Ever dreamt of landing your dream tech job? Well, chances are you’ll encounter a concept called Dynamic Programming (DP) during the interview process. Don’t worry, it’s not a superpower – it’s a powerful problem-solving technique used in software development. This blog will equip you with the answer of what are the top 10 most popular dynamic programming problems among interviewers, giving you a head start towards interview success!

## What is Dynamic Programming?

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Imagine you’re climbing a mountain. You could try every single path, but wouldn’t it be smarter to learn from past climbers and choose the most efficient route? Dynamic Programming works similarly. It breaks down complex problems into smaller, simpler subproblems.

The coolest part? It remembers the solutions to these subproblems to avoid redundant calculations, leading to the most optimal solution for the entire problem. This “remembering” is called memoization or tabulation depending on how it’s implemented.

Selection Criteria for “Top 10” Dynamic Programming Problems:To compile our list of the top 10 DP problems, we considered several factors: Frequency of occurrence in interviews. Complexity and depth of problem-solving. Versatility across different coding platforms. |

## What Are The Top 10 Most Popular Dynamic Programming Problems Among Interviewers?

### Easy Level

**Fibonacci Numbers:**You probably know this sequence – 0, 1, 1, 2, 3, 5… Each number is the sum of the two preceding ones. The DP challenge? Find the Nth Fibonacci number efficiently.

**Example:**Find the 5th Fibonacci number (0 + 1 = 1, 1 + 1 = 2, 1 + 2 = 3) – The answer is 3.

**Climbing Stairs:**Imagine you can take 1 or 2 steps at a time to reach the top floor. How many ways can you climb N stairs?

**Example:**Find the number of ways to climb 3 stairs (1 step 3 times, 2 steps and then 1 step) – The answer is 3.

### Medium Level

**Longest Common Subsequence (LCS):**Given two sequences, find the longest sequence of characters that appear in both, but not necessarily in the same order.

**Example:**Find the LCS of “AGGTAB” and “GXTXAYB” – The answer is “GTAB”

**Edit Distance:**Imagine you’re correcting typos. What’s the minimum number of edits (insertions, deletions, replacements) needed to transform one string into another?

**Example:**Find the edit distance between “kitten” and “sitting” (replace ‘k’ with ‘s’, insert ‘t’) – The answer is 3.

**Coin Change:**You have denominations of coins (e.g., 1 cent, 5 cents, 10 cents) and a target amount. In how many ways can you make the exact change using these coins?

**Example:**Find the number of ways to make 11 cents using coins of 1 cent and 5 cents – The answer is 3 (1 cent x 11, 5 cents x 2 + 1 cent x 1).

### Hard Level

**0-1 Knapsack Problem:**Imagine you’re a thief with a knapsack (bag) with a weight limit. You have items with different weights and values. How can you choose items to maximize the total value you can steal without exceeding the weight limit?

- This problem requires considering all possible combinations of items to find the optimal solution.

**Matrix Chain Multiplication:**You have a chain of matrices to multiply. The order you multiply them in affects the number of calculations needed. Find the most efficient order to minimize the total number of calculations.

- This problem involves breaking down the chain into smaller subproblems and finding the most efficient way to combine them.

**Longest Increasing Subsequence (LIS):**Given a sequence of numbers, find the longest subsequence where each number is greater than its preceding number.

**Example:**Find the LIS of [10, 9, 2, 5, 1, 7] – The answer is [2, 5, 7].

**Word Break:**Given a dictionary of words and a sentence, can the sentence be formed by breaking it down into words from the dictionary (considering spaces)?

**Example:**Check if “leetcode” can be formed using the dictionary [“leet”, “code”] – The answer is true.

**Rod Cutting:**You have a rod of fixed length that can be cut into smaller pieces. Each piece has a price. What’s the optimal way to cut the rod to maximize the total profit?

This expanded list provides a well-rounded mix of problems across different difficulty levels, ensuring there’s something for everyone to learn and practice.

## Solutions: 10 Most Popular Dynamic Programming Problems

### Easy Level

**Fibonacci Numbers:**

**Solution: **We can build a table dp where dp[i] stores the i-th Fibonacci number. Iterate from 1 to N, filling the table:

*dp[1] = 0**dp[2] = 1**dp[i] = dp[i-1] + dp[i-2] for i > 2*

**Time Complexity:**O(N)**Space Complexity:**O(N)

**Climbing Stairs:**

**Solution:** We can build a table dp where dp[i] stores the number of ways to climb i stairs. Iterate from 1 to N, filling the table:

*dp[1] = 1 (1 step)**dp[2] = 2 (1 step twice or 2 steps)**dp[i] = dp[i-1] + dp[i-2] for i > 2 (ways to reach i-th stair from i-1 or i-2 stairs)*

**Time Complexity:**O(N)**Space Complexity:**O(N)

### Medium Level

**Longest Common Subsequence (LCS):**

**Solution:** We can build a 2D table dp where dp[i][j] stores the length of the LCS for the first i characters in sequence 1 and the first j characters in sequence 2. Iterate through the table, filling it based on character matches:

*dp[i][j] = 0 if either i or j is 0 (no common subsequence)**dp[i][j] = dp[i-1][j-1] + 1 if characters at i and j are the same (add 1 to the previous LCS length)**dp[i][j] = max(dp[i-1][j], dp[i][j-1]) if characters differ (take the longer LCS from previous calculations)*

