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Data Structure Articles
Page 160 of 164
Box Stacking Problem
In this problem a set of different boxes are given, the length, breadth, and width may differ for different boxes. Our task is to find a stack of these boxes, whose height is as much as possible. We can rotate any box as we wish. But there is a rule to maintain.One can place a box on another box if the area of the top surface for the bottom box is larger than the lower area of the top box.Input and OutputInput: A list of boxes is given. Each box is denoted by (length, bredth, height). { (4, 6, 7), ...
Read MoreFleury's Algorithm
Fleury’s Algorithm is used to display the Euler path or Euler circuit from a given graph. In this algorithm, starting from one edge, it tries to move other adjacent vertices by removing the previous vertices. Using this trick, the graph becomes simpler in each step to find the Euler path or circuit.We have to check some rules to get the path or circuit −The graph must be a Euler Graph.When there are two edges, one is bridge, another one is non-bridge, we have to choose non-bridge at first.Choosing of starting vertex is also tricky, we cannot use any vertex as ...
Read MoreLongest Path in a Directed Acyclic Graph
One weighted directed acyclic graph is given. Another source vertex is also provided. Now we have to find the longest distance from the starting node to all other vertices, in the graph.We need to sort the nodes in topological sorting technique, and the result after the topological sort is stored into a stack. After that repeatedly popped from the stack and try to find the longest distance for each vertex.Input and OutputInput: The cost matrix of the graph. 0 5 3 -∞ -∞ -∞ -∞ 0 2 6 -∞ -∞ -∞ -∞ 0 7 4 2 -∞ -∞ ...
Read MoreHow to find if a graph is Bipartite?
A graph is said to be a bipartite graph, when vertices of that graph can be divided into two independent sets such that every edge in the graph is either start from the first set and ended in the second set, or starts from the second set, connected to the first set, in other words, we can say that no edge can found in the same set.Checking of a bipartite graph is possible by using the vertex coloring. When a vertex is in the same set, it has the same color, for another set, the color will change.Input and OutputInput: ...
Read MoreShortest path with exactly k Edges
One directed graph is provided with the weight between each pair of vertices, and two vertices u and v are also provided. Our task is to find the shortest distance from vertex u to vertex v, with exactly k number of edges. To solve this problem, we will start from vertex u and go to all adjacent vertices and recur for adjacent vertices using the k value as k - 1.Input and OutputInput: The cost matrix of the graph. 0 10 3 2 ∞ 0 ∞ 7 ∞ ∞ 0 6 ∞ ∞ ∞ 0 Output: Weight of the shortest ...
Read MoreCheck if a given graph is tree or not
In this problem, one undirected graph is given, we have to check the graph is tree or not. We can simply find it by checking the criteria of a tree. A tree will not contain a cycle, so if there is any cycle in the graph, it is not a tree.We can check it using another approach, if the graph is connected and it has V-1 edges, it could be a tree. Here V is the number of vertices in the graph.Input and OutputInput: The adjacency matrix. 0 0 0 0 1 0 0 0 0 1 0 0 0 ...
Read MoreConnectivity in a directed graph
To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. After completing the traversal, if there is any node, which is not visited, then the graph is not connected.For the directed graph, we will start traversing from all nodes to check connectivity. Sometimes one edge can have the only outward edge but no inward edge, so that node will be unvisited from any other starting node.In this case, the traversal algorithm is recursive DFS traversal.Input and OutputInput: Adjacency matrix of a graph 0 1 0 0 0 0 0 1 0 ...
Read MoreDetect Cycle in a an Undirected Graph
To detect if there is any cycle in the undirected graph or not, we will use the DFS traversal for the given graph. For every visited vertex v, when we have found any adjacent vertex u, such that u is already visited, and u is not the parent of vertex v. Then one cycle is detected. We will assume that there are no parallel edges for any pair of vertices.Input and Output: Adjacency matrix 0 1 0 0 0 1 0 1 1 0 0 1 0 0 1 0 1 ...
Read MoreBridges in a Graph
An edge in an undirected graph is said to be a bridge, if and only if by removing it, disconnects the graph, or make different components of the graph. In a practical approach, if some bridges are present in a network when the connection of bridges is broken, it can break the whole network.Input and OutputInput: The adjacency matrix of the graph. 0 1 1 1 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 Output: Bridges in given graph: Bridge 3--4 Bridge 0--3AlgorithmbridgeFind(start, visited, disc, low, ...
Read MoreWord Break Problem
In the input of this problem, one sentence is given without spaces, another dictionary is also provided of some valid English words. We have to find the possible ways to break the sentence in individual dictionary words.We will try to search from the left of the string to find a valid word when a valid word is found, we will search for words in the next part of that string.Input and OutputInput: A set of valid words as dictionary, and a string where different words are placed without spaces. Dictionary: {mobile, sam, sung, man, mango, icecream, and, go, i, love, ...
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