Dynamic Programming Interview Questions and Answers for 5 years experience

1. What is dynamic programming? Explain its core principles.

   • Answer: Dynamic programming is an algorithmic technique for solving optimization problems by breaking them down into smaller, overlapping subproblems, solving each subproblem only once, and storing their solutions to avoid redundant computations. Its core principles are:
     • Overlapping Subproblems: The problem can be broken down into smaller subproblems that are reused multiple times.
     • Optimal Substructure: An optimal solution to the problem can be constructed from optimal solutions to its subproblems.
     • Memoization (Top-Down): Storing the results of subproblems to avoid recalculation.
     • Tabulation (Bottom-Up): Building a table of solutions to subproblems in a systematic way.

2. Explain the difference between memoization and tabulation.

   • Answer: Both memoization and tabulation are techniques used in dynamic programming to avoid redundant computations. Memoization is a top-down approach where you start with the main problem and recursively solve subproblems, storing the results in a cache (usually a hash map or dictionary). Tabulation is a bottom-up approach where you build a table of solutions to subproblems, starting from the base cases and working your way up to the main problem. Memoization is generally easier to implement for complex problems, while tabulation can be more efficient in terms of space and time complexity in some cases.

3. How would you solve the 0/1 Knapsack problem using dynamic programming?

   • Answer: The 0/1 Knapsack problem can be solved using dynamic programming with a 2D table. The table `dp[i][w]` represents the maximum value that can be achieved with the first `i` items and a maximum weight capacity of `w`. The base case is `dp[0][w] = 0` for all `w`, and `dp[i][0] = 0` for all `i`. The recurrence relation is: `dp[i][w] = max(dp[i-1][w], dp[i-1][w - weight[i]] + value[i])` if `w >= weight[i]` `dp[i][w] = dp[i-1][w]` otherwise. The solution is found in `dp[n][W]`, where `n` is the number of items and `W` is the maximum weight capacity.

4. Describe a situation where dynamic programming wouldn't be the best approach.

   • Answer: Dynamic programming is not suitable for problems that lack overlapping subproblems or optimal substructure. For example, problems with highly non-linear relationships or those involving significant randomness might not benefit from dynamic programming. Greedy algorithms or other techniques might be more efficient.

5. Explain how you would use dynamic programming to find the longest common subsequence of two strings.

   • Answer: Use a 2D array dp[i][j] where dp[i][j] stores the length of the longest common subsequence of the first i characters of string 1 and the first j characters of string 2. The base case is dp[0][j] = 0 and dp[i][0] = 0 for all i and j. The recurrence relation is: dp[i][j] = dp[i-1][j-1] + 1 if string1[i-1] == string2[j-1] dp[i][j] = max(dp[i-1][j], dp[i][j-1]) otherwise. The solution is dp[m][n], where m and n are the lengths of the two strings.

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