Programming Articles - Page 1840 of 3366

Suppose we have a binary 2D matrix, now we have to find the beginning point and terminating point of all rectangles filled with 0s. We have to keep in mind that rectangles are separated and do not touch each other however they can touch the array boundary. A rectangle with only single element is also possible.So, if the input is like −10111011101111101100110110011011011101000011100011011101then the output will be [[0, 1, 0, 1], [0, 5, 0, 5], [1, 2, 1, 2], [2, 3, 2, 4], [3, 1, 5, 1], [3, 4, 6, 5], [5, 3, 6, 5], [7, 1, 7, 1], [7, 5, ... Read More

Suppose we have one undirected graph and a set of vertices; we have to find all reachable nodes from every vertex present in the given set.So, if the input is likethen the output will be [1, 2, 3] and [4, 5] as these are two connected components.To solve this, we will follow these steps −nodes := number of nodes in the graphDefine an array visited of size: nodes+1. And fill with 0Define one map mcomp_sum := 0for initialize i := 0, when i < n, update (increase i by 1), do −u := arr[i]if visited[u] is false, then −(increase comp_sum ... Read More

Suppose we have a string; we have to find all the palindromic sub-strings from that string. Here aa and aa are considered as two sub-strings, not one.So, if the input is like redivider, then the output will be ['r', 'e', 'd', 'i', 'v', 'ivi', 'divid', 'edivide', 'redivider', 'i', 'd', 'e', 'r']To solve this, we will follow these steps −v := a new listpos := 0.0while pos < size of s, dorad := pos - (pos as integer)while (pos + rad) < size of s and (pos - rad) >= 0 and (s[integer of (pos - rad)] is same as s[integer ... Read More

Suppose we have an array A of numbers, we have to find all indices of this array so that after deleting the ith element from the array, the array will be a good array. We have to keep in mind that −Good array is an array with an element that equals to the sum of all other elements.1-based indexing will be used here.So, if the input is like [10, 4, 6, 2], then the output will be [1, 4] as when we remove A[1], the array will look like [4, 6, 2] and it is good, as 6 = 4+2. ... Read More

Suppose we have a string with lowercase ASCII characters, we have to find all distinct continuous palindromic substrings of it.So, if the input is like "bddaaa", then the output will be [a, aa, aaa, b, d, dd]To solve this, we will follow these steps −m := a new mapn := size of smatrix := make two rows of n number of 0ss := "@" concatenate s concatenate "#"for j in range 0 to 1, dotemp := 0matrix[j, 0] := 0i := 1while i

Suppose we have a number n; we have to check a lower case stringof length n+1 so that the character at any position should be lexicographically bigger than its immediate next character.So, if the input is like 15, then the output will be ponmlkjihgfedcba.To solve this, we will follow these steps −temp_str := blank stringextra := n mod 26if extra >= 1, thenfor i in range 26 -(extra + 1) to 25, dotemp_str := temp_str + str[i]count := n / 26 (integer division)for i in range 1 to count + 1, dofor j in range 0 to 25, dotemp_str := ... Read More

Suppose we have an array with N numbers, we have to check whether the 3 elements such that b[i]< b[j] < b[k] and i < j < k in linear (O(n)) time. If there are multiple such triplets, then print any one of them.So, if the input is like [13, 12, 11, 6, 7, 3, 31], then the output will be [6, 7, 31]To solve this, we will follow these steps −n := size of Amaximum := n-1, minimum := 0smaller := an array of size 1000, and fill with 0smaller[0] := -1for i in range 1 to n, doif ... Read More

Suppose we have a number N, we have to find a positive number M such that gcd(N^M, N&M) is as large as possible and m < n. We will also return the largest gcd thus obtained.So, if the input is like 20, then the output will be 31To solve this, we will follow these steps −if bit_count(n) is same as 0, thenfor i in range 2 to int(square root of (n)) + 1, doif n mod i is same as 0, thenreturn int(n / i)otherwise, val := 0p :=dupn := nwhile n is non-zero, doif (n AND 1) is same ... Read More

ConceptWith respect of a given set of elements, determine which permutation of these elements would result in worst case of Merge Sort?We know, asymptotically, merge sort always consumes O(n log n) time, but the cases that need more comparisons generally consume more time in practice. Now we basically require determining a permutation of input elements that would lead to largest number of comparisons when sorted implementing a typical Merge Sort algorithm.Example Consider the below set of elements as Sorted array 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26Resultant input array that will result ... Read More

ConceptWith respect of a given Balanced Binary Search Tree and a target sum, we write a function that returns true if there is a pair with sum equals to target sum, otherwise return false. In this case, expected time complexity is O(n) and only O(Logn) extra space can beimplemented. Here, any modification to Binary Search Tree is not permitted.We have to note that height of a Balanced BST is always O(Logn).ExampleMethodAccording to the Brute Force Solution, we consider each pair in BST and verify whether the sum equals to X. The time complexity of this solution will be O(n^2).Now a ... Read More