This problem asks to rearrange elements in a sequence in specific order. This operation is called sort, which is provided as a standard feature in most languages. For example, Python and C++ both provides a function named sort.

To solve the problem, concatenate $A$ and $B$ to define $C$, sort $C$, inspect the elements of $C$, and check if adjacent elements are both contained in $A$.

To determine if a value is contained in a sequence, it is useful to use the in operator in Python, and set operator in C++, for example.

Sample code (Python)

N, M = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = A + B
C.sort()
for i in range(N + M - 1):
    if C[i] in A and C[i + 1] in A:
        print("Yes")
        exit()
print("No")

N, M = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = A + B
C.sort()
for i in range(N + M - 1):
    if C[i] in A and C[i + 1] in A:
        print("Yes")
        exit()
print("No")

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;
int main() {
    int N, M;
    cin >> N >> M;
    vector<int> A(N), B(M);
    for (auto& x : A) cin >> x;
    for (auto& x : B) cin >> x;
    auto C = A;
    for (auto x : B) C.push_back(x);
    sort(C.begin(), C.end());
    set<int> st(A.begin(), A.end());
    for (int i = 0; i < N + M - 1; ++i) {
        if (st.contains(C[i]) && st.contains(C[i + 1])) {
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
}

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

int main() {
    int N, M;
    cin >> N >> M;
    vector<int> A(N), B(M);
    for (auto& x : A) cin >> x;
    for (auto& x : B) cin >> x;

    auto C = A;
    for (auto x : B) C.push_back(x);
    sort(C.begin(), C.end());

    set<int> st(A.begin(), A.end());

    for (int i = 0; i < N + M - 1; ++i) {
        if (st.contains(C[i]) && st.contains(C[i + 1])) {
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
}