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Data Structure Articles
Page 158 of 164
Minimum Cost Polygon Triangulation
When nonintersecting diagonals are forming a triangle in a polygon, it is called the triangulation. Our task is to find a minimum cost of triangulation.The cost of triangulation is the sum of the weights of its component triangles. We can find the weight of each triangle by adding their sides, in other words, the weight is the perimeter of the triangle.Input and OutputInput: The points of a polygon. {(0, 0), (1, 0), (2, 1), (1, 2), (0, 2)} Output: The total cost of the triangulation. Here the cost of the triangulation is 15.3006.AlgorithmminCost(polygon, n)Here cost() will be used to calculate ...
Read MoreMinimum Initial Points to Reach Destination
To start from the top-left corner of a given grid, one has to reach the bottom-right corner. Each cell in the grid contains a number, the number may positive or negative. When the person reaches a cell (i, j) the number of tokens he has, may be increased or decreased along with the values of that cell. We have to find the minimum number of initial tokens are required to complete the journey.There are some rules −We can either move to the right or to the bottom.We cannot move to a cell (i, j) if our total token is less ...
Read MoreGenerate Fibonacci Series
The Fibonacci sequence is like this, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ……In this sequence, the nth term is the sum of (n-1)'th and (n-2)'th terms.To generate we can use the recursive approach, but in dynamic programming, the procedure is simpler. It can store all Fibonacci numbers in a table, by using that table it can easily generate the next terms in this sequence.Input and OutputInput: Take the term number as an input. Say it is 10 Output: Enter number of terms: 10 10th fibinacci Terms: 55AlgorithmgenFiboSeries(n)Input: max number of terms.Output − The nth Fibonacci ...
Read MoreHow to print maximum number of A's using given four keys
Let us consider, we will try to write the letter ‘A’, using the keyboard. Our goal is to use only four keys and try to write maximum ‘A’s on the text field. The keys are ‘A’, ‘C’, ‘V’ and ‘Ctrl’.To write the maximum number of A, we will use Ctrl + A to select All, Ctrl + C to copy and Ctrl + V to paste.Input and OutputInput: Number of keystrokes, say 7 Output: Maximum Number of A's with 7 keystrokes is: 9 Press A three times. Then Ctrl+A, Ctrl+C, Ctrl+V, Ctrl+VAlgorithmkeyNumbers(keyStrokes)Input: number of keystrokes.Output: Maximum number of letters using these ...
Read MoreLargest Independent Set Problem
The Independent Set is the subset of all binary tree nodes when there is no edge between any two nodes in that subset. Now from a set of elements, we will find the longest independent set. i.e. If the elements are used to form a binary tree, then all largest subset, where no elements in that subset are connected to each other.Input and OutputInput: A binary tree. Output: Size of the Largest Independent Set is: 5AlgorithmlongSetSize(root)In this algorithm Binary tree will be formed, each node of that tree will hold data and setSize.Input − Root node of the binary tree.Output − ...
Read MoreLargest Sum Contiguous Subarray
An array of integers is given. We have to find the sum of all elements which are contiguous, whose sum is largest, that will be sent as output.Using dynamic programming we will store the maximum sum up to current term. It will help to find the sum for contiguous elements in the array.Input and OutputInput: An array of integers. {-2, -3, 4, -1, -2, 1, 5, -3} Output: Maximum Sum of the Subarray is: 7AlgorithmmaxSum(array, n)Input − The main array, the size of the array.Output − maximum sum.Begin tempMax := array[0] currentMax = tempMax for i := ...
Read MoreLongest Bitonic Subsequence
A sequence is said to be bitonic if it is first increasing and then decreasing. In this problem, an array of all positive integers is given. We have to find a subsequence which is increasing first and then decreasing.To solve this problem, we will define two subsequences, they are the Longest Increasing Subsequence and the Longest Decreasing Subsequence. The LIS array will hold the length of increasing subsequence ending with array[i]. The LDS array will store the length of decreasing subsequence starting from array[i]. Using these two arrays, we can get the length of longest bitonic subsequence.Input and OutputInput: A ...
Read MoreLongest consecutive path from a given starting character
A matrix of different characters is given. Starting from one character we have to find the longest path by traversing all characters which are greater than the current character. The characters are consecutive to each other.To find the longest path, we will use the Depth First Search algorithm. During DFS, some subproblems may arise multiple times. To avoid the computation of that, again and again, we will use a dynamic programming approach.Input and OutputInput: The matrix as shown above. And the starting point. Here the starting point is e. Output: Enter Starting Point (a-i): e Maximum consecutive path: 5AlgorithmfindLongestLen(i, j, ...
Read MoreLongest Increasing Subsequence
Longest Increasing Subsequence is a subsequence where one item is greater than its previous item. Here we will try to find Longest Increasing Subsequence length, from a set of integers.Input and OutputInput: A set of integers. {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15} Output: The length of longest increasing subsequence. Here it is 6. The subsequence is 0, 2, 6, 9, 13, 15.AlgorithmlongestSubSeq(subarray, n)Input − The sub array and the size of sub array.Output − Longest increasing sub sequence length.Begin define array length of size n initially set 0 ...
Read MoreCollect maximum points in a grid using two traversals
There is a matrix with points in each cell, how to get maximum points from that grid using two traversals.There is some condition to satisfy −The first traversal starts from the top left cell in the grid and should go to the bottom left corner. And in the second traversal starting from top right corner to bottom right cornerFrom one cell we can only move to bottom, bottom left of the current cell and bottom right of the current cells only.If one traversal already gets some points from a cell, In the next traversal no points will be ...
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