Coding Interview Question: Matrix Product

Matrix Product: Coding Interview Question

Question

Given a matrix, find the path from top left to bottom right with the greatest product by moving only down and right.

eg.

Once you think that you’ve solved the problem, click below to see the solution.

 

As always, remember that practicing coding interview questions is as much about how you practice as the question itself. Make sure that you give the question a solid go before skipping to the solution. Ideally if you have time, write out the solution first by hand and then only type it into your computer to verify your work once you've verified it manually. To learn more about how to practice, check out this blog post.


Solution

How was that problem? You can check out the solution in the video below.

Here is the source code for the solution shown in the video (Github):

Did you get the right answer to this coding interview question? Please share your thoughts in the comments below.

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  • yogi ozark

    I tried this but got stuck at the entries in the list but the remainder part seems working right..

    import java.util.List;

    import java.util.Map;

    import java.util.Map.Entry;

    import java.util.TreeMap;

    import java.util.ArrayList;

    import java.util.Iterator;

    public class Solution

    {

    static void PathHelperUtility(int r, int c, int[][] M, int product, ArrayList list, Map tMap)

    {

    if( c >= M[0].length || r >= M.length )

    return;

    product = product * M[r][c];

    list.add(M[r][c]);

    PathHelperUtility(r, c+1, M, product, list, tMap);

    PathHelperUtility(r+1, c, M, product, list, tMap);

    if ( c == M[0].length – 1 && r == M.length – 1)

    {

    if(!tMap.containsKey(product)) {

    tMap.put(product, list);

    }

    //index = 0;

    //list = new ArrayList();

    }

    return;

    }

    static void dumpTreeMap(Map tMap) {

    Iterator it = tMap.entrySet().iterator();

    while(it.hasNext()) {

    Map.Entry me = (Entry) it.next();

    System.out.print(“Product: ” + me.getKey());

    System.out.println(” Path: ” + me.getValue().toString());

    }

    }

    public static void main(String[] args) {

    // TODO Auto-generated method stub

    int T[][] = { {-1, 2, 3} , {4, 5, -6}, {7, 8, 9} };

    int p = T[0][0], row = 1, col = 0;

    ArrayList l = new ArrayList();

    l.add(p);

    Map treeMap = new TreeMap();

    PathHelperUtility(row, col, T, p, l, treeMap);

    dumpTreeMap(treeMap);

    }

    }

    • I’m a little confused about which part of the solution is causing you trouble

  • Максим Ерёменко

    Hi Sam,
    Thanks for this lesson. I enjoyed it!

    May I suggest a little addition to your code?

    You have implemented a case with negative numbers, but in the end, function returns only max value from maxCache and does not compare it with minCache (What if the best result is placed there?).

    I think the last lines can be like this

    ——-
    int maxCacheValue = maxCache[maxCache.length – 1][maxCache[0].length – 1];
    int minCacheValue = minCache[minCache.length – 1][minCache[0].length – 1];

    return Math.max(maxCacheValue, minCacheValue);

    • That’s a great point, but remember that the maxCacheValue is always going to be greater than the minCacheValue, so this is actually unnecessary. This is because the maxCache is always set with the maximum and the minCache is always set with the minimum of the same set of numbers.