Sir please compile all coding problems in one playlist

Nice explanation. Today only i solved this question

السلام عليكم ورحمة الله وبركاته.. شكرا على المجهود الرائع الذي تبذله في هذه القناة.. اردت أن اسأل: هل من الممكن أن نستخدم البحث الثنائي للبحث عن عدد الذي عند قسمة عدد آخر معلوم عليه يكون باقي القسمه يساوي صفر؟ و ماهو معيار الانتقال بين الاعداد التي نختارها لذلك؟ ارجو الرد رجاءا.. شكرا جزيلا.

We can have even more concise code here. class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root == null || p == root || q == root) return root; TreeNode node1 = lowestCommonAncestor(root.left, p, q); TreeNode node2 = lowestCommonAncestor(root.right, p, q); if(node1 != null && node2 != null) return root; if(node2 != null) return node2; return node1; } }

2244. Minimum Rounds to Complete All Tasks, in this we are going to collect all the tasks of the same difficulty and count their frequency, map is the perfect datastructure for that. After this we know that any tasks whose count freq is greater than 3 it can be either done in values/3 + 1 days or values/3 days, as something which has remainder 2 it can be done in 3k + 2 days, or if it has remainder as 2 with values/3, then if values/3 == k, we can do the same thing as 3*(k-1) + 2 + 2, which again is k+1.