Part 5: Dynamic Programming and Interview Success

My Dynamic Programming Journey

Dynamic Programming (DP) was my nemesis. I avoided DP problems during preparation, thinking I'd never master them. Then I had an interview where the optimal solution required DP. I struggled, failed, and realized I couldn't avoid it anymore.

That failure forced me to truly understand DP. What I discovered: DP is just recursion with memory. Once this clicked, everything changed.

Understanding Dynamic Programming

What is DP?

From my experience, DP solves problems by:

1. Breaking them into smaller subproblems

2. Solving each subproblem once

3. Storing results to avoid redundant work

When to Use DP

I look for these characteristics:

• Optimal substructure: Optimal solution contains optimal solutions to subproblems

• Overlapping subproblems: Same subproblems solved multiple times

DP vs Recursion vs Greedy

# Problem: Fibonacci

# 1. Naive Recursion - O(2ⁿ) exponential!
def fib_recursive(n: int) -> int:
    """Extremely slow for large n"""
    if n <= 1:
        return n
    return fib_recursive(n - 1) + fib_recursive(n - 2)

# 2. DP Top-Down (Memoization) - O(n)
def fib_memo(n: int) -> int:
    """Cache results"""
    memo = {}
    
    def helper(n):
        if n <= 1:
            return n
        if n in memo:
            return memo[n]
        
        memo[n] = helper(n - 1) + helper(n - 2)
        return memo[n]
    
    return helper(n)

# 3. DP Bottom-Up (Tabulation) - O(n)
def fib_dp(n: int) -> int:
    """Build table from bottom"""
    if n <= 1:
        return n
    
    dp = [0] * (n + 1)
    dp[1] = 1
    
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    
    return dp[n]

# 4. Space Optimized - O(1) space
def fib_optimized(n: int) -> int:
    """Only need last two values"""
    if n <= 1:
        return n
    
    prev2, prev1 = 0, 1
    
    for _ in range(2, n + 1):
        current = prev1 + prev2
        prev2 = prev1
        prev1 = current
    
    return prev1

# Performance comparison
print(fib_dp(40))  # Fast
# print(fib_recursive(40))  # Takes forever!

My DP Framework

5-Step DP Approach

1. Define the state: What does dp[i] represent?

2. Find the base case: What are the simplest inputs?

3. State transition: How to compute dp[i] from previous states?

4. Determine order: Bottom-up or top-down?

5. Optimize space: Can we reduce space complexity?

Common DP Patterns

Pattern 1: Linear DP (1D)

Problem: Climbing Stairs

My first successful DP problem:

def climb_stairs(n: int) -> int:
    """
    Each step can be 1 or 2 stairs
    How many ways to reach top?
    
    My approach:
    dp[i] = ways to reach step i
    dp[i] = dp[i-1] + dp[i-2]
    (Can come from 1 step below or 2 steps below)
    
    Time: O(n), Space: O(n) → O(1) optimized
    """
    if n <= 2:
        return n
    
    # dp[i] = number of ways to reach step i
    dp = [0] * (n + 1)
    dp[1] = 1  # One way to step 1
    dp[2] = 2  # Two ways to step 2: (1,1) or (2)
    
    for i in range(3, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    
    return dp[n]

# Space optimized
def climb_stairs_optimized(n: int) -> int:
    """Only need last two values"""
    if n <= 2:
        return n
    
    prev2, prev1 = 1, 2
    
    for _ in range(3, n + 1):
        current = prev1 + prev2
        prev2 = prev1
        prev1 = current
    
    return prev1

# Test
print(climb_stairs(5))  # 8

Problem: House Robber

def rob(nums: list[int]) -> int:
    """
    Can't rob adjacent houses
    Maximum money can rob?
    
