For Example: N=5, M=4
A=[4,2,10,5,9]
B=[4,5,6,15]
so the one of longest sorted array is [2,4,5,6,9,15]
this is my Approach but want to know is there any efficient approach available. I can assure it is not from any online coding contest, it is being asked to me in an interview so I want to know the efficient approach if any.
#include<bits/stdc++.h> 
using namespace std; 
int mx=0;
void generateUtil(int A[], int B[], int C[], int i, int j, int m, int n, 
                int len, bool flag) 
{ 
    if (flag) 
    { 
        if (len) 
            mx=max(mx,len+1); 
        for (int k = i; k < m; k++) 
        { 
            if (!len) 
            { 
                C[len] = A[k];
                generateUtil(A, B, C, k+1, j, m, n, len, !flag); 
            } 
            { 
                if (A[k] > C[len]) 
                { 
                    C[len+1] = A[k]; 
                    generateUtil(A, B, C, k+1, j, m, n, len+1, !flag); 
                } 
            } 
        } 
    } 
    { 
        for (int l = j; l < n; l++) 
        { 
            if (B[l] > C[len]) 
            { 
                C[len+1] = B[l]; 
                generateUtil(A, B, C, i, l+1, m, n, len+1, !flag); 
            } 
        } 
    } 
} 
void generate(int A[], int B[], int m, int n) 
{ 
    int C[m+n];
    generateUtil(A, B, C, 0, 0, m, n, 0, true); 
} 
int main() 
{ 
    int n,m,i,j;
    cin>>n>>m;
    int A[n],B[m];
    for(i=0;i<n;i++)
    {
     cin>>A[i];
    }
    for(j=0;j<n;j++)
    {
     cin>>B[j];
    }
    generate(A, B, n, m); 
    cout<<mx<<"\n";
    return 0; 
} 
 
     
    