Programmer Analyst

Declare an array of particular size.
Define a variable “Top” and initialize it as -1.
push(x): insert element in the array by increasing top by one.
pop(): check whether top is not equal to -1 if it is so, return top and decrease its value by one.
size(): return top+1.

Brute force: We will use 2 nested loops, the outer loop will iterate through a text string and the inner loop will iterate through the pattern string. If we find a match, we will print the index and move on to find further patterns, whether they are present in the text or not. We will slide the pattern string over the text string one by one. If we get a match, we will print the index and slide by 1 to check the remaining string.

Worked fine as constraints were small but it can be solved using the Rabin-Karp Algorithm.

Here, we first take the input array and insert all elements of the array into a BST.

After that, we take a variable K.

Then, I have to find the Kth largest and smallest element in the BST.

So, I created two functions: one is kthlargest and the other is kthsmallest.

The first function gives us the Kth largest element of that BST, and the second function gives us the Kth smallest element of that BST.

Both functions have two arguments: one is root and another is K. In the kthlargest element function, I call that function in reverse inorder traversal. That means, first right subtree, then the root, and lastly the left subtree. I decrement K by 1 in each function because the maximum value of that BST is the right-most subtree value. Thus, we decrement K by 1 in each function. If at any instance we find K equal to 0, then we simply return the root (which is our desired value). The question is why we reduce K by 1? Because we traverse like reverse inorder traversal, so if we construct an array by traversal, the 0 indexed value will be the rightmost value and the next value will be the next function which terminated after that.

Now, for the kthsmallest element, it's the same as the kthlargest element except for one change: here we follow inorder traversal because we will get the element in sorted order. Therefore, we call the left subtree first, then the root, and then the right subtree. We decrement K by 1 in each function. If at any instance we find K equal to 0, then we return the root. If any function finds that root is equal to NULL, then return NULL.

Just write simple javascript function which return true or false and within it check things such as length of email and it contains @ and . extension or not.

Tip 1: Total weight is 20kg, so 50% will be 10kg. Each family gets 10kg. 

Tip 2: Each member gets 20% of 10kg, which means 2kg each. Assuming all rice is distributed, 10/2 = 5. 

Tip 3: Each family had five members.