Server Side Programming Articles - Page 1531 of 2650

Suppose we have a list of numbers called nums, we will define a function that returns the largest sum of non-adjacent numbers. Here the numbers can be 0 or negative.So, if the input is like [3, 5, 7, 3, 6], then the output will be 16, as we can take 3, 7, and 6 to get 16.To solve this, we will follow these steps−if size of nums

The rank of a matrix is defined as the maximum number of linearly independent vectors in rows or columns. If we have a matrix with dimensions R x C, having R number of rows and C number of columns, and if R is less than C then the rank of the matrix would be R. To find the rank of a matrix in R, we can use rankMatrix function in Matrix package.Loading Matrix package −library(Matrix)Example Live DemoM1

If we have multiple vectors then the sum of the corresponding elements can be found by using rowSums functions and the vectors can be combined by using cbind so that R can easily read the corresponding elements. But if there are NA values in one or more vectors then we also need to add na.rm=TRUE argument.Example Live Demoset.seed(100) x1

Suppose we have an array A. We have to find the contiguous sublist which has the maximum sum, and also return its sum. So if the array A is like A = [-2, 1, -3, 4, -1, 2, 1, -5, 4], then the sum will be 6. And the subarray will be [4, -1, 2, 1].To solve this we will try to use Dynamic programming approach.define an array dp same as the size of A, and fill it with 0dp[0] := A[0]for i := 1 to size of A – 1dp[i] := maximum of dp[i – 1] + A[i] and ... Read More

Suppose we have a list of numbers nums, now consider a circular list of nums where the start and end of nums are neighbors. We have to find the maximum sum of a non-empty sublist in the circular list.So, if the input is like nums = [2, 3, -7, 4, 5], then the output will be 14, as we can take the sub list [4, 5, 2, 3] which sums to 14.To solve this, we will follow these steps −max_sum := negative infinity, cur_max := 0min_sum := positive infinity, cur_min := 0for each num in nums, docur_max := maximum of ... Read More

When we create a sample using inbuilt or imported data set then the numbering for selected rows as in the original data set, therefore, the numbering becomes unordered. To change this unordered numbering to a sequence say starting from one to the total number of rows in the sample, we can use 1:nrow(“sample_object_name”).Consider the below data frame −Example Live Demoset.seed(999) x

Suppose we have a list of numbers called nums and an integer k, we have to select elements from nums to create a list of size k such that the difference between the largest integer in the list and the smallest integer is as low as possible. And we will return this difference.So, if the input is like nums = [3, 11, 6, 2, 9], k = 3, then the output will be 4, as the best list we can make is [2, 3, 6].To solve this, we will follow these steps −sort the list numsls := a new listfor ... Read More

The two plots created by using ggplot2 can be arranged in a vertical manner with the help gridExtra package, we simply needs to use grid.arrange function to do so. For example, if we have two plots created by using ggplot2 and saved in objects p1 and p2 then they can be vertically arranged as grid.arrange(p1,p2)Consider the below data frame −Exampleset.seed(151) x

If we have small number of rows then we might want to create bar plot for rows instead of using columns as categories. This can be done by using barplot function but we need to convert the data frame to matrix and take the transpose of it. For example, if we have a data frame data_frame with 4 rows and 4 columns, then the barplot with rows as categories can be created as barplot(t(as.matrix(data_frame)),beside=TRUE)Consider the below data frame −Example Live Demox1

Suppose we have a list of strings words, we have to group all anagrams together and return the size of the largest grouping.So, if the input is like words = ["xy", "yx", "xyz", "zyx", "yzx", "wwwww"], then the output will be 3, as ["xyz", "zyx", "yzx"] is the largest grouping.To solve this, we will follow these steps −lookup := a new map, initially emptyres := 0for each i in words, dop := sort i in lexicographical wayif p is in lookup, increase count, otherwise 1res := maximum of res and lookup[p]return resLet us see the following implementation to get better ... Read More