    My approach:
    dp[i] = max money robbing houses 0 to i
    dp[i] = max(
        dp[i-1],              # Don't rob house i
        dp[i-2] + nums[i]     # Rob house i
    )
    
    Time: O(n), Space: O(1)
    """
    if not nums:
        return 0
    if len(nums) == 1:
        return nums[0]
    
    # prev2 = dp[i-2], prev1 = dp[i-1]
    prev2, prev1 = nums[0], max(nums[0], nums[1])
    
    for i in range(2, len(nums)):
        current = max(prev1, prev2 + nums[i])
        prev2 = prev1
        prev1 = current
    
    return prev1

# Test
print(rob([2, 7, 9, 3, 1]))  # 12 (rob 2, 9, 1)

Pattern 2: 2D DP (Grid)

Problem: Unique Paths

def unique_paths(m: int, n: int) -> int:
    """
    Find paths from top-left to bottom-right
    Can only move right or down
    
    My approach:
    dp[i][j] = paths to reach cell (i, j)
    dp[i][j] = dp[i-1][j] + dp[i][j-1]
    
    Time: O(m*n), Space: O(m*n) → O(n) optimized
    """
    # Create 2D DP table
    dp = [[1] * n for _ in range(m)]
    
    # First row and column are all 1 (only one way)
    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
    
    return dp[m - 1][n - 1]

# Space optimized (only need previous row)
def unique_paths_optimized(m: int, n: int) -> int:
    """Use 1D array"""
    dp = [1] * n
    
    for i in range(1, m):
        for j in range(1, n):
            dp[j] += dp[j - 1]
    
    return dp[n - 1]

# Test
print(unique_paths(3, 7))  # 28

Problem: Minimum Path Sum

def min_path_sum(grid: list[list[int]]) -> int:
    """
    Find path with minimum sum from top-left to bottom-right
    
    My approach:
    dp[i][j] = min sum to reach cell (i, j)
    dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])
    
    Time: O(m*n), Space: O(1) - modify in place
    """
    m, n = len(grid), len(grid[0])
    
    # Initialize first row
    for j in range(1, n):
        grid[0][j] += grid[0][j - 1]
    
    # Initialize first column
    for i in range(1, m):
        grid[i][0] += grid[i - 1][0]
    
    # Fill rest of grid
    for i in range(1, m):
        for j in range(1, n):
            grid[i][j] += min(grid[i - 1][j], grid[i][j - 1])
    
    return grid[m - 1][n - 1]

# Test
grid = [[1, 3, 1], [1, 5, 1], [4, 2, 1]]
print(min_path_sum(grid))  # 7

Pattern 3: Subsequence DP

Problem: Longest Increasing Subsequence

This problem appeared in 2 of my interviews:

def length_of_lis(nums: list[int]) -> int:
    """
    Find length of longest increasing subsequence
    
    My approach:
    dp[i] = length of LIS ending at index i
    For each j < i:
        if nums[j] < nums[i]:
            dp[i] = max(dp[i], dp[j] + 1)
    
    Time: O(n²), Space: O(n)
    """
    if not nums:
        return 0
    
    n = len(nums)
    dp = [1] * n  # Each element is LIS of length 1
    
    for i in range(1, n):
        for j in range(i):
            if nums[j] < nums[i]:
                dp[i] = max(dp[i], dp[j] + 1)
    
    return max(dp)

# Optimized with binary search - O(n log n)
def length_of_lis_optimal(nums: list[int]) -> int:
    """
    Using tails array and binary search
    tails[i] = smallest tail of all LIS of length i+1
    """
    from bisect import bisect_left
    
    tails = []
    
    for num in nums:
        pos = bisect_left(tails, num)
        
        if pos == len(tails):
            tails.append(num)
        else:
            tails[pos] = num
    
    return len(tails)

# Test
print(length_of_lis([10, 9, 2, 5, 3, 7, 101, 18]))  # 4

Problem: Longest Common Subsequence

def longest_common_subsequence(text1: str, text2: str) -> int:
    """
    Find length of longest common subsequence
    
    My approach:
    dp[i][j] = LCS of text1[0:i] and text2[0:j]
    
    If text1[i-1] == text2[j-1]:
        dp[i][j] = dp[i-1][j-1] + 1
    Else:
        dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    
    Time: O(m*n), Space: O(m*n)
    """
    m, n = len(text1), len(text2)
    
    # Create DP table with extra row/column for base case
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if text1[i - 1] == text2[j - 1]:
                # Characters match
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                # Take max of skipping either character
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
    
    return dp[m][n]

# Test
print(longest_common_subsequence("abcde", "ace"))  # 3

Pattern 4: 0/1 Knapsack

Problem: Partition Equal Subset Sum

def can_partition(nums: list[int]) -> bool:
    """
    Can partition array into two subsets with equal sum?
    
