Number of moves required between the arrays to complete the traversal in sorted order

Given two sorted arrays, X[] of size N and Y[] of size M having unique values. The task is to count the total number of moves required between the arrays to traverse all the elements in both the arrays in ascending order if initially, the traversal starts from the X[] array.

Examples:

Input: X[] = {1}, Y[] = {2, 3, 4}
Output: 1 
Explanation: Only 1 move is required after traversing the X array and then move to the 0 index of the Y array and traverse its rest of the values.

Input: X[] = {1, 3, 4}, Y[] = {2, 5, 6}
Output: 3

Approach: The given problem can be solved using the two-pointer technique. Follow the below steps to solve the problem:

• Initialize two pointers, say i as 0 and j as 0 pointing to the X[] and Y[] array respectively.
• Initialize another variable total_moves as 0 to store the number of moves required.
• Since traversal always starts from the X[] array, so first compare the values at the current index. There arise two cases:
• If currently present at X[] array:
  • If X[i] < Y[j], just increment index i.
  • If X[i] > Y[j], increment total_moves and index j.
• If currently present at Y[] array:
  • If Y[j] < X[i], increment index j.
  • If Y[j] > X[i], increment total_moves and index i.
• Repeat the above steps until any one of the array traversals is finished.
• Once the above loop finishes, total_moves would be incremented in the following two conditions:
  • If traversal finishes at X array and j < M, then increment total_moves by 1.
  • If traversal finishes at Y array and i < N, then increment total_moves by 1.
• Print the value of total_moves as the result.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the number of moves
// required between the arrays to complete
// the traversal in sorted order
int numberofMoves(int X[], int Y[], int n, int m)
{

    // Variable to check if present
    // at X array or not
    bool present_at_X = true;

    // Store total number of moves
    int total_moves = 0;

    // Initialize i and j pointing to X and
    // Y array respectively
    int i = 0, j = 0;

    // Loop until one array is
    // completely traversed
    while (i < n && j < m) {

        // If currently present at X array,
        // and X[i]<Y[j], increment index i
        if (present_at_X == true && X[i] < Y[j]) {
            i++;
        }

        // If present at X array and Y[j]<X[i]
        else if (present_at_X == true && Y[j] < X[i]) {

            // Increment total_moves and update
            // present at X variable and index j
            total_moves++;
            present_at_X = false;
            j++;
        }

        // If present at Y array and
        // Y[j] < X[i], increment j
        else if (present_at_X == false && Y[j] < X[i]) {
            j++;
        }

        // If present at Y array and
        // X[i] < Y[j]
        else if (present_at_X == false && X[i] < Y[j]) {

            // Increment total_moves and index i
            // and update present at X variable
            total_moves++;
            present_at_X = true;
            i++;
        }
    }

    // Check if present at X array and j < m,
    // so increment total_moves by 1
    if (present_at_X == true && j < m) {
        total_moves++;
        present_at_X = false;
    }

    // If finished traversal at Y array
    // and X's array traversal is left,
    // increment total_moves by 1
    if (present_at_X == false && i < n) {
        total_moves++;
        present_at_X = true;
    }

    // Return the result
    return total_moves;
}

// Driver Code
int main()
{

    // Given Input
    int X[] = { 1, 3, 4 };
    int n = sizeof(X) / sizeof(X[0]);
    int Y[] = { 2, 5, 6 };
    int m = sizeof(Y) / sizeof(Y[0]);

    // Function Call
    cout << numberofMoves(X, Y, n, m);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the number of moves
// required between the arrays to complete
// the traversal in sorted order
static int numberofMoves(int[] X, int Y[], int n, int m)
{

    // Variable to check if present
    // at X array or not
    boolean present_at_X = true;

    // Store total number of moves
    int total_moves = 0;

    // Initialize i and j pointing to X and
    // Y array respectively
    int i = 0, j = 0;

    // Loop until one array is
    // completely traversed
    while (i < n && j < m) {

        // If currently present at X array,
        // and X[i]<Y[j], increment index i
        if (present_at_X == true && X[i] < Y[j]) {
            i++;
        }

        // If present at X array and Y[j]<X[i]
        else if (present_at_X == true && Y[j] < X[i]) {

            // Increment total_moves and update
            // present at X variable and index j
            total_moves++;
            present_at_X = false;
            j++;
        }

        // If present at Y array and
        // Y[j] < X[i], increment j
        else if (present_at_X == false && Y[j] < X[i]) {
            j++;
        }

        // If present at Y array and
        // X[i] < Y[j]
        else if (present_at_X == false && X[i] < Y[j]) {

            // Increment total_moves and index i
            // and update present at X variable
            total_moves++;
            present_at_X = true;
            i++;
        }
    }

    // Check if present at X array and j < m,
    // so increment total_moves by 1
    if (present_at_X == true && j < m) {
        total_moves++;
        present_at_X = false;
    }

    // If finished traversal at Y array
    // and X's array traversal is left,
    // increment total_moves by 1
    if (present_at_X == false && i < n) {
        total_moves++;
        present_at_X = true;
    }

    // Return the result
    return total_moves;
}

// Driver Code
public static void main(String[] args)
{
    // Given Input
    int X[] = { 1, 3, 4 };
    int n = X.length;
    int Y[] = { 2, 5, 6 };
    int m = Y.length;

    // Function Call
    System.out.print(numberofMoves(X, Y, n, m));
}
}

// This code is contributed by sanjoy_62.

