The problem can be solved with DP (Dynamic Programming).

Definition of DP

We define \(dp\) and \(ep\) by

\(dp[i] :=\) considering \((X_1,X_2,\ldots,X_i)\), can we let \(X_i = A_i\)?
\(ep[i] :=\) considering \((X_1,X_2,\ldots,X_i)\), can we let \(X_i = B_i\)?

Initial values of DP

\(dp[1]\) means “considering \((X_1)\), can we let \(X_1=A_1\)?”, which is True.
\(ep[1]\) is True in the same way.

Transitions of DP

We will determine \(dp[i]\) for \(i\geq 2\).

• If \(dp[i-1]=True\) and \(|A_{i-1}- A_i| \leq K\), then \(dp[i]=True\);
• If \(ep[i-1]=True\) and \(|B_{i-1}- A_i| \leq K\) then \(dp[i]=True\);
  
• Otherwise \(dp[i]=False\)

\(ep[i]\) can be found in the same way.

Answer for DP

If \(dp[N]=True\) or \(ep[N]=True\), then the answer is Yes; otherwise, the answer is No.

Therefore, the problem has been solved. The time complexity is \(O(N)\).

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


signed main(){
    ll n,k;cin>>n>>k;
    vector<ll>a(n+1),b(n+1);
    for(ll i=1;i<=n;i++)cin>>a[i];
    for(ll i=1;i<=n;i++)cin>>b[i];
    vector<ll>dp(n+1,false),ep(n+1,false);

    //Initial value
    dp[1]=ep[1]=true;

    //Transitions
    for(ll i=2;i<=n;i++){
        if(dp[i-1]){
            if(abs(a[i-1]-a[i])<=k)dp[i]=true;
            if(abs(a[i-1]-b[i])<=k)ep[i]=true;
        }
        if(ep[i-1]){
            if(abs(b[i-1]-a[i])<=k)dp[i]=true;
            if(abs(b[i-1]-b[i])<=k)ep[i]=true;
        }
    }

    //Answer
    if(dp[n]||ep[n])cout<<"Yes"<<endl;
    else cout<<"No"<<endl;

    return 0;
}