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Articles by Arnab Chakraborty
Page 119 of 377
Shortest Completing Word in Python
Suppose we have a dictionary words, and we have to find the minimum length word from a given dictionary words, it has all the letters from the string licensePlate. Now such a word is said to complete the given string licensePlate. Here, we will ignore case for letters. And it is guaranteed an answer exists. If there are more than one answers, then return answer that occurs first in the array.The license plate there may be same letter occurring multiple times. So when a licensePlate of "PP", the word "pile" does not complete the licensePlate, but the word "topper" does.So, ...
Read MoreMinimum Unique Word Abbreviation in C++
Suppose we have a string such as "word" and that contains the following abbreviations: ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]. We also have a target string and a set of strings in a dictionary, we have to find an abbreviation of this target string with the smallest possible length such that it does not conflict with abbreviations of the strings in the dictionary. Here each number or letter in the abbreviation is considered as length = 1. So as an example, the abbreviation "a32bc" is of length = 4.So, if ...
Read MoreFind minimum time to finish all jobs with given constraints in C++
ConceptWith respect of a given array of jobs with different time requirements, there exists k identical assignees available and we are also provided how much time an assignee consumesto do one unit of the job. Our task is to determine the minimum time to complete all jobs with following constraints.The first constraint is that an assignee can be assigned only contiguous jobs.Here, for example, an assignee can be assigned jobs at position 1 and 2, but not at position 3, in an array.The second constraint is that two assignees cannot share (or co-assigned) a job, that means, a job cannot ...
Read MorePrime Number of Set Bits in Binary Representation in Python
Suppose we have two integers L and R, we have to find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary form.So, if the input is like L = 6 and R = 10, then the output will be 4, as there are 4 numbers 6(110), 7(111), 9(1001), 10(1010), all have prime number of set bits.To solve this, we will follow these steps −count := 0for j in range L to R, doif set bit count of j is in [2, 3, 5, 7, 11, 13, 17, 19], thencount ...
Read MoreFind multiplication of sums of data of leaves at same levelss in C++
ConceptWith respect of a given Binary Tree, return following value for it.With respect of every level, calculate sum of all leaves if there are leaves at this level. Else ignore it.Calculate multiplication of all sums and return it.InputRoot of following tree 3 / \ 8 6 \ 10Output80First level doesn’t have leaves. Second levelhas one leaf 8 and third level also has one leaf 10. So result is 8*10 = 80InputRoot of following tree 3 ...
Read MoreUnique Morse Code Words in Python
Suppose we have a list of words, here each word can be written as a concatenation of the Morse code of each letter. For example, the word "cba" can be written as "-.-..--...", this is the concatenation "-.-." | "-..." | ".-"). This kind of concatenation is called the transformation of a word.We know that International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on.Here is the list of all 26 letters of the ...
Read MoreWord Squares in C++
Suppose we have a set of words (all are unique), we have to find all word squares and we can build from them. Here a sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < maximum of numRows and numColumns. So as an example, the word sequence ["ball", "area", "lead", "lady"] will construct a word square because each word reads the same both horizontally and vertically.ballarealeadladySo, if the input is like ["area", "lead", "wall", "lady", "ball"], then the output will be [[ "wall", "area", "lead", "lady"], ...
Read MoreFind n-th lexicographically permutation of a strings in C++
ConceptWith respect of a given string of length m containing lowercase alphabets only, our task to determine the n-th permutation of string lexicographically.Inputstr[] = "pqr", n = 3OutputResult = "qpr"ExplanationAll possible permutation in sorted order − pqr, prq, qpr, qrp, rpq, rqpInputstr[] = "xyx", n = 2OutputResult = "xyx"ExplanationAll possible permutation in sorted order − xxy, xyx, yxxMethodHere we use some Mathematical concept for solving this problem.The concept is based on following facts.Here, the total number of permutation of a string generated by N characters (all distinct) is N!Now, the total number of permutation of a string generated by N ...
Read MoreEncode N-ary Tree to Binary Tree in C++
Suppose we have a N-ary tree. We have to encode that tree into one binary. We also have to make deserializer to deserialize the binary tree to N-ary tree.So, if the input is likethen the output will beTo solve this, we will follow these steps −Define a function encode(), this will take root, if root is valid, then −return nullnode = new tree node with value of rootif size of children of root is not 0, then −left of node := encode(children[0] of root)curr = left of nodefor initialize i := 1, when i < size of children of root, ...
Read MoreFind nth term of a given recurrence relation in C++
ConceptAssume bn be a sequence of numbers, which is denoted by the recurrence relation b1=1 and bn+1/bn=2n. Our task is to determine the value of log2(bn) for a given n.Input6Output15Explanationlog2(bn) = (n * (n - 1)) / 2 = (6*(6-1))/2 = 15Input200Output19900Methodbn+1/bn = 2nbn/bn-1 = 2n-1...b2/b1 = 21, We multiply all of above in order to attain(bn+1/bn).(bn/n-1)……(b2/b1) = 2n + (n-1)+……….+1So, bn+1/b1 = 2n(n+1)/2Because we know, 1 + 2 + 3 + ………. + (n-1) + n = n(n+1)/2So, bn+1 = 2n(n+1)/2 . b1; Assume the initial value b1 = 1So, bn+1 = 2sup>n(n+1)/2Now substituting (n+1) for n, we get, ...
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