Server Side Programming Articles - Page 1525 of 2646

Suppose we have a string s, we have to find the minimum number of characters needed to be inserted so that the string becomes a palindrome.So, if the input is like s = "mad", then the output will be 2, as we can insert "am" to get "madam".To solve this, we will follow these steps −Define a function dp(). This will take i, jif i >= j, thenreturn 0if s[i] is same as s[j], thenreturn dp(i + 1, j - 1)otherwise, return minimum of dp(i + 1, j) and dp(i, j - 1) + 1From the main method, do the ... Read More

Suppose we have a binary tree, and we also have two numbers a and b, we have to find the value of the lowest node that has a and b as descendants. We have to keep in mind that a node can be a descendant of itself.So, if the input is likea = 6, b = 2, then the output will be 4To solve this, we will follow these steps −Define a method solve() this will take root and a, bif root is null, thenreturn -1if value of root is either a or b, thenreturn value of rootleft := solve(left ... Read More

Suppose we have a lowercase string s and another value k. Now consider an operation where we perform a run-length encoding on a string by putting repeated successive characters as a count and character. So if the string is like "aaabbc" would be encoded as "3a2bc". Here we do not put "1c" for "c" since it only appears once successively. So we can first remove any k consecutive characters in s, then find the minimum length possible of the resulting run-length encoding.So, if the input is like s = "xxxxxyyxxxxxzzxxx", k = 2, then the output will be 6, as ... Read More

Suppose we have a binary tree, we have to find the sum of the longest path from the root to a leaf node. If there are two same long paths, return the path with larger sum.So, if the input is likethen the output will be 20.To solve this, we will follow these steps −Define a function rec() . This will take currif curr is null, thenreturn(0, 0)bigger := maximum of rec(left of curr) , rec(right of curr)return a pair (bigger[0] + 1, bigger[1] + value of curr)From the main method do the following −ret := rec(root)return the 1th index of ... Read More

Suppose we have a string s, we have to find the length of the longest substring that contains at most 2 distinct characters.So, if the input is like s = "xyzzy", then the output will be 4, as "yzzy" is the longest substring with at most 2 unique characters.To solve this, we will follow these steps−start := 0c := a mapans := 0for end in range 0 to size of s, doc[s[end]] := c[s[end]] + 1while size of c > 2, doc[s[start]] := c[s[start]] - 1if c[s[start]] is 0, thendelete c[s[start]]start := start + 1ans := maximum of ans and ... Read More

Suppose we have a list of numbers called nums, we have to find the length of the longest sublist where 2 * minimum of sublist > maximum of sublist.So, if the input is like nums = [10, 2, 6, 6, 4, 4], then the output will be 4, as the sublist [6, 6, 4, 4] is the longest sublist that holds the criteria as 2 * 4 > 6.To solve this, we will follow these steps−ret := 0define two double ended queues minq and maxql := 0, r := 0while r < size of nums, don := nums[r]while minq and ... Read More

Suppose we have a string S. We have to find the length of longest palindromic substring in S. We are assuming that the length of the string S is 1000. So if the string is “BABAC”, then the longest palindromic substring is “BAB” and length is 3.To solve this, we will follow these steps −Define one square matrix of order same as the length of string, and fill it with FalseSet the major diagonal elements as true, so DP[i, i] = True for all i from 0 to order – 1start := 0for l in range 2 to length of ... Read More

There is no function in R to calculate the population variance but we can use the population size and sample variance to find it. We know that the divisor in population variance is the population size and if we multiply the output of var(it calculates sample variance) function with (population size – 1)/population size then the output will be population variance.Example Live Demoset.seed(141) x1

Generally, if we extract a single column from an R data frame then it is extracted as a vector but we might want it in data frame form so that we can apply operations of a data frame on it. Therefore, we can use single square brackets for the extraction with T (TRUE) or (FALSE) values and drop = FALSE so that the output becomes a data frame.Consider the below data frame −Example Live Demoset.seed(999) x1

If we have repeated values in a vector that means we have ties in the vector, therefore, the indexes of values will help us to identify the positions of a particular value in the vector. We can use which function with min function to find the positions of minimum values in a vector, if there exists more than one minimum then the output will show all the relevant positions.Example Live Demox1