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Suppose we have a list of numbers. We have to find length of longest bitonic subsequence. As we knot a sequence is said to be bitonic if it's strictly increasing and then strictly decreasing. A strictly increasing sequence is bitonic. Or a strictly decreasing sequence is bitonic also.So, if the input is like nums = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15], size of sequence 16., then the output will be 7.To solve this, we will follow these steps −increasingSubSeq := new array of given array size, and fill with 1for ... Read More
There is no function in R to find the cube root of negative values, hence we need to create that. The code to create the function is as shown below −CubeRoot
Often, we need to compare continuous variables using boxplots and thus side-by-side boxplots are required. Creating side-by-side boxplot in base R can be done with the help of creating space for graphs with the help of par(mfrow=). In this function, we can define the number of graphs and the sequence of these graphs, thus creation of side-by-side boxplot will become easy.Consider the below vectors −set.seed(100) x
Suppose we have a a list of color strings, these may contain "red", "green" and "blue", we have to partition the list so that the red come before green, and green come before blue.So, if the input is like colors = ["blue", "green", "blue", "red", "red"], then the output will be ['red', 'red', 'green', 'blue', 'blue']To solve this, we will follow these steps −green := 0, blue := 0, red := 0for each string in strs, doif string is same as "red", thenstrs[blue] := "blue"blue := blue + 1strs[green] := "green"green := green + 1strs[red] := "red"red := red + ... Read More
Suppose we have a list of numbers called nums, and another number k, we have to check whether the list can be split into lists where each list contains k values and the values are consecutively increasing.So, if the input is like nums = [4, 3, 2, 4, 5, 6], k = 3, then the output will be True, as we can split the list into [2, 3, 4] and [4, 5, 6]To solve this, we will follow these steps −Define one mapfor each key it in mincrease m[it] by 1ok := truewhile (size of m is not 0 and ... Read More
Suppose we have a sorted linked list node of size n, we have to create a binary search tree by Taking the value of the k = floor of (n / 2) the smallest setting it as the root. Then recursively constructing the left subtree using the linked list left of the kth node. And recursively constructing the right subtree using the linked list right of the kth node.So, if the input is like [2, 4, 5, 7, 10, 15], then the output will beTo solve this, we will follow these steps−Define a method solve(), this will take nodeif node ... Read More
Suppose we have a binary tree, the level of its root is 1, the level of its children is 2, and so on.We have to find the smallest level X such that the sum of all the values of nodes at level X is minimum. So if the tree is like −Output will be 2 as the sum is 4 – 10 = -6, which is minimum.To solve this, we will follow these steps −level := 1, sum := value of r, ansLevel := level, ansSum := sumdefine a queue q, insert given node r into qwhile q is not ... Read More
Suppose we have a binary tree. We have to traverse this tree using level order traversal fashion. So if the tree is likeThe traversal sequence will be like: [1, 2, 3, 5, 4]To solve this, we will follow these steps −define queue que to store nodesinsert root into the que.while que is not empty, doitem := item present at front position of queueprint the value of itemif left of the item is not null, then insert left of item into queif right of the item is not null, then insert right of item into quedelete front element from queLet us ... Read More
Suppose we have a binary search tree, we have to convert it to a singly linked list using levelorder traversal.So, if the input is likethen the output will be [5, 4, 10, 2, 7, 15, ]To solve this, we will follow these steps −head := a new linked list nodecurrNode := headq := a list with value rootwhile q is not empty, docurr := delete first element from qif curr is not null, thennext of currNode := a new linked list node with value of currcurrNode := next of currNodeinsert left of curr at the end of qinsert right curr ... Read More
Suppose we have binary tree, we have to show the values of each level by alternating from going left-to-right and right-to-left.So, if the input is likethen the output will be [5, -10, 4, -2, -7, 15]To solve this, we will follow these steps −if root is null, thenreturn a new lists1 := a list initially insert roots2 := a new listres := a new listwhile s1 is not empty or s2 is not empty, dowhile s1 is not empty, donode := delete last element from s1if left of node is not null, theninsert left of node at the end of ... Read More