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6 Dynamic Programming problems and solutions for your next coding interview

The Educative Team

The Educative Team

Dev Learning Daily

This article is based on Grokking Dynamic Programming Patterns for Coding Interviews , an interactive interview preparation course for developers. If you’ve gotten some value from this article, check out the course for many more problems and solutions like these.

A lot of programmers dread dynamic programming (DP) questions in their coding interviews. It’s easy to understand why. They’re hard!

For one, dynamic programming algorithms aren’t an easy concept to wrap your head around. Any expert developer will tell you that DP mastery involves lots of practice. It also requires an ability to break a problem down into multiple components, and combine them to get the solution.

Another part of the frustration also involves deciding whether or not to use DP to solve these problems. Most problems have more than one solution. How do you figure out the right approach? Memoize or recurse? Top-down or bottom-up?

So, we’ll unwrap some of the more common DP problems you’re likely to encounter in an interview, present a basic (or brute-force) solution, then offer one DP technique (written in Java) to solve each problem. You’ll be able to compare and contrast the approaches, to get a full understanding of the problem and learn the optimal solutions.

Sound good? Let’s go.

0–1 Knapsack Problem

Given the weights and profits of ’N’ items, put these items in a knapsack which has a capacity ‘C’. Your goal: get the maximum profit from the items in the knapsack. Each item can only be selected once.

A common example of this optimization problem involves which fruits in the knapsack you’d include to get maximum profit. Here’s the weight and profit of each fruit:

Items: { Apple, Orange, Banana, Melon } Weight: { 2, 3, 1, 4 } Profit: { 4, 5, 3, 7 } Knapsack capacity: 5

Let’s try to put different combinations of fruits in the knapsack, such that their total weight is not more than 5.

Apple + Orange (total weight 5) => 9 profit Apple + Banana (total weight 3) => 7 profit Orange + Banana (total weight 4) => 8 profit Banana + Melon (total weight 5) => 10 profit

This shows that Banana + Melon is the best combination, as it gives us the maximum profit and the total weight does not exceed the capacity.

The Problem:

Given two integer arrays to represent weights and profits of ’N’ items, find a subset of these items that will give us maximum profit such that their cumulative weight is not more than a given number ‘C’. Each item can only be selected once, so either you put an item in the knapsack or not.

The Simple Solution:

A basic brute force solution could be to try all combinations of the given items (as we did above), allowing us to choose the one with maximum profit and a weight that doesn’t exceed ‘C’. Take the example with four items (A, B, C, and D). To try all the combinations, the algorithm would look like:

for each item ‘i’ create a new set which includes item ‘i’ if the total weight does not exceed the capacity, and recursively process the remaining items create a new set without item ‘i’, and recursively process the remaining items return the set from the above two sets with higher profit

Here’s the code:

The time complexity of the above algorithm is exponential O(2^n), where ‘n’ represents the total number of items. This is also shown from the above recursion tree. We have a total of ‘31’ recursive calls — calculated through (2^n) + (2^n) -1, which is asymptotically equivalent to O(2^n).

The space complexity is O(n). This space is used to store the recursion stack. Since our recursive algorithm works in a depth-first fashion, we can’t have more than ‘n’ recursive calls on the call stack at any time.

One DP approach: Top-Down with Memoization

We can use an approach called memoization to overcome the overlapping sub-problems. Memoization is when we store the results of all the previously solved sub-problems and return the results from memory if we encounter a problem that’s already been solved.

Since we have two changing values (capacity and currentIndex) in our recursive function knapsackRecursive(), we can use a two-dimensional array to store the results of all the solved sub-problems. You’ll need to store results for every sub-array (i.e. for every possible index ‘i’) and for every possible capacity ‘c’.

What is the time and space complexity of the above solution? Since our memoization array dp[profits.length][capacity+1] stores the results for all the subproblems, we can conclude that we will not have more than N*C subproblems (where ’N’ is the number of items and ‘C’ is the knapsack capacity). This means that our time complexity will be O(N*C).

The above algorithm will be using O(N*C) space for the memoization array. Other than that we will use O(N) space for the recursion call-stack. So the total space complexity will be O(N*C + N), which is asymptotically equivalent to O(N*C).