**Time Complexity:**O(mn) (m, n being lengths of sequences)**Space Complexity:**O(mn)

**Edit Distance:**

**Solution:** We can build a 2D table dp where dp[i][j] stores the minimum edit distance to transform the first i characters in sequence 1 to the first j characters in sequence 2. Iterate through the table, filling it based on insertion, deletion, or replacement:

*dp[i][0] = i (i insertions needed to transform an empty string to sequence 1)**dp[0][j] = j (j deletions needed to transform sequence 2 to an empty string)**dp[i][j] = min(dp[i-1][j] + 1 (insertion), dp[i][j-1] + 1 (deletion), dp[i-1][j-1] (replacement if characters differ))*

**Time Complexity:**O(mn)**Space Complexity:**O(mn)

**Coin Change:**

**Solution: **We can build a 1D table dp where dp[i] stores the number of ways to make a change for amount i using the given denominations. Iterate from 0 to the target amount, filling the table:

*dp[0] = 1 (one way to make 0 change: no coins)**dp[i] = 0 for negative amounts (impossible)**dp[i] = sum(dp[i – denomination]) for valid amounts, considering each denomination (add ways to make change for remaining amount after using each denomination)*

**Time Complexity:**O(mn) (m being target amount, n being number of denominations)**Space Complexity:**O(m)

**0-1 Knapsack Problem:**

**Solution:** We can build a 2D table dp where dp[i][j] stores the maximum value achievable using items from the first i items with a weight limit of j. Iterate through the table:

*dp[i][0] = 0 (no value with weight limit 0)**dp[0][j] = 0 (no value with no items)**dp[i][j] = max(dp[i-1][j], value of item i + dp[i-1][j – weight of item i])*

This considers two options:

#1: Exclude item i (use the previous value)

#2: Include item i (add its value to the max value achievable with the remaining weight)

**Time Complexity:**O(nW) (n being number of items, W being weight limit)**Space Complexity:**O(nW)

**Matrix Chain Multiplication:**

**Solution: **We can build a 2D table dp where dp[i][j] stores the minimum number of scalar multiplications needed to multiply matrices from i to j in the chain. We also built another table p to store the order of splitting the chain at the minimum cost point. Use a divide-and-conquer approach:

*dp[i][i] = 0 (no multiplication needed for a single matrix)*

Solve subproblems dp[i][j] recursively based on the split point k (i <= k < j):

*dp[i][j] = min(dp[i][k] + dp[k+1][j] + p[i][k] * p[k+1][j])*

This finds the minimum cost for multiplying sub-chains before and after the split point.

*p[i][j] is determined based on the minimum cost split point k.*

**Time Complexity:**O(n^3) (n being number of matrices)**Space Complexity:**O(n^2)

**Longest Increasing Subsequence (LIS):**

**Solution:** We can build two tables: dp to store the length of the LIS ending at each index and prev to store the previous element in the LIS for that index. Iterate through the sequence:

*dp[i] = 1 (LIS of length 1 for each element initially)*

For each element i, iterate through elements before it (j):

*If arr[j] < arr[i] and dp[j] + 1 > dp[i]:*

Update dp[i] to the longer LIS length (including current element)

Update prev[i] to j (track the previous element in the LIS)

**Time Complexity:**O(n^2)**Space Complexity:**O(n) (only dp table needed for final LIS length)

**Word Break:**

**Solution:** We can build a boolean table dp where dp[i] indicates if the first i characters can be formed from the dictionary words. Use dynamic programming with memoization:

*dp[0] = True (empty string can be formed)*

Iterate from 1 to N (length of the sentence):

*dp[i] = False (initialize)*

For each word in the dictionary:

If the word matches the substring from 0 to i-1 and dp[i – len(word)] is True:

Set dp[i] to True (current word can be formed using previous valid substring)

**Time Complexity:**O(mn) (m being sentence length, n being number of words)**Space Complexity:**O(m) (only dp table needed for final result)

**Rod Cutting:**

**Solution:** We can build a 1D table dp where dp[i] stores the maximum profit achievable by cutting a rod of length i. Iterate from 1 to the rod length:

*dp[i] = 0 (no profit for a rod of length 0)*

For each valid cut length j (less than i):

*dp[i] = max(dp[i], price[j] + dp[i – j])*

This considers two options:

#1: Don’t cut the rod (profit remains 0)

#2: Cut the rod at length j and add the price for that cut to the maximum profit achievable for the remaining rod length (i – j).

Choose the maximum profit (dp[i]) after considering all cut lengths.

**Time Complexity:**O(n^2) (n being rod length)**Space Complexity:**O(n) (only dp table needed for final profit)

## Tips For Practicing DP Problems

**Start with simpler problems:**Gradually build your confidence by attempting easier problems before tackling more complex ones.**Visualize the problem:**Use diagrams or tables to represent the subproblems and solutions.**Break down the problem:**Identify the base cases and recurrence relation for the DP solution.**Write pseudocode**Explain your thought process and logic as you code the solution.or code comments: **Test your code with various inputs:**Ensure your solution works correctly for different scenarios.

By following these tips and consistently practicing these problems, you’ll be well on your way to mastering dynamic programming and impressing your interviewers!

## Conclusion

Dynamic Programming is a crucial skill for acing technical interviews, and mastering the top 10 DP problems can significantly enhance your problem-solving abilities.

By understanding the fundamentals, practicing regularly, and exploring advanced techniques, you can confidently tackle DP problems in interviews and beyond. Keep practicing, stay curious, and happy coding! I hope you get the necessary answer to the question: what are the top 10 most popular dynamic programming problems among interviewers?