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Steps
setup

Initialize augmented array and DP table

We add 1 at both ends of the input array to simplify edge cases, resulting in nums = [1,3,1,5,8,1]. Then, we initialize a 6x6 DP table filled with zeros to store maximum coins for intervals.

💡 Augmenting the array with 1s at both ends allows uniform calculation of coins without worrying about boundaries. Initializing dp with zeros sets the base for bottom-up computation.
Line:nums = [1] + nums + [1] n = len(nums) dp = [[0] * n for _ in range(n)]
💡 The problem is transformed into finding max coins for intervals in the augmented array, and dp table will store these results.
📊
Burst Balloons - Watch the Algorithm Execute, Step by Step
Watching this visualization helps you understand how the bottom-up DP approach breaks down the problem into smaller intervals and builds up the solution, which is difficult to grasp from code alone.
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·Active fill★Answer cell
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Result: 30
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Result: 30
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Result: 70
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Result: 70
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Result: 70
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Result: 70
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Result: 78
record
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Result: 167

Key Takeaways

✓ The DP table is filled by increasing interval length, ensuring smaller subproblems are solved before larger ones.

This order is hard to see from code alone but is critical for correctness.

✓ Each dp[left][right] entry represents the maximum coins obtainable by bursting balloons strictly between left and right.

Understanding this interval definition clarifies why the augmented array has 1s at both ends.

✓ Choosing which balloon to burst last in an interval is the key decision that leads to the optimal solution.

The visualization shows how trying all candidates and taking the max builds the solution.

Practice

(1/5)
1. Consider the following Python code implementing the space-optimized DP solution for Maximal Square. Given the input matrix below, what is the final returned value? Matrix: [ ["1", "0", "1"], ["1", "1", "1"], ["1", "1", "0"] ] 
def maximalSquare(matrix):
    if not matrix or not matrix[0]:
        return 0
    rows, cols = len(matrix), len(matrix[0])
    dp = [0] * (cols + 1)
    max_side = 0
    prev = 0
    for i in range(rows):
        for j in range(1, cols + 1):
            temp = dp[j]
            if matrix[i][j-1] == '1':
                dp[j] = 1 + min(dp[j], dp[j-1], prev)
                max_side = max(max_side, dp[j])
            else:
                dp[j] = 0
            prev = temp
    return max_side * max_side
easy
A. 0
B. 4
C. 9
D. 1

Solution

  1. Step 1: Trace dp array updates for each row

    Row 0: dp updates to [0,1,0,1], max_side=1; Row 1: dp updates to [0,1,1,1], then dp[3]=1+min(1,1,1)=2, max_side=2; Row 2: dp updates to [0,1,2,0], max_side remains 2.

  2. Step 2: Calculate final area

    max_side=2, so area = 2*2 = 4.

  3. Final Answer:

    Option B -> Option B

  4. Quick Check:

    Max square side 2 -> area 4 [OK]
Hint: Track dp updates row-wise to find max side [OK]
Common Mistakes:
  • Off-by-one in indexing dp array
  • Confusing prev and temp updates
  • Returning max_side instead of area
2. You are given an array of piles of stones. Two players alternately take stones from either end of the array, aiming to maximize their total stones. Both players play optimally. Which algorithmic approach guarantees finding the maximum difference in stones the first player can achieve over the second?
easy
A. Dynamic programming using interval-based state representation
B. Sorting piles and picking stones from the largest piles first
C. Simple recursion without memoization
D. Greedy approach picking the largest pile at each turn

Solution

  1. Step 1: Understand the problem structure

    The problem involves two players picking from ends of an array, which naturally forms intervals that shrink over time.

  2. Step 2: Identify the suitable algorithmic pattern

    Dynamic programming with intervals captures the state (i, j) representing the current subarray, allowing optimal substructure and overlapping subproblems to be exploited.

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Interval DP is the classic approach for two-player games on arrays [OK]
Hint: Two-player games on arrays -> interval DP [OK]
Common Mistakes:
  • Thinking greedy works because largest pile seems best
  • Using recursion without memoization causes TLE
3. You are given a printer that can print a continuous substring of identical characters in one turn. Given a string, you want to find the minimum number of turns needed to print the entire string. Which algorithmic approach guarantees an optimal solution for this problem?
easy
A. Dynamic Programming over intervals with string compression and merging states
B. Greedy approach printing each character separately from left to right
C. Simple recursion without memoization or compression
D. Sliding window technique to find longest repeated substrings

Solution

  1. Step 1: Understand problem constraints

    The problem requires minimizing turns to print a string where each turn prints a continuous block of identical characters.

  2. Step 2: Identify suitable algorithmic pattern

    Greedy or sliding window approaches fail to merge overlapping prints optimally. Pure recursion is inefficient. Interval DP with string compression and merging states captures overlapping subproblems and optimizes turns.

  3. Final Answer:

    Option A -> Option A

  4. Quick Check:

    Interval DP with compression is known optimal [OK]
Hint: Interval DP with compression merges overlapping prints [OK]
Common Mistakes:
  • Assuming greedy printing each char separately is optimal
4. You need to find the number of distinct ways to move from the top-left corner to the bottom-right corner of an m x n grid, moving only down or right at each step. Which algorithmic approach guarantees an efficient and optimal solution for this problem?
easy
A. Pure brute force recursion exploring all possible paths without memoization
B. Greedy algorithm that always moves towards the direction with fewer remaining steps
C. Dynamic Programming that builds solutions from smaller subproblems using a grid-based state representation
D. Divide and conquer by splitting the grid into quadrants and combining results

Solution

  1. Step 1: Understand problem constraints

    The problem requires counting all unique paths with only right and down moves, which naturally forms overlapping subproblems.

  2. Step 2: Identify suitable approach

    Dynamic Programming efficiently solves overlapping subproblems by storing intermediate results, unlike greedy or pure recursion which are either incorrect or inefficient.

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    DP uses subproblem reuse -> optimal and efficient [OK]
Hint: Counting paths with overlapping subproblems -> DP [OK]
Common Mistakes:
  • Thinking greedy can find all paths
  • Using pure recursion without memoization
5. Suppose the Dungeon Game is modified so that you can move right, down, or diagonally down-right at each step. Which of the following changes to the bottom-up DP solution correctly adapts to this variant?
hard
A. Change the dp recurrence to use min(dp[i+1][j], dp[i][j+1]) only, ignoring diagonal moves
B. Change the dp recurrence to min(dp[i+1][j], dp[i][j+1], dp[i+1][j+1]) and proceed similarly
C. Use a forward DP from start to end since diagonal moves break backward DP assumptions
D. Add an extra dimension to dp to track diagonal moves separately

Solution

  1. Step 1: Understand new moves

    Allowing diagonal moves adds one more possible next cell from (i,j): (i+1,j+1).

  2. Step 2: Update dp recurrence

    Minimum health needed at (i,j) depends on minimum among dp[i+1][j], dp[i][j+1], and dp[i+1][j+1].

  3. Step 3: Keep backward DP approach

    Backward DP still works; just include diagonal cell in min calculation.

  4. Final Answer:

    Option B -> Option B

  5. Quick Check:

    Including diagonal in min ensures correct health calculation [OK]
Hint: Add diagonal cell to min in dp recurrence [OK]
Common Mistakes:
  • Ignoring diagonal moves
  • Switching to forward DP unnecessarily
  • Adding extra dp dimensions without need