Unbounded Knapsack Problem

Given the weights and profits of ’N’ items, put these items in a knapsack with a capacity ‘C’. Your goal: get the maximum profit from the items in the knapsack. The only difference between the 0/1 Knapsack problem and this problem is that we are allowed to use an unlimited quantity of an item.

Using the example from the last problem, here are the weights and profits of the fruits:

Items: { Apple, Orange, Melon } Weight: { 1, 2, 3 } Profit: { 15, 20, 50 } Knapsack capacity: 5

Try different combinations of fruits in the knapsack, such that their total weight is not more than 5.

5 Apples (total weight 5) => 75 profit 1 Apple + 2 Oranges (total weight 5) => 55 profit 2 Apples + 1 Melon (total weight 5) => 80 profit 1 Orange + 1 Melon (total weight 5) => 70 profit

2 apples + 1 melon is the best combination, as it gives us the maximum profit and the total weight does not exceed the capacity.

Given two integer arrays representing weights and profits of ’N’ items, find a subset of these items that will give us maximum profit such that their cumulative weight is not more than a given number ‘C’. You can assume an infinite supply of item quantities, so each item can be selected multiple times.

The Brute Force Solution:

A basic brute force solution could be to try all combinations of the given items to choose the one with maximum profit and a weight that doesn’t exceed ‘C’. Here’s what our algorithm will look like:

for each item ‘i’ create a new set which includes one quantity of item ‘i’ if it does not exceed the capacity, and recursively call to process all items create a new set without item ‘i’, and recursively process the remaining items return the set from the above two sets with higher profit

The only difference between the 0/1 Knapsack optimization problem and this one is that, after including the item, we recursively call to process all the items (including the current item). In 0/1 Knapsack, we recursively call to process the remaining items.

The time and space complexity of the above algorithm is exponential O(2^n), where ‘n’ represents the total number of items.

There’s a better solution!

One DP solution: Bottom-Up Programming

Let’s populate our ‘dp[][]’ array from the above solution, working in a bottom-up fashion. We want to “find the maximum profit for every sub-array and for every possible capacity”.

For every possible capacity ‘c’ (i.e., 0 <= c <= capacity), there are two options:

  • Exclude the item. We will take whatever profit we get from the sub-array excluding this item: dp[index-1][c]
  • Include the item if its weight is not more than the ‘c’. We’ll include its profit plus whatever profit we get from the remaining capacity: profit[index] + dp[index][c-weight[index]]

Take the maximum of the above two values:

dp[index][c] = max (dp[index-1][c], profit[index] + dp[index][c-weight[index]])

Longest Palindromic Subsequence

Given a sequence, find the length of its Longest Palindromic Subsequence (or LPS). In a palindromic subsequence, elements read the same backward and forward.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Input: “abdbca” Output: 5 Explanation: LPS is “abdba”.
Input: = “cddpd” Output: 3 Explanation: LPS is “ddd”.
Input: = “pqr” Output: 1 Explanation: LPS could be “p”, “q” or “r”.

A basic brute-force solution could be to try all the subsequences of the given sequence. We can start processing from the beginning and the end of the sequence. So at any step, there are two options:

  • If the element at the beginning and the end are the same, we increment our count by two and make a recursive call for the remaining sequence.
  • We can skip the element either from the beginning or the end to make two recursive calls for the remaining subsequence.

If option one applies, it will give us the length of LPS. Otherwise, the length of LPS will be the maximum number returned by the two recurse calls from the second option.

One DP Approach: Top-Down with Memoization

We can use an array to store the already solved subproblems.

The two changing values to our recursive function are the two indexes, startIndex and endIndex. We can then store the results of all the subproblems in a two-dimensional array.

The Fibonacci Problem

Write a function to calculate the nth Fibonacci number.

Fibonacci numbers are a series of numbers in which each number is the sum of the two preceding numbers. The first few Fibonacci numbers are 0, 1, 2, 3, 5, 8, and so on.

