Data Modelling & AI Data Structure & Algorithm

Check if two sorted arrays can be merged to form a sorted array with no adjacent pair from the same array

23 June 2025

2

Given two sorted arrays A[] and B[] of size N, the task is to check if it is possible to merge two given sorted arrays into a new sorted array such that no two consecutive elements are from the same array.

Examples:

Input: A[] = {3, 5, 8}, B[] = {2, 4, 6}
Output: Yes

Explanation: Merged array = {B[0], A[0], B[1], A[1], B[2], A[2]}
Since the resultant array is sorted array, the required output is Yes.

Input: A[] = {12, 4, 2, 5, 3}, B[] = {32, 13, 43, 10, 8}
Output: No

Approach: Follow the steps below to solve the problem:

Initialize a variable, say flag = true to check if it is possible to form a new sorted array by merging the given two sorted arrays such that no two consecutive elements are from the same array.
Initialize a variable, say prev to check if the previous element of the merge array are from the array A[] or the array B[]. If prev == 1 then the previous element are from the array A[] and if prev == 0 then the previous element are from the array B[].
Traverse both the array using variables, i and j and check the following conditions:
- If A[i] < B[j] and prev != 0 then increment the value of i and update the value of prev to 0.
- If B[j] < A[i[ and prev != 1 then increment the value of j and update the value of prev to 1.
- If A[i] == B[j] and prev != 1 then increment the value of j and update the value of prev to 1.
- If A[i] == B[j] and prev != 0 then increment the value of i and update the value of prev to 0.
- If none of the above condition satisfy then update flag = false.
Finally, print the value of flag.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
bool checkIfPossibleMerge(int A[], int B[], int N)
{
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are from
    // the array A[] or from the array B[]
    int prev = -1;
 
    // Check if it is possible to merge the two
    // given sorted arrays with given conditions
    int flag = 1;
 
    // Traverse both the arrays
    while (i < N && j < N) {
 
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0) {
 
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1) {
 
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j]) {
 
            // If previous element
            // are not from B[]
            if (prev != 1) {
 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else {
 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else {
 
            // Update flag
            flag = 0;
            break;
        }
    }
 
    return flag;
}
 
// Driver Code
int main()
{
    int A[3] = { 3, 5, 8 };
    int B[3] = { 2, 4, 6 };
    int N = sizeof(A) / sizeof(A[0]);
 
    if (checkIfPossibleMerge(A, B, N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
static boolean checkIfPossibleMerge(int[] A, int[] B,
                                    int N)
{
     
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are from
    // the array A[] or from the array B[]
    int prev = -1;
 
    // Check if it is possible to merge the two
    // given sorted arrays with given conditions
    boolean flag = true;
 
    // Traverse both the arrays
    while (i < N && j < N) 
    {
         
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0)
        {
             
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1)
        {
             
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j]) 
        {
             
            // If previous element
            // are not from B[]
            if (prev != 1)
            {
                 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else
            {
                 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else
        {
             
            // Update flag
            flag = false;
            break;
        }
    }
    return flag;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] A = { 3, 5, 8 };
    int[] B = { 2, 4, 6 };
    int N = A.length;
 
    if (checkIfPossibleMerge(A, B, N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
 
# Function to check if it is possible 
# to merge the two given arrays with 
# given conditions
def checkIfPossibleMerge(A, B, N):
     
