Dynamic Programming amazing tricks! — Part I — DP strategies

Mariano Crosetti
11 min readOct 18, 2022

Never before seen tricks and perspective so that you can think and solve Dynamic Programming Problems!


This guide will be helpful for you if you are:

  • preparing coding interviews.
  • training for competitive programming.
  • just wanna learn how to come up with ideas to solve dynamic programming problems.

I will assume that you know how to code (and know recursive functions).

My goal is not to explain classic DP problems (knapsack, coin change, matrix chain multiplication, etc) but to explain the tricks that are behind many “originals” DP problems you can find in an interview or in a problem set, so you gain a “dynamic programming intuition”.

Finally, remember you will only truly gain it once you practice the ideas, so try solving many problems with these ideas.

Introduction — Computing Fibonacci

The Fibonacci sequence is defined by the recurrence:

Computing it is very easy:

But this has a problem: the complexity is exponential O(1.41ⁿ). This is clear by seeing the recursion call tree:

The problem is clear: we call many times the same function with the same argument. We are calculating the same thing many times!

We can add a cache to avoid this problem: we create an array with an entry for each possible argument of the function. We conveniently initialize it in -1 (indicating “non-calculated value”)

As we have O(N) states (possible argument values), then now, as the calculation is done only once for each state, the time complexity it’s linear! Wow!

Other DP problem: min path on 2D board

Given a N x M matrix, filled with positive numbers, we want to calculate the min cost path from left-up to the down-right corner by only moving down and right. The cost is the sum of the number on the path.

In this example, the min cost path is the colored which has a cost of 29.

If we think the problem “let’s calculate the min cost path from a cell (x,y) to the down-right corner (cell (N-1, M-1))” we can do it by the following recursive function:

DANGER: The function is clearly exponential!

The general case f(i,j) does two function calls! This might be exponential! But notice that the possible argument values are justN x M states. We can apply the same trick we did with Fibonacci and “remember” the calculated values:

The memset instruction allows us to easily initialize a chunk of memory with -1 but we could use a for loop.

If you don’t like recursion you can calculate f(i,j) in a (2-dimensional) array iteratively:

Notice that we are calculating row by row from the last row to the first one. And in each row, we do the calculations from the rightmost to the leftmost position. This works as a state (i,j) always will use the value of a right state (i,j+1) or down state (i+1,j), that will be already calculated.

We could also calculate column by column. In that case, each column would be calculated from downmost to the uppermost position.

Top-Down vs Bottom-Up DP

The first implementation (which uses a cache + recursive call) is called the “Top-Down” approach (or memoization). The second one is called “Bottom-Up” (also tabulation). Advantages and disadvantages are:

  • Top-Down is easy to write from a recursive math definition.
  • Bottom-Up is faster because recursion has a considerable computing cost (it does many stack operations).
  • Top-Down is faster in cases where there are many table’s states that are never calculated (it only calculates what it needs, it calculates “on-demand”).

Trick: From math to top-down DP

If you have a math definition of a recursive function you can implement it, and add memoization by:

  • Creating an cache array with as many states as the function’s arguments.
  • Initialitzing it in a non-usable variable indicating “non computed yet”.
  • Checking on each function call if the value was already calculated. If it’s not calculate it using the formula and store its value on the cache.

As each state is computed only once, the resultant complexity is:

Trick: from Top-Down DP to Bottom-Up

If you have a Top-Down implementation and want to create a Bottom-Up (for example for speed-up reasons) you only have to copy the body of the function to a group of nested for loops and ensure that:

  • They visit all the states.
  • For each state they visit, the “used” states (the one used in the formula of the one we are calculating) were already visited.

Reconstructing the path

Now let’s assume we want to reconstruct (any) min path by calculating a sequence of “D” (down) and “R” (right) steps.

In this case, the colored path would be represented by the string “RDRRDD”.

