Check if two sorted arrays can be merged to form a sorted array with no adjacent pair from the same array
Last Updated :
23 Jul, 2025
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 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 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# 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
<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>
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;
}
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");
}
}
}
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")
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");
}
}
}
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");
}
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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