We need to print the longest sequence of nos.
Input = [2,3,4,5,8,0]
Output= 3,[2,3,4,5]

Input=[2,3,4,5,8,0,10,9,8,7,6,5]
Output= 5,[10,9,8,9,6,5]

Below is the code that I have written, can someone please help me with optimising the code

I have used 2 flags, flag1 refers to the forward seq, and flag2 for the reverse seq, kepping in mind the below mentioned test case

In = [2,3,4,3,2,1]
Out = 3, [4,3,2,1]

I got this question in one of the interview, and I took a lot of time to solve this almost 50% of my interview time, any tips on attempting such questions at a faster pace if faced during an interview.

Here is my solution, please help me optimize it.

            l=[2,3,4,3,2,1]
            
            flag1=0
            flag2=0
            index=0
            seq=0
            seq_dict={}
            n=len(l)
            for i in range(n-1):
                
                if (l[i]+1==l[i+1]):
                    if (flag2==1):
                        index=i
                        seq_dict.update({seq:[index,flag1,flag2]})
                        seq=0
                        flag2=0 
                    elif(flag1==0 and i==n-2):
                        index=i+1
                        flag1=1
                        seq+=1
                        seq_dict.update({seq:[index,flag1,flag2]})   
                    flag1=1 
                    seq+=1
                
                elif (l[i]-1==l[i+1]):
                    if (flag1==1):
                        index=i-1
                        seq_dict.update({seq:[index,flag1,flag2]})
                        seq=0
                        flag1=0
                    elif(flag2==1 and i==n-2):
                        index=i+1
                        flag2=1
                        seq+=1
                        seq_dict.update({seq:[index,flag1,flag2]})  
                    flag2=1
                    seq+=1
            
                else:
                    if(seq!=0 and (flag1==1 or flag2==1)):
                        index=i
                        seq_dict.update({seq:[index,flag1,flag2]})  
                    seq=0
                    flag1=0
                    flag2=0
                    index=0   
            
            final=[]
            
            m=max(seq_dict)
            print(m)
            end=seq_dict[m][0]
            while(m>=0):
                final.append(l[end-m])
                m-=1
            print(final)