We can define the Fibonacci numbers as:

Fib(n) = Fib(n-1) + Fib(n-2) for n > 1 Given that: Fib(0) = 0, and Fib(1) = 1

The Basic Solution:

A basic solution could be to have a recursive implementation of the above mathematical formula.

One DP Approach: Bottom-Up

Let’s try to populate our ‘dp[]’ array from the above solution, working in a bottom-up fashion. Since every Fibonacci number is the sum of previous two numbers, we can use this fact to populate our array.

Here is the code for our bottom-up dynamic programming approach:

We can optimize the space used in our previous solution. We don’t need to store all the Fibonacci numbers up to ‘n’, since we only need two previous numbers to calculate the next Fibonacci number. We can now further improve our solution:

The above solution has time complexity of O(n) but a constant space complexity of O(1).

Longest Common Substring

Given two strings ‘s1’ and ‘s2’, find the length of the longest substring common in both the strings.

Input: s1 = “abdca” s2 = “cbda” Output: 2 Explanation: The longest common substring is “bd”.
Input: s1 = “passport” s2 = “ppsspt” Output: 3 Explanation: The longest common substring is “ssp”.

A basic brute-force solution could be to try all substrings of ‘s1’ and ‘s2’ to find the longest common one. We can start matching both the strings one character at a time, so we have two options at any step:

  • If the strings have a matching character, we can recursively match for the remaining lengths and keep track of the current matching length.
  • If the strings don’t match, we can start two new recursive calls by skipping one character separately from each string.

The length of the Longest common Substring (LCS) will be the maximum number returned by the three recurse calls in the above two options.

The time complexity of the above algorithm is exponential O(2^(m+n)), where ‘m’ and ’n’ are the lengths of the two input strings. The space complexity is O(n+m), this space will be used to store the recursion stack.

The three changing values to our recursive function are the two indexes (i1 and i2) and the ‘count’. Therefore, we can store the results of all subproblems in a three-dimensional array. (Another alternative could be to use a hash-table whose key would be a string (i1 + “-” i2 + “-” + count)).

Longest Common Subsequence:

Given two strings ‘s1’ and ‘s2’, find the length of the longest subsequence which is common in both the strings.

Input: s1 = “abdca” s2 = “cbda” Output: 3 Explanation: The longest substring is “bda”.
Input: s1 = “passport” s2 = “ppsspt” Output: 5 Explanation: The longest substring is “psspt”.

A basic brute-force solution could be to try all subsequences of ‘s1’ and ‘s2’ to find the longest one. We can match both the strings one character at a time. So for every index ‘i’ in ‘s1’ and ‘j’ in ‘s2’ we must choose between:

  • If the character ‘s1[i]’ matches ‘s2[j]’, we can recursively match for the remaining lengths.
  • If the character ‘s1[i]’ does not match ‘s2[j]’, we will start two new recursive calls by skipping one character separately from each string.

Since we want to match all the subsequences of the given two strings, we can use a two-dimensional array to store our results. The lengths of the two strings will define the size of the array’s two dimensions. So for every index ‘i’ in string ‘s1’ and ‘j’ in string ‘s2’, we can choose one of these two options:

  • If the character s1[i] matches s2[j], the length of the common subsequence would be one, plus the length of the common subsequence till the ‘i-1’ and ‘j-1’ indexes in the two respective strings.
  • If the character s1[i] doesn’t match s2[j], we will take the longest subsequence by either skipping ith or jth character from the respective strings.

So our recursive formula would be:

if si[i] == s2[j] dp[i][j] = 1 + dp[i-1][j-1] else dp[i][j] = max(dp[i-1)[j], dp[i][j-1])

The time and space complexity of the above algorithm is O(m*n), where ‘m’ and ’n’ are the lengths of the two input strings.

Keep Learning!

This is just a small sample of the dynamic programming concepts and problems you may encounter in a coding interview.

For more practice, including dozens more problems and solutions for each pattern, check out Grokking Dynamic Programming Patterns for Coding Interviews on Educative.

Originally published at blog.educative.io on January 15, 2019.

The Educative Team

Written by The Educative Team

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Dynamic programming algorithm for assigning tasks

I'm struggling with this problem. I tried solving it via simple recursion, but the time it takes for the large cases is huge and I'd like to improve it by writing a dynamic programming algorithm.

