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Dynamic Programming Articles
Page 12 of 14
Maximum profit by buying and selling a share at most twice
In a trading, one buyer buys and sells the shares, at morning and the evening respectively. If at most two transactions are allowed in a day. The second transaction can only start after the first one is completed. If stock prices are given, then find the maximum profit that the buyer can make.Input and OutputInput: A list of stock prices. {2, 30, 15, 10, 8, 25, 80} Output: Here the total profit is 100. As buying at price 2 and selling at price 30. so profit 28. Then buy at price 8 and sell it again at price 80. So ...
Read MoreMaximum Sum Increasing Subsequence
Maximum Sum Increasing subsequence is a subsequence of a given list of integers, whose sum is maximum and in the subsequence, all elements are sorted in increasing order.Let there is an array to store max sum increasing subsequence, such that L[i] is the max sum increasing subsequence, which is ending with array[i].Input and OutputInput: Sequence of integers. {3, 2, 6, 4, 5, 1} Output: Increasing subsequence whose sum is maximum. {3, 4, 5}.AlgorithmmaxSumSubSeq(array, n)Input: The sequence of numbers, number of elements.Output: Maximum sum of the increasing sub sequence.Begin define array of arrays named subSeqLen of size n. add ...
Read MoreMaximum sum rectangle in a 2D matrix
A matrix is given. We need to find a rectangle (sometimes square) matrix, whose sum is maximum.The idea behind this algorithm is to fix the left and right columns and try to find the sum of the element from the left column to right column for each row, and store it temporarily. We will try to find top and bottom row numbers. After getting the temporary array, we can apply the Kadane’s Algorithm to get maximum sum sub-array. With it, the total rectangle will be formed.Input and OutputInput: The matrix of integers. 1 2 -1 -4 -20 -8 -3 4 ...
Read MoreMin Cost Path
A matrix of the different cost is given. Also, the destination cell is provided. We have to find minimum cost path to reach the destination cell from the starting cell (0, 0).Each cell of the matrix represents the cost to traverse through that cell. From a cell, we cannot move anywhere, we can move either to the right or to the bottom or to the lower right diagonal cell, to reach the destination.Input and OutputInput: The cost matrix. And the destination point. In this case the destination point is (2, 2). 1 2 3 4 8 2 1 5 3 ...
Read MoreMinimum 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 := ...
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