Length;

    // Function call
    Console.WriteLine(InvalidOrder(A, B, N));
  }
}

// This code is contributed by ksam24000

 <script>
        // JavaScript code for the above approach

        // Function to count the number of elements
        // that occur before any of the element
        // present in its prefix in array A[]
        function InvalidOrder(a, b, n) {
            let i = 0, j = 0, temp, validcount = 0;

            // Loop to count the elements
            // that are in valid order
            while (i < n && j < n) {
                temp = j;
                while (temp < n && a[i] != b[temp]) {
                    temp++;
                }
                if (temp != n) {

                    validcount++;
                    j = temp + 1;
                }
                i++;
            }

            // Return the answer
            return (n - validcount);
        }

        // Driver code
        let A = [3, 5, 1, 2, 4];
        let B = [4, 3, 1, 5, 2];
        let N = A.length;

        // Function call
        document.write(InvalidOrder(A, B, N));

 // This code is contributed by Potta Lokesh

    </script>

2

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