# Project Euler #11 Largest Product in a Grid | Cache-optimized + sliding window (C++14)

Source: HackerRank & ProjectEuler.net

Problem: Largest Product in a Grid
In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10(26)38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95(63)94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17(78)78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35(14)00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

Input
Input consists of 20 lines each containing 20 integers.

Output

Limits
0 ≤ each integer in the grid ≤ 100

Sample
Input

89 90 95 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

Output

73812150

My solution (C++14)

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <numeric>

// 2D grid represented by 1D vector for cache optimization
auto getGrid(int rows, int columns) {
auto values = rows * columns;
std::vector<int> grid(values);
std::copy_n(std::istream_iterator<int>(std::cin), values, grid.begin());
return grid;
}

class LargestProductInAGrid {
public:
LargestProductInAGrid(std::vector<int> &grid, int rows, int columns, int nAdjacents) : grid_(grid), rows_(rows),
columns_(columns),

auto largestProductInAGrid() {
long long largestProduct = 0;

for (auto row = 0; row < rows_; row++) {
largestProduct = std::max(largestProduct, largestProductInARow(row));
}

for (auto column = 0; column < columns_; column++) {
largestProduct = std::max(largestProduct, largestProductInAColumn(column));
largestProduct = std::max(largestProduct, largestProductInARightDiagonal(column));
largestProduct = std::max(largestProduct, largestProductInALeftDiagonal(column));
}

return largestProduct;
}

private:
long long largestProductInARow(int row) {
int low = row * columns_, end = low + columns_;
long long currentProduct = 0, highestProduct = 0;

while (low + nAdjacents_ - 1 < end) {
if (currentProduct == 0) {
currentProduct = std::accumulate(grid_.begin() + low, grid_.begin() + low + nAdjacents_, 1LL,
std::multiplies<>());
} else {
currentProduct /= grid_[low - 1];
currentProduct *= grid_[low + nAdjacents_ - 1];
}

if (currentProduct == 0) {
auto zero = std::find(grid_.begin() + low, grid_.begin() + low + nAdjacents_, 0);
low += zero - (grid_.begin() + low) + 1;
} else {
low++;
}

highestProduct = std::max(highestProduct, currentProduct);
}

return highestProduct;
}

int oneDIndex(int row, int column) {
return row * columns_ + column;
}

long long largestProductInAColumn(int column) {
int row = 0;
long long currentProduct = 0, highestProduct = 0;

while (row + nAdjacents_ - 1 < rows_) {
if (currentProduct == 0) {
currentProduct = 1;
for (int i = row; i < row + nAdjacents_; i++) {
currentProduct *= grid_[oneDIndex(i, column)];
}
} else {
currentProduct /= grid_[oneDIndex(row - 1, column)];
currentProduct *= grid_[oneDIndex(row + nAdjacents_ - 1, column)];
}

if (currentProduct == 0) {
while (grid_[oneDIndex(row, column)] != 0) {
row++;
}
}

row++;

highestProduct = std::max(highestProduct, currentProduct);
}

return highestProduct;
}

long long largestProductInARightDiagonal(int startingColumn) {
int row = 0, column = startingColumn;
long long currentProduct = 0, highestProduct = 0;

while (row + nAdjacents_ - 1 < rows_ && column + nAdjacents_ - 1 < columns_) {
if (currentProduct == 0) {
currentProduct = 1;
for (int i = row; i < row + nAdjacents_; i++) {
currentProduct *= grid_[oneDIndex(i, column + i)];
}
} else {
currentProduct /= grid_[oneDIndex(row - 1, column - 1)];
currentProduct *= grid_[oneDIndex(row + nAdjacents_ - 1, column + nAdjacents_ - 1)];
}

if (currentProduct == 0) {
while (grid_[oneDIndex(row, column)] != 0) {
row++, column++;
}
}

row++, column++;

highestProduct = std::max(highestProduct, currentProduct);
}

return highestProduct;
}

long long largestProductInALeftDiagonal(int startingColumn) {
int row = 0, column = startingColumn;
long long currentProduct = 0, highestProduct = 0;

while (row + nAdjacents_ - 1 < rows_ && column - nAdjacents_ + 1 >= 0) {
if (currentProduct == 0) {
currentProduct = 1;
for (int i = row; i < row + nAdjacents_; i++) {
currentProduct *= grid_[oneDIndex(i, column - i)];
}
} else {
currentProduct /= grid_[oneDIndex(row - 1, column + 1)];
currentProduct *= grid_[oneDIndex(row + nAdjacents_ - 1, column - nAdjacents_ + 1)];
}

if (currentProduct == 0) {
while (grid_[oneDIndex(row, column)] != 0) {
row++, column--;
}
}

row++, column--;

highestProduct = std::max(highestProduct, currentProduct);
}

return highestProduct;
}

std::vector<int> grid_;
int rows_;
int columns_;
};

int main() {
const int rows = 20, columns = 20, nAdjacents = 4;
auto grid = getGrid(rows, columns);
std::cout << solution.largestProductInAGrid() << std::endl;
}

Analysis
Time complexity: $$\O(r * c)\$$
Space complexity: $$\O(1)\$$, not counting initial data

In the past, I solved this problem when I was learning to program using the brute force method. This time, I wanted to optimize it for performance while maintaining a reasonable standard of readability.

• Welcome to Code Review. I have rolled back your last edit. Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers. – Heslacher Mar 28 at 14:38

You are using long long for your products. The maximum possible product of 4 numbers between 0 & 100 is $$\100^4\$$, which fits in 27 bits. Using a long would be sufficient, as well as probably faster.

When traversing rows, columns, right diagonals and left diagonals, your oneDIndex() increases by 1, 20, 21, and 19, respectively. The structure of your product_line maximum search is roughly the same in all 4 cases. You could combine these functions into one general purpose utility function to handle all of the 4 cases, with appropriate parametrization of the starting points, stopping points and strides.

(Since this is a programming challenge, I will leave the realization of the above algorithm to the reader.)

• Probably could template the largestProduct's type. My thinking was that 5+ elements would overflow with a long. – Eric Mar 28 at 12:44
• I implemented your suggestions here: codereview.stackexchange.com/questions/216426/… – Eric Mar 28 at 14:57

There's a lot more duplication here than I'd expect. The processing of rows, columns, forward diagonals and reverse diagonals are all very similar to each other, particularly once the coordinates are linearised to an index into grid:

• rows have a step of 1 from one element to the next
• columns have a step of columns (or, in general, the "stride" of the array)
• diagonals have a step of columns ± 1 depending on their direction.

The other difference is the start and end margins to prevent advancing beyond the 2D bounds of the array; these are easily adjusted according to the direction.

We ought to be more robust when reading inputs in getGrid(). We could test std::cin after reading from it, or simply set it before reading to throw exceptions on failures.