Python3

# Python3 program for the above approach

# Function to find the number of moves
# required between the arrays to complete
# the traversal in sorted order
def numberofMoves(X, Y, n, m):

    # Variable to check if present
    # at X array or not
    present_at_X = True

    # Store total number of moves
    total_moves = 0

    # Initialize i and j pointing to X and
    # Y array respectively
    i, j = 0, 0

    # Loop until one array is
    # completely traversed
    while (i < n and j < m):

        # If currently present at X array,
        # and X[i]<Y[j], increment index i
        if (present_at_X == True and X[i] < Y[j]):
            i += 1

        # If present at X array and Y[j]<X[i]
        elif (present_at_X == True and Y[j] < X[i]):

            # Increment total_moves and update
            # present at X variable and index j
            total_moves += 1
            present_at_X = False
            j += 1

        # If present at Y array and
        # Y[j] < X[i], increment j
        elif (present_at_X == False and Y[j] < X[i]):
            j += 1

        # If present at Y array and
        # X[i] < Y[j]
        elif (present_at_X == False and X[i] < Y[j]):

            # Increment total_moves and index i
            # and update present at X variable
            total_moves += 1
            present_at_X = True
            i += 1

    # Check if present at X array and j < m,
    # so increment total_moves by 1
    if (present_at_X == True and j < m):
        total_moves += 1
        present_at_X = False

    # If finished traversal at Y array
    # and X's array traversal is left,
    # increment total_moves by 1
    if (present_at_X == False and i < n):
        total_moves += 1
        present_at_X = True

    # Return the result
    return total_moves

# Driver Code
if __name__ == '__main__':

    # Given Input
    X = [ 1, 3, 4 ]
    n = len(X)
    Y = [ 2, 5, 6 ]
    m = len(Y)

    # Function Call
    print(numberofMoves(X, Y, n, m))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach

using System;

public class GFG{

    // Function to find the number of moves
// required between the arrays to complete
// the traversal in sorted order
static int numberofMoves(int[] X, int[] Y, int n, int m)
{

    // Variable to check if present
    // at X array or not
    bool present_at_X = true;

    // Store total number of moves
    int total_moves = 0;

    // Initialize i and j pointing to X and
    // Y array respectively
    int i = 0, j = 0;

    // Loop until one array is
    // completely traversed
    while (i < n && j < m) {

        // If currently present at X array,
        // and X[i]<Y[j], increment index i
        if (present_at_X == true && X[i] < Y[j]) {
            i++;
        }

        // If present at X array and Y[j]<X[i]
        else if (present_at_X == true && Y[j] < X[i]) {

            // Increment total_moves and update
            // present at X variable and index j
            total_moves++;
            present_at_X = false;
            j++;
        }

        // If present at Y array and
        // Y[j] < X[i], increment j
        else if (present_at_X == false && Y[j] < X[i]) {
            j++;
        }

        // If present at Y array and
        // X[i] < Y[j]
        else if (present_at_X == false && X[i] < Y[j]) {

            // Increment total_moves and index i
            // and update present at X variable
            total_moves++;
            present_at_X = true;
            i++;
        }
    }

    // Check if present at X array and j < m,
    // so increment total_moves by 1
    if (present_at_X == true && j < m) {
        total_moves++;
        present_at_X = false;
    }

    // If finished traversal at Y array
    // and X's array traversal is left,
    // increment total_moves by 1
    if (present_at_X == false && i < n) {
        total_moves++;
        present_at_X = true;
    }

    // Return the result
    return total_moves;
}

// Driver Code
    static public void Main ()
    {

        // Given Input
    int[] X = { 1, 3, 4 };
    int n = X.Length;
    int[] Y = { 2, 5, 6 };
    int m = Y.Length;

    // Function Call
    Console.WriteLine(numberofMoves(X, Y, n, m));
    }
}

// This code is contributed by unknown2108.

Javascript

<script>

// JavaScript program for the above approach


// Function to find the number of moves
// required between the arrays to complete
// the traversal in sorted order
function numberofMoves(X,Y,n,m)
{

    // Variable to check if present
    // at X array or not
    let present_at_X = true;

    // Store total number of moves
    let total_moves = 0;

    // Initialize i and j pointing to X and
    // Y array respectively
    let i = 0, j = 0;

    // Loop until one array is
    // completely traversed
    while (i < n && j < m) {

        // If currently present at X array,
        // and X[i]<Y[j], increment index i
        if (present_at_X == true && X[i] < Y[j]) {
            i++;
        }

        // If present at X array and Y[j]<X[i]
        else if (present_at_X == true && Y[j] < X[i]) {

            // Increment total_moves and update
            // present at X variable and index j
            total_moves++;
            present_at_X = false;
            j++;
        }

        // If present at Y array and
        // Y[j] < X[i], increment j
        else if (present_at_X == false && Y[j] < X[i]) {
            j++;
        }

        // If present at Y array and
        // X[i] < Y[j]
        else if (present_at_X == false && X[i] < Y[j]) {

            // Increment total_moves and index i
            // and update present at X variable
            total_moves++;
            present_at_X = true;
            i++;
        }
    }

    // Check if present at X array and j < m,
    // so increment total_moves by 1
    if (present_at_X == true && j < m) {
        total_moves++;
        present_at_X = false;
    }

    // If finished traversal at Y array
    // and X's array traversal is left,
    // increment total_moves by 1
    if (present_at_X == false && i < n) {
        total_moves++;
        present_at_X = true;
    }

    // Return the result
    return total_moves;
}

// Driver Code


    // Given Input
   var X = [1, 3, 4 ];
   var n =X.length;
   var Y = [2, 5, 6 ];
   var m = Y.length;

    // Function Call
    document.write(numberofMoves(X, Y, n, m));

// This code is contributed by Potta Lokesh

</script>

Output

3

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