    # Stores index of
    # the array A[]
    i = 0
 
    # Stores index of
    # the array  B[]
    j = 0
 
    # Check if the previous element 
    # are from the array A[] or from
    # the array B[]
    prev = -1
 
    # Check if it is possible to merge 
    # the two given sorted arrays with
    # given conditions
    flag = 1
 
    # Traverse both the arrays
    while (i < N and j < N):
 
        # If A[i] is less than B[j] and
        # previous element are not from A[]
        if (A[i] < B[j] and prev != 0):
 
            # Update prev
            prev = 0
 
            # Update i
            i += 1
 
        # If B[j] is less than A[i] and
        # previous element are not from B[]
        elif (B[j] < A[i] and prev != 1):
 
            # Update prev
            prev = 1
 
            # Update j
            j += 1
 
        # If A[i] equal to B[j]
        elif (A[i] == B[j]):
 
            # If previous element
            # are not from B[]
            if (prev != 1):
                 
                # Update prev
                prev = 1
 
                # Update j
                j += 1
 
            # If previous element is
            # not from A[]
            else:
 
                # Update prev
                prev = 0
 
                # Update i
                i += 1
 
        # If it is not possible to merge two
        # sorted arrays with given conditions
        else:
 
            # Update flag
            flag = 0
            break
 
    return flag
 
# Driver Code
if __name__ == '__main__':
 
    A = [ 3, 5, 8 ]
    B = [ 2, 4, 6 ]
    N = len(A)
 
    if (checkIfPossibleMerge(A, B, N)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
static bool checkIfPossibleMerge(int[] A, int[] B,
                                 int N)
{
     
    // Stores index of
    // the array A[]
    int i = 0;
 
    // Stores index of
    // the array  B[]
    int j = 0;
 
    // Check if the previous element are 
    // from the array A[] or from the 
    // array B[]
    int prev = -1;
 
    // Check if it is possible to merge
    // the two given sorted arrays with 
    // given conditions
    bool flag = true;
 
    // Traverse both the arrays
    while (i < N && j < N) 
    {
         
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0) 
        {
             
            // Update prev
            prev = 0;
 
            // Update i
            i++;
        }
 
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1)
        {
             
            // Update prev
            prev = 1;
 
            // Update j
            j++;
        }
 
        // If A[i] equal to B[j]
        else if (A[i] == B[j])
        {
             
            // If previous element
            // are not from B[]
            if (prev != 1) 
            {
                 
                // Update prev
                prev = 1;
 
                // Update j
                j++;
            }
 
            // If previous element is
            // not from A[]
            else
            {
                 
                // Update prev
                prev = 0;
 
                // Update i
                i++;
            }
        }
 
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else
        {
             
            // Update flag
            flag = false;
            break;
        }
    }
    return flag;
}
 
// Driver Code
public static void Main()
{
    int[] A = { 3, 5, 8 };
    int[] B = { 2, 4, 6 };
    int N = A.Length;
 
    if (checkIfPossibleMerge(A, B, N))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by akhilsaini

Javascript

<script>
 
// Javascript program to implement
// the above approach
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
function checkIfPossibleMerge(A, B, N)
{
      
    // Stores index of
    // the array A[]
    let i = 0;
  
    // Stores index of
    // the array  B[]
    let j = 0;
  
    // Check if the previous element are from
    // the array A[] or from the array B[]
    let prev = -1;
  
    // Check if it is possible to merge the two
    // given sorted arrays with given conditions
    let flag = true;
  
    // Traverse both the arrays
    while (i < N && j < N)
    {
          
        // If A[i] is less than B[j] and
        // previous element are not from A[]
        if (A[i] < B[j] && prev != 0)
        {
              
            // Update prev
            prev = 0;
  
            // Update i
            i++;
        }
  
        // If B[j] is less than A[i] and
        // previous element are not from B[]
        else if (B[j] < A[i] && prev != 1)
        {
              
            // Update prev
            prev = 1;
  
            // Update j
            j++;
        }
  
        // If A[i] equal to B[j]
        else if (A[i] == B[j])
        {
              
            // If previous element
            // are not from B[]
            if (prev != 1)
            {
                  
                // Update prev
                prev = 1;
  
                // Update j
                j++;
            }
  
            // If previous element is
            // not from A[]
            else
            {
                  
                // Update prev
                prev = 0;
  
                // Update i
                i++;
            }
        }
  
        // If it is not possible to merge two
        // sorted arrays with given conditions
        else
        {
              
            // Update flag
            flag = false;
            break;
        }
    }
    return flag;
}
 
    // Driver Code
     
    let A = [ 3, 5, 8 ];
    let B = [ 2, 4, 6 ];
    let N = A.length;
  
    if (checkIfPossibleMerge(A, B, N))
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    }
 
// This code is contributed by splevel62.
</script>

Output

Yes

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

Approach 2: Dynamic Programming:

The given problem can be solved using a dynamic programming approach. We can create a 2D boolean table dp[][] of size (N+1)x(N+1) where dp[i][j] represents if it is possible to merge the first i elements of array A and the first j elements of array B with given conditions.

