Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
Articles on Trending Technologies
Technical articles with clear explanations and examples
Partition Array into Disjoint Intervals in C++
Suppose we have an array A, we have to partition it into two subarrays left and right such that −Every element in left subarray is less than or equal to every element in right subarray.left and right subarrays are non-empty.left subarray has the smallest possible size.We have to find the length of left after such a partitioning. It is guaranteed that such a partitioning exists.So if the input is like [5, 0, 3, 8, 6], then the output will be 3, as left array will be [5, 0, 3] and right subarray will be [8, 6].To solve this, we will ...
Read MoreSort an Array in C++
Suppose we have an array of integers; we have to sort them in ascending order. So if the array is like [5, 2, 3, 1], then the result will be [1, 2, 3, 5]To solve this, we will follow these steps −Make one method called partition, this will take array, low and highset pivot := lowfor i in range low to high – 1if nums[i] < nums[high], then swap(nums[i] and nums[pivot]), increase pivot by 1swap nums[pivot] and nums[high]Define a method called sortArr(), this will take array, low and highif low >= high, then returnpartitionIndex := partition(nums, low, high)sortArr(nums, low, partitionIndex ...
Read MoreHow to implement a Set interface in JShell in Java 9?
JShell is a command-line tool in Java 9 that has been used to execute simple statements like expressions, classes, interfaces, methods, and etc.A Set is an interface in Java that specifies a contract for collections having unique elements. If object1.equals(object2) returns true, then only one of object1 and object2 have a place in Set implementation.In the below code snippet, we have to use the Set.of() method. The collection returned by the Set.of() method is immutable, so it doesn't support the add() method. If we trying to add an element, throws UnsupportedOperationException. If we want to create a HashSet collection instead, which supports the ...
Read MorePossible Bipartition in C++
Suppose we have a set of N people (they are numbered 1, 2, ..., N), we would like to split everyone into two subgroups of any size. Now each person may dislike some other people, and they should not go into the same group. So, if dislikes[i] = [a, b], it indicates that it is not allowed to put the people numbered a and b into the same group. We have to find if it is possible to split everyone into two groups in this way.So if the input is like N = 4 and dislike = [[1, 2], [1, ...
Read MoreSpiral Matrix III in C++
Suppose we have a 2 dimensional grid with R rows and C columns, we start from (r0, c0) facing east. Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column. We will walk in a clockwise spiral shape to visit every position in this grid. When we are at the outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.). We have to find a list of coordinates representing the positions ...
Read MoreLength of Longest Fibonacci Subsequence in C++
Suppose we have a sequence X_1, X_2, ..., X_n is fibonacci-like if −n >= 3X_i + X_{i+1} = X_{i+2} for all i + 2 = 3 otherwise return 0.Let us see the following implementation to get better understanding −Example Live Demo#include using namespace std; class Solution { public: int lenLongestFibSubseq(vector & A) { int ret = 0; unordered_map m; int n = A.size(); vector < vector > dp(n, vector (n)); for(int i = 0; i < n; i++){ ...
Read MoreDifferences between Jdeps and Jdeprscan tools in Java 9?
Jdeps tool can be used to analyze the dependencies of our classes. The running of the "jdeps -jdkinternals jararchive.jar" command prints a list of all classes that use Java internal API. Jdeps tool returns a detailed description of the dependencies while Jdeprscan is another useful tool particularly used in combination with the "-for-removal" flag. This tool shows us all uses of deprecated API by a given jar archive, and only deprecated uses of jdk methods can be shown and can't use this tool to check for deprecation in third party jar.Jdeps tool:"jdeps" is a class dependency analyzer tool can be used for package level ...
Read MoreMaximum Level Sum of a Binary Tree in C++
Suppose we have the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on. We have to return the smallest level X such that the sum of all the values of nodes at level X is maximal. So if the tree is like −Then the output will be 2, The level 1 sum = 1, level 2 sum is 7 + 0 = 7, level 2 sum is 7 + (-8) = -1, so max is of level 2, so output is 2.To solve this, we will follow ...
Read MoreConstruct Binary Tree from Preorder and Postorder Traversal in Python
Suppose we have two traversal sequences Preorder and Postorder, we have to generate the binary tree from these two sequences. So if the sequences are [1, 2, 4, 5, 3, 6, 7], [4, 5, 2, 6, 7, 3, 1], then the output will beTo solve this, we will follow these steps −ans := make a tree node by taking value pre[0], stack := empty stack, and insert ansi := 1 and j := 0while i < length of pre and j < length of postif stack top value = post[j], then increase j by 1, pop from stack, and go ...
Read MoreCan I Win in C++
Suppose in a game called "100 games, " two players take turns adding, to a running total, any integer from 1 through 10. The player who first causes the running total to reach or exceed 100, he/she wins. So what if we change the game so that players cannot re-use integers?For example, if two players take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.So suppose given an integer maxChoosableInteger and another integer desired total, determine if the first player to move can force a win, assuming both players play ...
Read More