If we already coded the Top-Down solution, we can easily reconstruct the solution using the following backtracking function:

Essentially what we do is to start in the first state and move across transitions we know are optimal. For doing this, we use the already provided Top-Down function to calculate the cost of each state. As we have memoization the cost of doing this is amortized (can be considered O(1) in the backtracking as it will be in the memory after running the calculation of the first state).

Trick: Recipe for reconstructing the solution to any DP problem

This procedure can be applied to any DP problem that requires reconstructing the solution:

  • Solve the problem with a recursive math function
  • Implement it and add memoization. We have a Top-Down!
  • Make backtracking by copying the Top-Down implementation and checking on each transition if it’s optimal. If it is, then move the backtracking recursively through this transition.
  • DON’T FORGET THE RETURNS after the recursive calls: we don’t want to visit all the states, only the one on (any) optimal solution.

Reconstructing the k-th solution

Consider the list of “D”,“R” strings that represent min paths lexicographically sorted. Calculate the K-th element of that list (for a given K on the problem input).

At a glance, this looks like a really hard problem. But it’s easy and there is a recipe (I love recipes!) that will allow you to “reconstruct the k-th solution for any DP”. First, let’s solve it for this problem.

Let’s calculate the number of minimal paths from any given position (x,y) to the end (down-right corner). We can do it by doing other DP:

Here we use the original DP, f, to check if a transition is optimal. Given that, we sum the amount of that transition to the current solution or not.

Now, for reconstructing the k-th solution we just modify the backtracking we already had:

The key idea behind that is that we want the k-th lexicographical solution to include K in the backtracking: if we are in a certain state of the backtracking we are deciding if moving “D” (down) or “R” (right). If we want the k-th lexicographically and the “D” transition is optimal (moving to (i+1, j) is optimal) we can use the information of how many optimal paths are in that transition, amountMin(i+1, j). If they are less than k it indicates we need to move downward and calculate the k-th on (i+1, j). If not, we need to “pass” all these optimal solutions, take other transitions and calculate the k-amountMin(i+1, j) — th solution (we subtract the number of solutions we are “passing”).

Trick Reconstructs the k-th solution for any DP problem

This procedure can be done in any DP problem. The recipe is:

  • Solve the problem with a recursive math function.
  • Implement it and add memoization. We have a Top-Down!
  • Calculate the number of optimal solutions for each state. You can do it by writing other DP based on the code of the Top-Down and using it for checking if a transition is optimal. If it’s optimal we need to accumulate its results.
  • Make the reconstruction backtracking including K on the arguments. Inspect the transitions that comes lexicographically first. If they are optimal decide if we need to choose this transition or pass all the solutions that imply taking this transition. We decide this by comparing K and the number of solutions over this transition. Remember that if we pass, we need to subtract the number of passed solutions from k. For that we calculated the number of optimal solution for each state.

Reducing memory complexity from O(N²) to O(N)

Analyzing again the order that our Bottom-Up implementation calculates the table, we can notice that when we are filling the i-th row, we only need the already calculated values on this same row and the values from the (i+1)-th row. We don’t need, for example, the values on (i+2)-th row (or further). Then we can do:

There we are overwriting the (i+2)-th values with the i-th (as we will not need them anymore!).

This way we are reducing by N factor the memory complexity! Wow!

Trick: Saving memory by overwriting states on DP

We can apply the last technique in any DP!

  • Have a Bottom-Up implementation (remember that in general, we can easily translate any recursive math definition in Bottom-Up implementation if we are cautious about the order we iterate the states).
  • For overwriting certain states we can use % (modulo operator) over all the usages of certain dimensions. Of course, we will modify the array declaration to declare less memory.
  • Notice here we are using %2 because we are only maintaining a “strip” of memory of size 2. This can be easily generalized to greater strip sizes (imagine the problem where we can take steps of size two).
  • We have to be careful to apply the trick to the dimension that it’s iterated by the outer for loop from the Bottom-Up. A good tip is to make a picture to visualize the order we are filling the table for the purpose of choosing which dimension to overwrite.
  • REMEMBER the % when you are showing/using the result!