The base case is dp[0][0] = true, as we can merge 0 elements from both arrays.
For each (i,j) such that i>0 and j>0, we can calculate dp[i][j] as follows:
If A[i-1] == B[j-1], then we can merge both the elements into the resulting array, and the previous element can be from either array. So, dp[i][j] = dp[i-1][j-1].
If A[i-1] < B[j-1], then we can only merge the element A[i-1] into the resulting array if the previous element is from array B. So, dp[i][j] = dp[i][j-1].
If A[i-1] > B[j-1], then we can only merge the element B[j-1] into the resulting array if the previous element is from array A. So, dp[i][j] = dp[i-1][j].
Finally, the answer is dp[N][N]. If dp[N][N] is true, it means it is possible to merge the entire arrays A and B with given conditions.

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible to merge
// the two given arrays with given conditions
bool checkIfPossibleMerge(int A[], int B[], int N)
{
    // Create a 2D boolean table
    bool dp[N+1][N+1];
 
    // Base case
    dp[0][0] = true;
 
    // Initialize the first row and column
    for (int i = 1; i <= N; i++) {
        dp[i][0] = (A[i-1] > A[i-2]) && dp[i-1][0];
        dp[0][i] = (B[i-1] > B[i-2]) && dp[0][i-1];
    }
 
    // Fill the remaining table
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= N; j++) {
            if (A[i-1] == B[j-1]) {
                dp[i][j] = dp[i-1][j-1];
            }
            else if (A[i-1] < B[j-1]) {
                dp[i][j] = dp[i][j-1] && (A[i-1] > B[j-2]);
            }
            else {
                dp[i][j] = dp[i-1][j] && (B[j-1] > A[i-2]);
            }
        }
    }
 
    return dp[N][N];
}
 
// Driver Code
int main()
{
    int A[3] = { 3, 5, 8 };
    int B[3] = { 2, 4, 6 };
    int N = sizeof(A) / sizeof(A[0]);
 
    if (checkIfPossibleMerge(A, B, N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

public class MergeArrays {
 
    // Function to check if it is possible to merge
    // the two given arrays with given conditions
    static boolean checkIfPossibleMerge(int[] A, int[] B, int N) {
        // Create a 2D boolean table
        boolean[][] dp = new boolean[N + 1][N + 1];
 
        // Base case
        dp[0][0] = true;
 
        // Initialize the first row and column
        for (int i = 1; i <= N; i++) {
            dp[i][0] = (A[i - 1] > A[i - 2]) && dp[i - 1][0];
            dp[0][i] = (B[i - 1] > B[i - 2]) && dp[0][i - 1];
        }
 
        // Fill the remaining table
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= N; j++) {
                if (A[i - 1] == B[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (A[i - 1] < B[j - 1]) {
                    dp[i][j] = dp[i][j - 1] && (A[i - 1] > B[j - 2]);
                } else {
                    dp[i][j] = dp[i - 1][j] && (B[j - 1] > A[i - 2]);
                }
            }
        }
 
        return dp[N][N];
    }
 
    // Driver Code
    public static void main(String[] args) {
        int[] A = {3, 5, 8};
        int[] B = {2, 4, 6};
        int N = A.length;
 
        if (checkIfPossibleMerge(A, B, N)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python3

def checkIfPossibleMerge(A, B, N):
    # Create a 2D boolean table
    dp = [[False]*(N+1) for _ in range(N+1)]
 
    # Base case
    dp[0][0] = True
 
    # Initialize the first row and column
    for i in range(1, N+1):
        dp[i][0] = (A[i-1] > A[i-2]) and dp[i-1][0]
        dp[0][i] = (B[i-1] > B[i-2]) and dp[0][i-1]
 
