Greedy solution

We will try to optimize it a bit. Let's start with a relatively simple greedy approach.

Greedy algorithms

Greedy algorithms can be described as algorithms that decide the action on the optimal option at the moment.

We can try to adjust the naïve solution. The most problematic part are the recursive calls. Let's apply the greedy approach there:

public static int longestSlideDown(int[][] pyramid, int row, int col) {
    if (row == pyramid.length - 1) {
        // BASE: We're at the bottom
        return pyramid[row][col];
    }

    if (col + 1 >= pyramid[row + 1].length
            || pyramid[row + 1][col] > pyramid[row + 1][col + 1]) {
        // If we cannot go right or it's not feasible, we continue to the left.
        return pyramid[row][col] + longestSlideDown(pyramid, row + 1, col);
    }

    // Otherwise we just move to the right.
    return pyramid[row][col] + longestSlideDown(pyramid, row + 1, col + 1);
}

OK, if we cannot go right or the right path adds smaller value to the sum, we simply go left.

Time complexity

We have switched from adding the maximum to following the “bigger” path, so we improved the time complexity tremendously. We just go down the pyramid all the way to the bottom. Therefore we are getting:

$$\mathcal{O}(rows)$$

We have managed to convert our exponential solution into a linear one.

Running the tests

However, if we run the tests, we notice that the second test failed:

Test #1: passed
Test #2: failed

What's going on? Well, we have improved the time complexity, but greedy algorithms are not the ideal solution to all problems. In this case there may be a solution that is bigger than the one found using the greedy algorithm.

Imagine the following pyramid:

      1
     2  3
   5  6  7
  8  9 10 11
99 13 14 15 16

We start at the top:

1. Current cell: 1, we can choose from 2 and 3, 3 looks better, so we choose it.
2. Current cell: 3, we can choose from 6 and 7, 7 looks better, so we choose it.
3. Current cell: 7, we can choose from 10 and 11, 11 looks better, so we choose it.
4. Current cell: 11, we can choose from 15 and 16, 16 looks better, so we choose it.

Our final sum is: 1 + 3 + 7 + 11 + 16 = 38, but in the bottom left cell we have a 99 that is bigger than our whole sum.

tip

Dijkstra's algorithm is a greedy algorithm too, try to think why it is correct.