Server Side Programming Articles - Page 1532 of 2650

When we have a single long string or a vector of string values and the values within the string are separated by some special characters then splitting the values can help us to properly understand those strings. This could happen in situations when the string data is recorded with mistakes or have some other purpose. We can do the splitting using strsplit function.Example Live Demox1

Suppose we have a string s of lowercase alphabet characters, and another number k, we have to find the minimum number of required changes in the string so that the resulting string has at most k distinct characters. In this case The change is actually changing a single character to any other character.So, if the input is like s = "wxxyyzzxx", k = 3, then the output will be 1, as we can remove the letter "w" to get 3 distinct characters (x, y, and z).To solve this, we will follow these steps −count := a map of each character ... Read More

To plot all the values of an R data frame, we can use matplot function. This function plots all the values based on the columns of an R data frame and represent them by the column number. For example, if we have five columns in an R data frame then matplot will represent the first column by 1, second column by 2, third column by 3 and so on.Consider the below data frame −Example Live Demoset.seed(555) v1

The proportion of row values can be calculated if we divide each row value with the sum of all values in a particular row. Therefore, the total sum of proportions will be equal to 1. This can be done by dividing the data frame with the row sums and for this purpose we can use the below syntax −Syntaxdata_frame_name/rowSums(data_frame_name)Consider the below data frame −Example Live Demoset.seed(111) x1

Suppose we have a list of sorted unique numbers called nums and an integer k, we have to find the kth missing number from the first element of the given list.So, if the input is like nums = [5, 6, 8, 10, 11], k = 1, then the output will be 9, as 9 is the second (index 1) missing number.To solve this, we will follow these steps −for i in range 1 to size of nums, dodiff := nums[i] - nums[i - 1] - 1if k >= diff, thenk := k - diffotherwise, return nums[i - 1] + k ... Read More

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