There are n given students and n given tasks. Each student is represented by a 1d array of length n of 0 s and 1 s. A[i] == 0 means that this student can't accomplish i 'th task, and A[i] == 1 means that this student can accomplish i 'th task. The goal is to determine how many different ways there are to assign tasks to students in a way, that all tasks can be accomplished and single student completes only 1 task.

I'll appreciate any hints on how to approach this problem in a dynamic programming way.

  • dynamic-programming

abhishek_naik's user avatar

  • Example input output for more clearer explanation? –  nice_dev Commented Sep 20, 2019 at 13:44
  • 2 and what are the bounds for input? –  v78 Commented Sep 20, 2019 at 13:53
  • If you will give the recursion solution I'm sure it will be simple to transfer it to a DP solution. –  Yonlif Commented Sep 20, 2019 at 15:07
  • en.wikipedia.org/wiki/Knuth%27s_Algorithm_X ? –  גלעד ברקן Commented Sep 20, 2019 at 18:51

2 Answers 2

your problem really look like a matching problem with bipartite graph (Graph Theory), where a student is a node and a task is an other one (from the other ensemble). Edges represent compatibility between them. I don't have any solution in mind but you will be able to find some good stuff about matching problem and dynamic programming.

Theo Rabut's user avatar

Initialisation condition is dp[0]=1 so the state of dp is (mask), where x represents that person upto x have been assigned a task which is also equal to setbits of mask, and mask is a binary number, whose each bit represents if that task has been assigned or not.

if i'th bit is not set and x'th person can do that work

Time complexity is O(N*2^N) and Space Complexity O(2^N)

Arjun Singh's user avatar

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assignment work available dp

Toddle Help Center for the IB curriculum

Toddle provides an intuitive workflow for DP students to track the tasks assigned to them, submit their responses and view the assessment done by the teacher.

This article will give you a detailed walkthrough of the following:

Task submission

View the assessed task, navigating to pending tasks.

via Homepage

On the homepage, you will see a ‘Class stream to-do’ list of your assignments from all your classes. This task list is divided into three tabs - Upcoming tasks, Overdue tasks and the tasks which have no due date.

Click on ‘View all’ to get an elaborate list of your tasks.

Journal (14).jpg

Here, you have a full-page view of your tasks, their due date and status. You can also filter them out based on classes.

Journal (1).jpg

via Class stream

You can also view class-specific tasks by selecting a class on your homepage. Navigate to the ‘Class stream’ page from the collapsible menu on the left where you can view tasks of the selected class, distributed among three tabs - To-do, Scheduled and Done.

Journal (2).jpg

To submit your response, click on the task name either on the above screen or using the to-do list on the homepage. Once you have entered on the submission page, you can add your work in two ways:

Option 1: If the teacher has not provided a template for the task, you can click on the ‘Add work’ button and submit your response in the form of a photo, voice, video, audio, file, workbook or link. Depending on the chosen response type, you'll get options to directly capture/record content or upload from a device/one drive/Google Drive.

Journal (32).jpg

Option 2: If your teacher has provided a specific template for the assignment, it will show up on your submission page and you can click on the ‘Add response’ button to submit your work using the given template.

Journal (5).jpg

Once you have added your work, click on ‘Mark as done’ at the bottom to submit it.

Journal (33).jpg

If you would like to resubmit your work, click on ‘Unsubmit’. Please note that you will not be able to modify/resubmit your work once it is marked as evaluated by your teacher.

Journal (34).jpg

After the teacher has evaluated and shared your work, you will get a notification under the bell icon on the homepage, as shown below. You can view the teacher’s assessment by clicking on the notification or by directly going to the given assignment under the ‘Done’ tab on ‘Class Stream’, as shown below.

Journal (7).jpg

Please note that once the work is evaluated, you cannot make any changes to it. You will be able to download your work or post it to the class journal. You will also be able to view your teacher’s remarks, if any.

Journal (16).jpg

There are three kinds of assessment tools used in DP: Score-based, DP Criteria and Final Remarks/Comments. Your teacher can choose one or more of these assessment methods and you will accordingly see the evaluated assessment at your end.

