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matrixProduct.java
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matrixProduct.java
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/*
Given a matrix, find the path from top left to bottom right with the greatest product by moving only down and right.
eg.
[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
[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
/*
//******Bottom up solution
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];
}