## Question

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

eg.

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[1, 2, 3] [4, 5, 6] [7, 8, 9] 1 -> 4 -> 7 -> 8 -> 9 2016 [-1, 2, 3] [4, 5, -6] [7, 8, 9] -1 -> 4 -> 5 -> -6 -> 9 1080 |

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):

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public static int matrixProduct(int[][] matrix) { if (matrix.length == 0 || matrix[0].length == 0) return 0; // Create cache of min and max product to a given cell int[][] maxCache = new int[matrix.length][matrix[0].length]; int[][] minCache = new int[matrix.length][matrix[0].length]; // Fill caches. We start at the top left and iteratively find the greatest // at smallest path to each subsequent cell by considering the greatest and // smallest path to the cells above and to the left of the current cell for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[0].length; j++) { int maxVal = Integer.MIN_VALUE; int minVal = Integer.MAX_VALUE; // If you're in the top left corner, just copy to cache if (i == 0 && j == 0) { maxVal = matrix[i][j]; minVal = matrix[i][j]; } // If we're not at the top, consider the value above if (i > 0) { int tempMax = Math.max(matrix[i][j] * maxCache[i-1][j], matrix[i][j] * minCache[i-1][j]); maxVal = Math.max(tempMax, maxVal); int tempMin = Math.min(matrix[i][j] * maxCache[i-1][j], matrix[i][j] * minCache[i-1][j]); minVal = Math.min(tempMin, minVal); } // If we're not on the left, consider the value to the left if (j > 0) { int tempMax = Math.max(matrix[i][j] * maxCache[i][j-1], matrix[i][j] * minCache[i][j-1]); maxVal = Math.max(tempMax, maxVal); int tempMin = Math.min(matrix[i][j] * maxCache[i][j-1], matrix[i][j] * minCache[i][j-1]); minVal = Math.min(tempMin, minVal); } maxCache[i][j] = maxVal; minCache[i][j] = minVal; } } // Return the max value at the bottom right return maxCache[maxCache.length - 1][maxCache[0].length - 1]; } |

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