Can anyone tell the algorithm for finding the maximum path cost in a NxM matrix starting from top left corner and ending with bottom right corner with left ,rightdown movement is allowed in a matrix and contains negative cost. A cell can be visited any number of times and after visiting a cell its cost is replaced with 0.
Explanation is given in the image Explanation of Output. Since you have also negative costs then use bellman-ford. What you do is that you change sign of all the costs convert negative signs to positive and positive to negative then find the shortest path and this path will be the longest because you have changed the signs.
If the sign is never becoms negative then use dijkstra shrtest-path but before that make all values negative and this will return you the longest path with it's cost. You matrix is a direct graph.
In your image you are trying to find a path max or min from index 0,0 to n-1,n Here linkedlist represents only nodes which are connected not there cost. You have to add extra variable there which keep cost for each two nodes and it will look like this. Now instead of first linkedlist put this linkedlist in your array and you have a graph now to run shortest or longest path algorithm.
You can always find the longest path by converting the minus sign to plus and plus to minus and then run shortest path algorithm. Path founded will be longest. I answered this question and as you said in comments to look at point two. If that's a task then main idea of this assignment is ensure the Monotonocity. If there is path from A to B such that f S B then set cost of the f S A where S means source. Since moving up is not allowed, paths always look like a set of horizontal intervals that share at least 1 position for the down move.
Answers can be characterized as, say. An alternative encoding is to note only layer widths and offsets to each other; but you would still have to make sure that the last layer includes the exit cell. However, it may be necessary to expand intervals to ensure that overlap occurs; and there may be multiple highest-scoring intervals in a given layer.
I can only go down or right. In most cases, I don't have a problem. My problem is only when there is nowhere to go, and I need to take a step backwards. The problem is when it gets to matrix. I don't understand why it won't go back in the recursion.
It returns false when it should return true. You need to check you're not dropping off the matrix, and I find it slightly neater to check the 1's and 0's in the matrix after rather than before the move. Putting those together gives this code. Return 1 from that place only if that recursive call returns 1. Becuase in your code call for  its returning 0 and you are returning from that place itself with 0. Learn more.
Recursive path finding in matrix Ask Question. Asked 7 years, 2 months ago. Active 7 years, 2 months ago. Viewed 7k times. You show a 6x6 matrix so it should be possible to proceed beyond matrix. Index 5 should be the limit in each direction. Active Oldest Votes. Your recursion function is wrong in that it only tries first '1' neighbor cell it sees. The recursion erroneously takes a look only to smaller part of the matrix.
You have either to adapt the matrix size but then also its originor coordinates. Not both.Given an NxM N rows and M columns integer matrix with non-negative values What is the maximum sum from going top left 0, 0 to bottom right N-1, M-1? My example below includes both versions, and the first version will print the path too. Just go from source node to destination node and calculate the distance for each path,basically dfs will find out all the paths and all you need to do is to keep the minimum.
Sorry but bfs won't work as each node doesn't have equal weight. One should not disqualify the whole path based on the current adjacent cell values. The remainder of "disqualified path" may overweight the remainder of the "winning path". It's gotta be recursive. The easiest way is just BFS your way through a known size matrix.
You can do this by thinking of the matrix as a series of diagonals. First, its 0,0. One step away you can get 1,0 0,1. Two steps, you can get 2,0 0,2 1,1. Three steps, you get 3,0 2,1 1,2 0,3 etc, notice how it forms a diagonal across the matrix. At each step, you can calculate the max for each square on the diagonal according to the previous step. Eventually, you'll hit the destination and you can just read off the max value. The hidden trick in this question is the range of values.
This can be done using DP in O mn and with O 1 i. Used DP CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.
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O mn time and O 1 space.
Samsung interview question - Anonymous July 26, Flag Reply. I think minimum extra space required is o mn - rajdeeps. Add a Comment Add a Comment. Writing Code? Add Question. CareerCup is the world's biggest and best source for software engineering interview preparation.
The dark mode beta is finally here. Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. There is a garden of strawberry plants represented by a 2D, square array. Each plant each element has a number of strawberries. You start at the top left corner of the array, and you can only move to the right or down.
I need to design a recursive method to calculate the paths through the garden and then output which one yields the most strawberries. I think I have an understanding of really really simple recursion problems, but this problem has gone way over my head. I'm not really sure where to start or where to go as far as creating a recursive method.
