u/Dry_Resolution3449

The observation i had was just find the smallest partition possible for first 2 by counting number of 1 in 1st half and when it exceeds no of 2s and 3s change boolean variable to true and close the 2nd partition with the same concept of counting nos of 1 and 2. Basically make the 2nd partition as soon as it is possible so that there is atleast 1 element left for 3rd partition.

Code:

void solve(){


    int n;
    cin>>n;
    vector<int> v(n);
    for(auto& x: v) cin>>x;
    int count1=0;
    int count23=0;
    int count12=0;
    int count3=0;
    bool found=false;
    bool found2=false;
    int number;
    for(int i=0;i<n;i++){
        if(!found){
            if(v[i]==2||v[i]==3){
                count23++;
            }else{
                count1++;
                if(count1>count23){
                     found=true;
                     if(v[i+1]==3) i++;
                }else if(count1==count23 ) found=true;
            }
        }else if(!found2){
           
            if(v[i]==3) count3++;
            else{
                count12++;
                if(count12>=count3){
                       found2=true;
                       number=i;
                       break;
                } 
            }
        }
        
        
    }
  
    if(found && found2&& number!=n-1) cout<<"YES\n";
    else cout<<"NO\n";
    
}