A transparent polygon just represents the border lines and a hollow area; thus, we can only understand the area covered but it becomes a little difficult to understand the scales. Hence, this visualisation technique is not as useful as others that fills the area with a different color. But it could be used if the range of the data is not large.Consider the below data frame −Example Live Demoset.seed(123) x

Suppose we have a singly linked list, we have to check find the value of the kth last node (0-indexed). We have to solve this in single pass.So, if the input is like node = [5, 4, 6, 3, 4, 7], k = 2, then the output will be 3, as The second last (index 3) node has the value of 3.To solve this, we will follow these steps −klast := nodelast := nodefor i in range 0 to k, dolast := next of lastwhile next of last is not null, dolast := next of lastklast := next of klastreturn ... Read More

Suppose we have a list of stacks and an integer k. We have to find the maximum possible sum that can be achieved from popping off exactly k elements from any combination of the stacks.So, if the input is like stacks = [[50, -4, -15], [2], [6, 7, 8]], k = 4, then the output will be 39, as we can pop off all 3 elements from the first stack and pop the last element of the last stack to get -15 + -4 + 50 + 8 = 39.To solve this, we will follow these steps −Define a function ... Read More

Suppose we have a list of numbers called nums and another number k, we have to find a list of count of distinct numbers in each window of size k.So, if the input is like nums = [2, 2, 3, 3, 4], k = 2, then the output will be [1, 2, 1, 2], as the windows are [2, 2], [2, 3], [3, 3], and [3, 4].To solve this, we will follow these steps −c := make a dictionary of elements in nums and their frequenciesans := a new listfor i in range k to size of nums, doinsert size ... Read More

Suppose we have a binary matrix where 0 shows empty cell and 1 shows a block that forms a shape, now we have to find the perimeter of the shape. The shape will not hold any hole inside it.So, if the input is like0000000111001100111000000then the output will be 14.To solve this, we will follow these steps −d := 0perimeter := 0height := row count of matrixlength := column count of matrixfor each row in matrix, doc := 0for each val in row, doif val is same as 1, thensurround := 4if c is not same as length - 1, thenif ... Read More

Suppose we have two linked lists l1 and l2, we have to return one linked list by interleaving elements of these two lists starting with l1. If there are any leftover nodes in a linked list, they should be appended to the list.So, if the input is like l1 = [5, 4, 6, 3, 4, 7] l2 = [8, 6, 9], then the output will be [5, 8, 4, 6, 6, 9, 3, 4, 7]To solve this, we will follow these steps −ans := l1while l2 is not null, doif ans is not null, thenif next of ans is not ... Read More

Suppose we have a number n, we have to find the number of n-digit positive integers such that the digits are in strictly increasing order.So, if the input is like n = 3, then the output will be 84, as numbers are 123, 124, 125, ..., 678,789To solve this, we will follow these steps −if n < 9 is non-zero, thenreturn Combination (9Cn)otherwise,return 0Let us see the following implementation to get better understanding −Example Live Demofrom math import factorial as f class Solution:    def solve(self, n):       if n < 9:          return f(9) / f(n) / f(9 - n)       else:          return 0 ob = Solution() print(ob.solve(3))Input3Output84

Suppose we have a list of numbers called nums where each number shows the maximum number of jumps we can make; we have to check whether we can reach to the last index starting at index 0 or not.So, if the input is like nums = [2, 5, 0, 2, 0], then the output will be True, as we can jump from index 0 to 1, then jump from index 1 to end.To solve this, we will follow these steps−n := size of numsarr := an array of size n and fill with falsearr[n - 1] := Truefor i in ... Read More

We might want to create a subset of an R data frame using one or more values of a particular column. For example, suppose we have a data frame df that contain columns C1, C2, C3, C4, and C5 and each of these columns contain values from A to Z. If we want to select rows using values A or B in column C1 then it can be done as df[df$C1=="A"|df$C1=="B",].Consider the below data frame −Exampleset.seed(99) x1

The contingency table considers the numerical values for two categorical variables. Often, we require contingency table for counts, especially in non-parametric analysis but it is also possible that we want to use means for our analysis. Hence, we can use cast function from reshape package which solves the problem of creating contingency table easily.Consider the below data frame −Example Live Demoset.seed(99) x1