    My approach: Reduce to 0/1 knapsack
    - Target sum = total_sum / 2
    - Can we make target sum using subset?
    
    dp[i] = can we make sum i?
    
    Time: O(n * sum), Space: O(sum)
    """
    total = sum(nums)
    
    # If odd, can't split equally
    if total % 2 != 0:
        return False
    
    target = total // 2
    
    # dp[i] = can we make sum i?
    dp = [False] * (target + 1)
    dp[0] = True  # Can always make 0
    
    for num in nums:
        # Iterate backwards to avoid using same element twice
        for i in range(target, num - 1, -1):
            dp[i] = dp[i] or dp[i - num]
    
    return dp[target]

# Test
print(can_partition([1, 5, 11, 5]))  # True (11 and 11)
print(can_partition([1, 2, 3, 5]))   # False

Pattern 5: String DP

Problem: Edit Distance

This is a classic I was asked at Google:

def min_distance(word1: str, word2: str) -> int:
    """
    Minimum operations to convert word1 to word2
    Operations: insert, delete, replace
    
    My approach:
    dp[i][j] = min operations for word1[0:i] to word2[0:j]
    
    If word1[i-1] == word2[j-1]:
        dp[i][j] = dp[i-1][j-1]  # No operation needed
    Else:
        dp[i][j] = 1 + min(
            dp[i-1][j],    # Delete from word1
            dp[i][j-1],    # Insert to word1
            dp[i-1][j-1]   # Replace
        )
    
    Time: O(m*n), Space: O(m*n)
    """
    m, n = len(word1), len(word2)
    
    # Create DP table
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # Base cases
    for i in range(m + 1):
        dp[i][0] = i  # Delete all characters
    for j in range(n + 1):
        dp[0][j] = j  # Insert all characters
    
    # Fill DP table
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if word1[i - 1] == word2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = 1 + min(
                    dp[i - 1][j],      # Delete
                    dp[i][j - 1],      # Insert
                    dp[i - 1][j - 1]   # Replace
                )
    
    return dp[m][n]

# Test
print(min_distance("horse", "ros"))  # 3

Final Interview Preparation

The Week Before

My checklist:

Monday-Wednesday:

• Review notes on patterns

• Solve 2-3 problems per day

• Focus on weak areas

Thursday:

• Mock interview with friend

• Time yourself strictly

• Practice explaining solutions

Friday:

• Light review only

• Read through common problems

• Relax and get good sleep

Saturday (Interview Day):

• Light breakfast

• Review one easy problem for confidence

• Arrive early (virtual or in-person)

Mock Interview Best Practices

From conducting 20+ mock interviews:

1. Use real tools: Practice in actual interview environments

2. Time yourself: 45 minutes per problem

3. Talk out loud: Get comfortable verbalizing

4. Get feedback: Ask what you could improve

5. Record yourself: Review your performance

Interview Day Strategy

Before the Interview

✓ Test equipment (camera, mic, screen share)
✓ Have pen and paper ready
✓ Close all distractions
✓ Have water nearby
✓ Use the bathroom
✓ Take deep breaths

During the Interview

First 5 minutes:

• Listen carefully to problem

• Ask clarifying questions

• Verify understanding with examples

Next 5-10 minutes:

• Discuss approaches

• Analyze time/space complexity

• Get interviewer buy-in

Next 20-25 minutes:

• Implement solution

• Think out loud

• Write clean code

Last 5-10 minutes:

• Test with examples

• Discuss optimizations

• Ask questions

Communication Templates

Starting:

"Let me make sure I understand the problem correctly.
We need to [restate problem].
The input is [describe], and output should be [describe].
Is that correct?"

Clarifying:

"Can I assume [assumption]?"
"What should we return if [edge case]?"
"Are there any constraints on [aspect]?"

