How is the FlowerGarden pr0blem on TopCoder a DP-one?

Question

I'm reading this excellent tutorial by Dumitru on DP based problems here. And I'm trying to come up with a DP based approach for the FlowerGarden problem mentioned in the list of 1D DP problems.

I can only think of a non-DP solution that would involve initially sorting the flowers in an order and then reordering them based on different condition checks mentioned in the problem. That doesn't classify as DP, does it?

The editorial also doesn't mention anything about DP. Could anyone, by any chance, point me to a proper DP-based solution to this problem?

Thanks!

Edit:

I didn't realize the link would require registration. This is the problem:

Problem Statement You are planting a flower garden with bulbs to give you joyous flowers throughout the year. However, you wish to plant the flowers such that they do not block other flowers while they are visible.

You will be given a int[] height, a int[] bloom, and a int[] wilt. Each type of flower is represented by the element at the same index of height, bloom, and wilt. height represents how high each type of flower grows, bloom represents the morning that each type of flower springs from the ground, and wilt represents the evening that each type of flower shrivels up and dies. Each element in bloom and wilt will be a number between 1 and 365 inclusive, and wilt[i] will always be greater than bloom[i]. You must plant all of the flowers of the same type in a single row for appearance, and you also want to have the tallest flowers as far forward as possible. However, if a flower type is taller than another type, and both types can be out of the ground at the same time, the shorter flower must be planted in front of the taller flower to prevent blocking. A flower blooms in the morning, and wilts in the evening, so even if one flower is blooming on the same day another flower is wilting, one can block the other.

You should return a int[] which contains the elements of height in the order you should plant your flowers to acheive the above goals. The front of the garden is represented by the first element in your return value, and is where you view the garden from. The elements of height will all be unique, so there will always be a well-defined ordering.

Edit two:

Example 1:

height={5,4,3,2,1}

bloom={1,1,1,1,1}

wilt={365,365,365,365,365}

Returns: { 1, 2, 3, 4, 5 }

These flowers all bloom on January 1st and wilt on December 31st. Since they all may block each other, you must order them from shortest to tallest.

Example 2:

h={5,4,3,2,1}

b={1,5,10,15,20}

w={4,9,14,19,24}

Returns: { 5, 4, 3, 2, 1 } The same set of flowers now bloom all at separate times. Since they will never block each other, you can order them from tallest to shortest to get the tallest ones as far forward as possible.

Example 3: height={5,4,3,2,1}

bloom={1,5,10,15,20}

wilt={5,10,14,20,25}

Returns: { 3, 4, 5, 1, 2 } The difference here is that the third type of flower wilts one day earlier than the blooming of the fourth flower. Therefore, we can put the flowers of height 3 first, then the flowers of height 4, then height 5, and finally the flowers of height 1 and 2. Note that we could have also ordered them with height 1 first, but this does not result in the maximum possible height being first in the garden.

Answer 1

It's not a dynamic programming problem. It's a greedy algorithm problem.

This confused me too, since topcoder's own dynamic programming tutorial links to it as a practice problem in the “Elementary” section.

Sort the flowers by height, shortest to tallest. Start with an empty list of rows. For each flower (shortest to tallest), find the forward-most place where you can insert that flower such that it blocks no flowers behind it.

In Python:

def getOrdering(height, bloom, wilt):
    flowers = zip(height, bloom, wilt)
    flowers.sort()

    def flowersOverlap(f1, f2):
        # Overlap if each blooms before the other wilts.
        return f2[1] <= f1[2] and f1[1] <= f2[2]

    rows = [ ]
    for flower in flowers:
        rowIndex = len(rows)
        # Start at the back and march forward as long as
        # `flower` wouldn't block any flowers behind it.
        while rowIndex > 0 and not flowersOverlap(flower, rows[rowIndex - 1]):
            rowIndex -= 1
        rows[rowIndex:rowIndex] = [flower]

    return [flower[0] for flower in rows]

Answer 2

public int[] getOrdering(int[] height, int[] bloom, int[] wilt) {
    int[] optimal = new int[height.length];
    int[] optimalBloom = new int[bloom.length];
    int[] optimalWilt = new int[wilt.length];

    // init state
    optimal[0] = height[0];
    optimalBloom[0] = bloom[0];
    optimalWilt[0] = wilt[0];

    // run dynamic programming
    for(int i = 1; i < height.length; i ++) {
        int currHeight = height[i];
        int currBloom = bloom[i];
        int currWilt = wilt[i];

        int offset = 0; // by default, type i is to be put to 1st row
        for(int j = 0; j < i; j ++) {
            if(currWilt >= optimalBloom[j] && currWilt <= optimalWilt[j] ||
                    currBloom >= optimalBloom[j] && currBloom <= optimalWilt[j] ||
                    currWilt >= optimalWilt[j] && currBloom <= optimalBloom[j]) {  // life period overlap
                if(currHeight < optimal[j]) {  // life overlap, and type i is shorter than type j
                    offset = j;
                    break;
                } else {
                    offset = j + 1; // type i overlap with type j, and i is taller than j. Put i after j
                }
            } else {  // not overlap with current
                if(currHeight < optimal[j]) {
                    offset = j + 1; // type i not overlap with j, i is shorter than j, put i after j
                }
                // else keep offset as is considering offset is smaller than j
            }
        }

        // shift the types after offset
        for(int k = i - 1; k >= offset; k -- ) {
            optimal[k+1] = optimal[k];
            optimalBloom[k+1] = optimalBloom[k];
            optimalWilt[k+1] = optimalWilt[k];
        }
        // update optimal
        optimal[offset] = currHeight;
        optimalBloom[offset] = currBloom;
        optimalWilt[offset] = currWilt;
    }
    return optimal;
}

This is my tested working code.