    # Fill the remaining table
    for i in range(1, N+1):
        for j in range(1, N+1):
            if A[i-1] == B[j-1]:
                dp[i][j] = dp[i-1][j-1]
            elif A[i-1] < B[j-1]:
                dp[i][j] = dp[i][j-1] and (A[i-1] > B[j-2])
            else:
                dp[i][j] = dp[i-1][j] and (B[j-1] > A[i-2])
 
    return dp[N][N]
 
# Driver Code
if __name__ == '__main__':
    A = [3, 5, 8]
    B = [2, 4, 6]
    N = len(A)
 
    if checkIfPossibleMerge(A, B, N):
        print("No")
    else:
        print("Yes")

C#

using System;
 
class Program
{
    // Function to check if it is possible to merge
    // the two given arrays with given conditions
    static bool CheckIfPossibleMerge(int[] A, int[] B, int N)
    {
        // Create a 2D boolean table
        bool[,] dp = new bool[N + 1, N + 1];
 
        // Base case
        dp[0, 0] = true;
 
        // Initialize the first row and column
        for (int i = 1; i <= N; i++)
        {
            dp[i, 0] = (i > 1) && (A[i - 1] > A[i - 2]) && dp[i - 1, 0];
            dp[0, i] = (i > 1) && (B[i - 1] > B[i - 2]) && dp[0, i - 1];
        }
 
        // Fill the remaining table
        for (int i = 1; i <= N; i++)
        {
            for (int j = 1; j <= N; j++)
            {
                if (A[i - 1] == B[j - 1])
                {
                    dp[i, j] = dp[i - 1, j - 1];
                }
                else if (A[i - 1] < B[j - 1])
                {
                    dp[i, j] = dp[i, j - 1] && (A[i - 1] > B[j - 2]);
                }
                else
                {
                    dp[i, j] = dp[i - 1, j] && (B[j - 1] > A[i - 2]);
                }
            }
        }
 
        return dp[N, N];
    }
 
    // Driver Code
    static void Main()
    {
        int[] A = { 3, 5, 8 };
        int[] B = { 2, 4, 6 };
        int N = A.Length;
 
        if (CheckIfPossibleMerge(A, B, N))
        {
            Console.WriteLine("No");
        }
        else
        {
            Console.WriteLine("Yes");
        }
    }
}

Javascript

function checkIfPossibleMerge(A, B, N) {
    // Create a 2D boolean table
    let dp = new Array(N + 1).fill(false).map(() => new Array(N + 1).fill(false));
 
    // Base case
    dp[0][0] = true;
 
    // Initialize the first row and column
    for (let i = 1; i <= N; i++) {
        dp[i][0] = (A[i - 1] > A[i - 2]) && dp[i - 1][0];
        dp[0][i] = (B[i - 1] > B[i - 2]) && dp[0][i - 1];
    }
 
    // Fill the remaining table
    for (let i = 1; i <= N; i++) {
        for (let j = 1; j <= N; j++) {
            if (A[i - 1] === B[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            } else if (A[i - 1] < B[j - 1]) {
                dp[i][j] = dp[i][j - 1] && (A[i - 1] > B[j - 2]);
            } else {
                dp[i][j] = dp[i - 1][j] && (B[j - 1] > A[i - 2]);
            }
        }
    }
 
    return dp[N][N];
}
 
// Driver Code
let A = [3, 5, 8];
let B = [2, 4, 6];
let N = A.length;
 
if (checkIfPossibleMerge(A, B, N)) {
    console.log("No");
} else {
    console.log("Yes");
}

Output

Yes

Time Complexity: O(N^2)
Auxiliary Space: O(N)

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Check if two sorted arrays can be merged to form a sorted array with no adjacent pair from the same array

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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Android 16 QPR2 Beta 3 lands with a flurry of bug fixes

Google is working on dedicated ‘Bills’ and ‘Travel’ folders for Gmail

Mint Mobile’s big bet on 5G home internet might change everything

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Android 16 QPR2 Beta 3 lands with a flurry of bug fixes

Google is working on dedicated ‘Bills’ and ‘Travel’ folders for Gmail

Mint Mobile’s big bet on 5G home internet might change everything

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