Count pairs from 2 BST whose sum is equal to given value "X" | BST | Love Babbar DSA Sheet | Amazon🔥

[@raviashwin1157]

Finally a high 1080p quality video 🤩

[@voldemort576]

Bhai inorder traversal or searching me bhi recursive stack ki space lagagi!

[@pokeindia5361]

Wahi mera bhi doubt H!!?

[@pritishpattnaik4674]

great explanation , love ur content

[@CodeLibrary]

Thanks

[@shristisingh6539]

recursive stack ki space nhi lagegegi??

[@akshatkumar4842]

Sir mera ek doubt h. Agar tree skewed hua tb to height 'm' ho jayega aur fir time complexity O(n*m) ho jayega. To ye sahi solution hoga?

[@justarandomguy6106]

Make 1 or 2 maps , and then it will be in o of n

[@manu-singh]

awesome intro music

[@memefulland7140]

my approach: push bst1 and bst2 in two different vectors then merge those two vector in a vector v and sort the vector v then set i at 0 and j at v.size()-1 then simply count the pairs but why its getting wrong cAN ANYONE TELL ITS TIME AND SPACE COMPLEXITY

[@nishantahlawat5643]

Space complexity is o(n+m). Sorting time complexity is o(nlogn)

[@RamanKumar-mg8qs]

@@nishantahlawat5643 mergre can be done in o(m +n) and pair can be found in o (m +n);

[@warfarecreations379]

It's getting wrong because it can choose a pair in the same bst

[@tusharsharma6518]

because it may find the 2 elements from same bst

[@bhawnarajput5641]

Jab ques me clearly likha hai ki O(n) me krna hai to O(n*h) soln kyun btate ho. kyuki bss gfg pe accept hojata hai weak testcases ki wajah se isliye code ko optimise hi ni karte. 🙂

[@codingachinilgtifirbhikrrh9009]

i think ques m galat de rkha h even gfg k editorial m O(n) ka koi solution nahi h, so this is the optimized solution