Adding a flag to the state

Let’s have the same problem but with a change: now we have K “magic bishops” in our pockets. We can use a magic bishop to move as a bishop, formally: moving from (i,j) to (i+k, j+k) visiting all cells in the diagonal in between. Each time we do this we spend one of the K “magic bishops”.

It’s easy to carry K in the recursion state and adapt the definition to consider the possibility of moving as a bishop:

What’s the complexity here? Remember the complexity of any DP is:

The number of states is: N x N x K.

Now we have a for loop inside the calculating function so the complexity of calculating one state is O(K).

So the total complexity is O(N² K²).

But we can erase the for loop by adding a flag (we will call it “b”) to our implementation that indicates if “the last movement was a diagonal movement”.

Another interpretation of b is that indicates if “we were moving as a bishop”

Notice that now, the number of states is: N x N x K x 2

But the complexity of calculating one state is O(1)

So the total complexity is O(N² K)

We reduced the calculation complexity by a linear factor and only increased the state size by a constant.

Trick: Adding a flag to the state for codifying certain information about the past

This is a trick that appears in many DP problems: a flag that codifies certain information about “the past”. Notice that in general, the DP doesn’t carry all the information about “the past” of the solution. Because we are solving a subproblem from a certain position (or state) to the end.

So it’s very typical to add flags/numbers/states with the necessary information from the past.

Note: The fact that we are as agnostic of the past as possible it’s the source of the power of the DP: the overlapping of the subproblems. Each state could be associated to “many pasts”.

Trick: recovering a parameter

Motivational problem: Given a sequence of integers, let’s split them into two subsequences. We will define the cost of a split as the sum of the costs of the subsequences. Each of both costs are calculated by summing the square of the difference of consecutive numbers in the subsequence. Example:

27 2 30 1 2 31 = (27–30)² + (30–31)² + (2–1)² + (1–2)² = 12

We can solve it by the following recursive function:

r, b, i indicates lastRed, lastBlue and currentPosition respectively.

f is calculating the cost of splitting the sequence of [currentPosition, N) (from currentPosition to the end), considering that in the past lastRed was the latest position painted by red (and lastBlue was the latest position painted by blue).

It’s clear we have only two options for currentPosition: painting it with blue or red. These are the options represented in the min. If we paint it with red we will need to add the cost of the square of the difference with the value that was in lastRed. That’s why we need this information from the past (lastRed and lastBlue): because when painting currentPosition we need to match it with a value that painted in the past.

(We didn’t write the corner cases where currentPosition = n just because it’s trivially 0).

If we make a trivial implementation the complexity would be O(N³) (N³ states and O(1) complexity on the state calculation).

But observe that:

This constraint tells us that there are many states from the N³ we don’t really need. Notice that a Top-Down implementation will never visit a state that doesn’t hold the condition.

We can only use two of the three parameters in our memoization:

Also, we don’t need to calculate one parameter from another, as we can carry it on the recursion, but not using it for accessing the memorization:


You will only truly gain expertise once you practice the ideas, so try solving many problems with these ideas. Here there are problems that apply the knowledge from this class:

End of Part I

This is the end of Part I. I will publish a series of articles about “Dynamic Programming” teaching the thinking intuition behind the technique (in contrast to most of tutorials that focus on classic DP problems).

About me

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I will try to keep on posting regularly about SE, Web, and ML always with these objectives in mind:

  • keep it short
  • being meaningful
  • explain with examples



Mariano Crosetti

NLP & Computer Vision Engineer 🧠 Only posts that add VALUE to GROWTH as SWE AI / ML 📈 ICPC LATAM Champion 🏆 (defeated Berkeley, Stanford & ETH) Ex-@Google