Any help related to the code or helping me understand the concept behind this problem is greatly appreciated. Like dasblinkenlight said, the most efficient way to do this is using a memoization or dynamic programming technique. I tend to prefer dynamic programming, but I'll use pure recursion here.
The answer centers around the answer to one fundamental question: "If I'm in the square in row r and column c on my field, how can I evaluate the path from the top left to here such that the number of strawberries is maximized? The key to realize is that there's only two ways to get in the plot in row r and column c: either I can get there from above, using the plot in row r-1 and column c, or I can get there from the side, using the plot in row r and column c After that, you just need to make sure you know your base cases Call max r-1, c-1 to get your answer.
Notice there's a lot of inefficiency here; you'll do much better by using dynamic programming which I'll provide below or memoization which has already been defined. The thing to remember, though, is that both the DP and memoization techniques are simply more efficient ways that come from the recursive principles used here. In both cases, if you want to recreate the actual path, just keep a 2D table of booleans that corresponds with "Did I come from above or to the left"?
If the most strawberry path comes from above, put true, otherwise put false. That can allow you to retrace the patch after the calculation. Notice that this is still recursive in principal: at each step, we're looking back at our previous results. We just happen to be caching our previous results so we don't waste a bunch of work, and we're attacking the subproblems in an intelligent order so that we can always solve them.
For more on dynamic programming, see Wikipedia. You can do it using memoization. Here is Java-like pseudodoce memoRand C are assumed to be instance variables available to the max method. Following is my Python code.
Hope it can help. Learn more. Java - Maximum sum in path through a 2D array Ask Question. Asked 7 years, 9 months ago.
Active 2 years, 2 months ago. Viewed 19k times. Basically I have a problem that goes something similar to this: There is a garden of strawberry plants represented by a 2D, square array.There are only three possible moves from a cell Matrix[r][c]. Starting from any column in row 0, return the largest sum of any of the paths up to row N Input: The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N denoting the order of matrix. Output: Output the largest sum of any of the paths starting from any cell of row 0 to any cell of row N Print the output of each test case in a new line.
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Please note that Custom Input s should be mentioned in the same order format as stated in the problem description. Cancel Send. Sign In Sign Up. Remember me Forgot Password. Why Create an Account? Please enter your email address or userHandle.Largest "Path-Sum" in a triangle of numbers -- Contest Problem #1
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Describe Your Issue. Send Close. Ibrahim Nash.In the mathematical discipline of graph theorya matching or independent edge set in a graph is a set of edges without common vertices. Finding a matching in a bipartite graph can be treated as a network flow problem. A vertex is matched or saturated if it is an endpoint of one of the edges in the matching. Otherwise the vertex is unmatched. A maximal matching is a matching M of a graph G that is not a subset of any other matching.
A matching M of a graph G is maximal if every edge in G has a non-empty intersection with at least one edge in M. The following figure shows examples of maximal matchings red in three graphs. A maximum matching also known as maximum-cardinality matching  is a matching that contains the largest possible number of edges. There may be many maximum matchings. Every maximum matching is maximal, but not every maximal matching is a maximum matching.
The following figure shows examples of maximum matchings in the same three graphs. A perfect matching a. That is, every vertex of the graph is incident to exactly one edge of the matching. Every perfect matching is maximum and hence maximal.
In some literature, the term complete matching is used. In the above figure, only part b shows a perfect matching. A perfect matching is also a minimum-size edge cover. A perfect matching can only occur when the graph has an even number of vertices. A near-perfect matching is one in which exactly one vertex is unmatched. This can only occur when the graph has an odd number of vertices, and such a matching must be maximum. In the above figure, part c shows a near-perfect matching.Find the maximum path sum in matrix.
The maximum path is sum of all elements from first row to last row where you are allowed to move only down or diagonally to left or right.
You can start from any element in first row. To find max path sum first we have to find max value in first row of matrix. Store this value in res.
Now for every element in matrix update element with max value which can be included in max path. If the value is greater then res then update res.
In last return res which consists of max path sum value. This article is contributed by nuclode. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Writing code in comment? Please use ide.
Python 3 prorgam for finding max path in matrix. To calculate max path in matrix.
Maximum path sum in matrix
To find max val in first row. When all paths are possible. When diagonal right is not possible. When diagonal left is not possible. Store max path sum. Driver program to check findMaxPath. This code is contributed by Azkia Anam. WriteLine findMaxPath mat. Load Comments.