Explaining approach:

"I'm thinking of two approaches:
1. [Approach 1]: Time O(...), Space O(...)
2. [Approach 2]: Time O(...), Space O(...)

I think approach 2 is better because [reason].
Does that sound good?"

While coding:

"I'm creating a hash map to store [purpose]..."
"Now I'll iterate through the array..."
"Let me handle this edge case..."

Testing:

"Let me test with the example:
Input: [values]
Step 1: [trace execution]
Step 2: [trace execution]
Output: [result]
That matches the expected output."

What Interviewers Look For

From my experience interviewing candidates:

Technical Skills (50%)

• Problem-solving ability

• Code quality and correctness

• Knowledge of data structures

• Complexity analysis

Communication (30%)

• Explaining thought process

• Asking good questions

• Responding to hints

• Handling feedback

Collaboration (20%)

• Working with interviewer

• Openness to suggestions

• Handling disagreement

• Professional demeanor

Handling Different Outcomes

When You're Stuck

Don't:

• Sit in silence

• Give up immediately

• Get defensive

Do:

"I'm thinking through a few approaches.
Let me start with a brute force solution and optimize from there."

Or:

"I'm considering [approach], but I'm not sure about [specific part].
Could you provide a hint about [question]?"

When You Find a Bug

Don't:

• Panic

• Blame the interviewer

• Randomly change code

Do:

"Let me trace through this example...
Ah, I see the issue. When [condition],
my code does [wrong thing] instead of [right thing].
Let me fix that."

After the Interview

Immediate Actions

1. Write down the problem: While it's fresh

2. Note what went well: Build confidence

3. Identify improvements: Learn from experience

4. Follow up: Send thank you email

If You Get Rejected

From my 10+ rejections:

1. It's normal: Even great developers get rejected

2. Ask for feedback: Some companies provide it

3. Learn from it: Review what to improve

4. Keep practicing: Each interview makes you better

5. Apply again: Most companies allow reapplication

Key Takeaways

From my entire interview journey:

1. Preparation is learnable: Not talent, but practice

2. Communication matters: As much as coding

3. Patterns are powerful: Learn them deeply

4. Mock interviews work: Practice under pressure

5. Rejection is feedback: Use it to improve

6. Stay consistent: Regular practice beats cramming

7. Be yourself: Authenticity shows through

Final Advice

For Different Experience Levels

New Graduates:

• Focus on fundamentals

• Do 100+ easy/medium problems

• Join study groups

Career Switchers:

• Build strong foundation first

• Leverage domain knowledge in discussions

• Network for referrals

Experienced Developers:

• Don't underestimate prep

• System design is crucial

• Emphasize impact in behavioral

Resources Recap

Best Platforms:

• LeetCode (primary)

• NeetCode (patterns)

• AlgoExpert (structure)

Best Books:

• Cracking the Coding Interview

• Elements of Programming Interviews

Community:

• r/cscareerquestions

• Blind app

• Discord study groups

Your Next Steps

1. Today: Solve one problem from each pattern

2. This week: Create your study plan

3. This month: Solve 50 problems

4. Next month: Do 5 mock interviews

5. Apply: Don't wait for "perfect" readiness

Series Conclusion

We've covered:

• Part 1: Preparation strategy and mindset

• Part 2: Problem-solving framework and patterns

• Part 3: Arrays, strings, hash tables, linked lists

• Part 4: Trees, graphs, and heaps

• Part 5: Dynamic programming and interview success

You now have the knowledge. What matters now is practice.

My Promise

If you:

• Follow the framework

• Practice consistently

• Learn from mistakes

• Mock interview regularly

You will succeed in coding interviews.

Closing Thoughts

I failed many interviews before succeeding. Each failure taught me something. Your journey will be unique, but the fundamentals remain the same.

Remember: Interviews are a skill, not a reflection of your worth as a developer.

Good luck with your interviews! You've got this. 🚀

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This series represents my experience with 50+ technical interviews as both candidate and interviewer. I hope it accelerates your interview success!

Now go practice! Each problem you solve makes you stronger.