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Server Side Programming Articles - Page 1528 of 2646
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Suppose we have two lowercase strings X and Y, we have to find the length of their longest common substring.So, if the input is like X = "helloworld", Y = "worldbook", then the output will be 5, as "world" is the longest common substring and its length is 5.To solve this, we will follow these steps −Define an array longest of size: m+1 x n+1.len := 0for initialize i := 0, when i
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Suppose we have two strings text1 and text2, we have to find the length of their longest common subsequence. As we know a subsequence of a string is a new string generated from the original string with some characters deleted without changing the relative order of the remaining characters. (So for example "abe" is a subsequence of "abcde" but "adc" is not). A common subsequence of two strings is a subsequence that is common to both strings. So If there is no common subsequence, return 0. If the input is like “abcde”, and “ace”, then the result will be 3.To ... Read More
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A block-diagonal matrix means that a matrix added to another matrix at the end the last element. For example, if we have a matrix with nine values and the other matrix also has nine values then the second matrix will be added to the first matrix and the elements below first matrix will be zero and the elements above the second matrix will also be zero.Example Live DemoM1
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Suppose we have a list of numbers. We have to find length of longest bitonic subsequence. As we knot a sequence is said to be bitonic if it's strictly increasing and then strictly decreasing. A strictly increasing sequence is bitonic. Or a strictly decreasing sequence is bitonic also.So, if the input is like nums = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15], size of sequence 16., then the output will be 7.To solve this, we will follow these steps −increasingSubSeq := new array of given array size, and fill with 1for ... Read More
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Often, we need to compare continuous variables using boxplots and thus side-by-side boxplots are required. Creating side-by-side boxplot in base R can be done with the help of creating space for graphs with the help of par(mfrow=). In this function, we can define the number of graphs and the sequence of these graphs, thus creation of side-by-side boxplot will become easy.Consider the below vectors −set.seed(100) x
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In Python, a string is immutable data structure in which sequence of characters are enclosed in double("") or single quotes(''). In some cases, we need to find the length of the longest subsequence of characters that can be rearranged to form a palindrome. This type of subsequence is referred to as a palindromic anagram subsequence. To solve this, we need to count how many characters appear an even number of times and at most one character can appear an odd number of times. An anagram is a word or phrase formed by rearranging the letters of another word or phrase ... Read More
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In Python, a list is a ordered and mutable data structure where elements are enclosed in square braces []. In some scenarios, we may need to count how many elements to the right of each element in the list are smaller than it. Using Binary Search with bisect_left() Function The bisect_left() Function of the bisect method is used locate the insertion point for an element in a sorted list to maintain the list's sorted order. The insertion point returned by bisect_left() is the index where the element can be inserted without violating the sort order by placing it before any ... Read More
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In Python, datastructures such as string, list etc., may contain duplicates of a value. In few scenarios, we may need to find which element appears only once while all others elements appear multiple times. Let's go through different methods to find which element occurs exactly once in Python. Using Bitwise XOR The XOR is abbrivated as Exclusive OR is a bitwise operator which returns 1 if the corresponding bits of two operands are different otherwise, it returns 0 if they are same. In Python, we can use the ^ between two operands to perform Exclusive OR operation. Example Following is ... Read More
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Suppose we have a a list of color strings, these may contain "red", "green" and "blue", we have to partition the list so that the red come before green, and green come before blue.So, if the input is like colors = ["blue", "green", "blue", "red", "red"], then the output will be ['red', 'red', 'green', 'blue', 'blue']To solve this, we will follow these steps −green := 0, blue := 0, red := 0for each string in strs, doif string is same as "red", thenstrs[blue] := "blue"blue := blue + 1strs[green] := "green"green := green + 1strs[red] := "red"red := red + ... Read More