My Approach -

Answer is no if there exist i < j in range `minIndx of j`

to `maxIdx of j`

.

For This I have used segment tree.

Time Complexity according to me - O((n+q)*logn) Which Should pass TL.

what is your approach?? how you have used segment tree?

i think solution possible without segment tree .

Can you elaborate your condition for no solution. According to me it was that if there exist i<j<k such that a[i] = a[k] and a[j] is smaller then answer is no. Can you briefly explain your approach.

Yes, I’m using this condition. Along with checking if no q exist or not in given array. Or if it can be inserted or not.

I’m using segment tree. But I have not inserted 0 in tree. Instead using INF. And for given range l,r. I’m returning smallest no in this range. And using condition @vishesh345 's answer.

Thank you friends. I have got AC on 33rd submission xD. I just removed one linear loop for inserting q by using some more variables. And removed handling of special case of 1.