The score and the remarks given by the teacher are visible directly on this page. In this case, the teacher has used DP assessment criteria so you can view the ratings given for it by clicking on the ‘View internal assessment’ option as shown below. Please note that the name of this button will change depending on the assessment criteria used by the teacher.

Journal (15).jpg

Clicking on the assessment criteria will show you the score given for each criteria and the total marks. You can use the arrow icons to expand an individual section and view the corresponding strands and descriptors.

Journal (38).jpg

Please note that once the work is evaluated, you cannot make any changes to it. You will also be able to download your work from the assessment page.

This powerful feature will help students easily submit assessments and plan learning goals based on their progress.

We hope that you found what you were looking for. Explore other articles for more!

  • Branch and Bound Tutorial
  • Backtracking Vs Branch-N-Bound
  • 0/1 Knapsack
  • 8 Puzzle Problem
  • Job Assignment Problem
  • N-Queen Problem
  • Travelling Salesman Problem

Job Assignment Problem using Branch And Bound

Let there be N workers and N jobs. Any worker can be assigned to perform any job, incurring some cost that may vary depending on the work-job assignment. It is required to perform all jobs by assigning exactly one worker to each job and exactly one job to each agent in such a way that the total cost of the assignment is minimized.

jobassignment

Let us explore all approaches for this problem.

Solution 1: Brute Force  

We generate n! possible job assignments and for each such assignment, we compute its total cost and return the less expensive assignment. Since the solution is a permutation of the n jobs, its complexity is O(n!).

Solution 2: Hungarian Algorithm  

The optimal assignment can be found using the Hungarian algorithm. The Hungarian algorithm has worst case run-time complexity of O(n^3).

Solution 3: DFS/BFS on state space tree  

A state space tree is a N-ary tree with property that any path from root to leaf node holds one of many solutions to given problem. We can perform depth-first search on state space tree and but successive moves can take us away from the goal rather than bringing closer. The search of state space tree follows leftmost path from the root regardless of initial state. An answer node may never be found in this approach. We can also perform a Breadth-first search on state space tree. But no matter what the initial state is, the algorithm attempts the same sequence of moves like DFS.

Solution 4: Finding Optimal Solution using Branch and Bound  

The selection rule for the next node in BFS and DFS is “blind”. i.e. the selection rule does not give any preference to a node that has a very good chance of getting the search to an answer node quickly. The search for an optimal solution can often be speeded by using an “intelligent” ranking function, also called an approximate cost function to avoid searching in sub-trees that do not contain an optimal solution. It is similar to BFS-like search but with one major optimization. Instead of following FIFO order, we choose a live node with least cost. We may not get optimal solution by following node with least promising cost, but it will provide very good chance of getting the search to an answer node quickly.

There are two approaches to calculate the cost function:  

  • For each worker, we choose job with minimum cost from list of unassigned jobs (take minimum entry from each row).
  • For each job, we choose a worker with lowest cost for that job from list of unassigned workers (take minimum entry from each column).

In this article, the first approach is followed.

Let’s take below example and try to calculate promising cost when Job 2 is assigned to worker A. 

jobassignment2

Since Job 2 is assigned to worker A (marked in green), cost becomes 2 and Job 2 and worker A becomes unavailable (marked in red). 

jobassignment3

Now we assign job 3 to worker B as it has minimum cost from list of unassigned jobs. Cost becomes 2 + 3 = 5 and Job 3 and worker B also becomes unavailable. 

jobassignment4

Finally, job 1 gets assigned to worker C as it has minimum cost among unassigned jobs and job 4 gets assigned to worker D as it is only Job left. Total cost becomes 2 + 3 + 5 + 4 = 14. 

jobassignment5

Below diagram shows complete search space diagram showing optimal solution path in green. 

jobassignment6

Complete Algorithm:  

Below is the implementation of the above approach:

Time Complexity: O(M*N). This is because the algorithm uses a double for loop to iterate through the M x N matrix.  Auxiliary Space: O(M+N). This is because it uses two arrays of size M and N to track the applicants and jobs.

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