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Suppose we have a binary tree. We will collect and remove all leaves and repeat until the tree is empty.So, if the input is likethen the output will be [[4, 5, 3], [2], [1]]To solve this, we will follow these steps −Define one map szDefine one 2D array retDefine a function dfs(), this will take node, if node is null, then −sz[val of node] := 1 + maximum of dfs(left of node) and dfs(right of node)if size of ret < sz[val of node], then −Define an array tempinsert temp at the end of retinsert val of node at the end ... Read More
Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Different from the previous question where weight is increasing from root to leaf, now the weight is defined from bottom up. i.e., the leaf level integers have weight 1, and the root level integers have the largest weight.So, if the input is like [[1, 1], 2, [1, 1]], then the output will be 8, as four 1's at depth 1, one 2 ... Read More
Suppose we want to design a hit counter which counts the number of hits received in the past 5 minutes. There will be a function and that accepts a timestamp parameter in the second unit and we may assume that calls are being made to the system in chronological order (so, the timestamp is monotonically increasing). We also assume that the earliest timestamp starts at 1.It is possible that several hits arrive roughly at the same time.So we will call the hit() function to hit and getHits() function to get the count of hits.To solve this, we will follow these ... Read More
Suppose we have a 2D grid, here each cell is either a wall 'W', an enemy 'E' or that is empty '0', We have to find the maximum enemies we can kill using one bomb. The bomb kills all the enemies in the same row and column from the planted point until it hits the wall. And we can put bombs only on blank spaces.So, if the input is likethen the output will be 3, as placing bomb at the green place, it will kill three enemies.To solve this, we will follow these steps −ret := 0n := row count ... Read More
Suppose we have a sorted array of integer nums and integer values a, b and c. We have to apply a quadratic function of the form f(x) = ax^2 + bx + c to each element x in the array. And the final array must be in sorted order.So, if the input is like nums = [-4, -2, 2, 4], a = 1, b = 3, c = 5, then the output will be [3, 9, 15, 33]To solve this, we will follow these steps −Define function f(), that takes x, a, b, c −return ax^2 + bx + cFrom ... Read More
Suppose we have n points on a 2D plane, we have to check whether there is any line parallel to y-axis that reflect the given points symmetrically, in other words, check whether there exists a line that after reflecting all points over the given line the set of the original points is the same that the reflected ones.So, if the input is like points = [[1, 1], [-1, 1]]then the output will be trueTo solve this, we will follow these steps −Define one set okn := size of pointsminVal := infmaxVal := -inffor initialize i := 0, when i < ... Read More
Often, we get missing data for analysis and we need to replace those missing values with something. Sometimes, we might want to replace them with the corresponding values in other column especially in situations when both the columns depict similar characteristics. This type of replacement can be easily done with the help of mutate function of dplyr package as shown in the below examples.Example1Consider the below data frame:Live Demo> set.seed(214) > x1 x2 df1 df1Output x1 x2 1 4 75 2 8 24 3 5 38 4 4 38 5 7 NA 6 6 24 7 10 75 8 4 75 ... Read More
Matrix can be only generated if we pass an even number of elements for it. If we want to create a matrix using vector generated with rep function then the length of this vector must be divisible by 2. For example, if we have a vector x that is created with rep function and it’s length is 20 then the matrix say M of size 10x2 using that vector can be constructed by using matrix(x, ncol=2).Example 1Live Demo> x M1 M1Output[, 1] [, 2] [1, ] 10 10 [2, ] 4 4 [3, ] 7 7 [4, ] 3 3 ... Read More
To change the column name of a data frame in R, we can use setNames function. For example, if we have a data frame called df that contains column x and we want to change it to value “Ratings” which is stored in a vector called x then we can use the code df x y yOutput x 1 3 2 8 3 3 4 9 5 5 6 5 7 10 8 2 9 6 10 6 11 3 12 5 13 9 14 1 15 1 16 6 17 2 18 6 19 10 20 6Changing x in y to Ratings:Example> y yOutputRatings 1 3 2 8 3 3 4 9 5 5 6 5 7 10 8 2 9 6 10 6 11 3 12 5 13 9 14 1 15 1 16 6 17 2 18 6 19 10 20 6Let’s have a look at another example:ExampleLive Demo> S df_Salary df_SalaryOutput S 1 31827 2 24697 3 45790 4 45345 5 22294 6 30749 7 37721 8 33535 9 45941 10 24028 11 48927 12 33818 13 49152 14 43334 15 20294 16 29664 17 23358 18 20475 19 39355 20 40386Changing S in df_Salary to Salary:Example> df_Salary df_SalaryOutputSalary 1 31827 2 24697 3 45790 4 45345 5 22294 6 30749 7 37721 8 33535 9 45941 10 24028 11 48927 12 33818 13 49152 14 43334 15 20294 16 29664 17 23358 18 20475 19 39355 20 40386
There is no limitation to installation of packages in R but base R also has some packages associated to it. Hence, there is no need to install or load them in R console every time. We can directly use any of the base R package functions to perform the analysis. If we want to get the list of these package then we can use the code as shown below:Example> installed.packages(priority="base")OutputPackage LibPath Version Priority base "base" "C:/Program Files/R/R-4.0.2/library" "4.0.2" "base" compiler "compiler" "C:/Program Files/R/R-4.0.2/library" "4.0.2" "base" datasets "datasets" "C:/Program Files/R/R-4.0.2/library" "4.0.2" "base" graphics "graphics" "C:/Program Files/R/R-4.0.2/library" "4.0.2" "base" grDevices "grDevices" "C:/Program ... Read More