Integer Break

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Steps

setup

Initialize dp array

Create a dp array of size n+1 initialized with zeros, then set dp[1] = 1 as the base case.

💡 Setting dp[1] = 1 establishes the smallest subproblem's solution, which is needed to build larger solutions.

Line:

dp = [0] * (n + 1)
dp[1] = 1

💡 The dp array will store maximum products for all integers from 0 to n, starting with a known base case.

fill_row

Start filling dp for i=2

Begin computing dp[2]. Initialize max_product to 0 before trying all breaks.

💡 We start from i=2 because dp[1] is base case; dp[2] depends on smaller subproblems.

Line:

for i in range(2, n + 1):
    max_product = 0

💡 We prepare to find the maximum product for integer 2 by trying all possible breaks.

fill_cells

Try break j=1 for i=2

Try breaking 2 into 1 and 1. Compute product = max(1, dp[1]) * max(1, dp[1]) = 1 * 1 = 1. Update max_product to 1.

💡 Trying all breaks j < i helps find the best product by combining smaller solutions.

Line:

for j in range(1, i):
    max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Using dp[1] = 1 allows us to compute dp[2] by breaking into two 1s, product 1.

fill_row

Set dp[2] = max_product

Assign dp[2] = 1, the maximum product found for integer 2.

💡 Storing the best product for dp[2] allows future computations to use this subproblem's solution.

Line:

dp[i] = max_product

💡 dp[2] is now known and can be used to compute dp for larger integers.

fill_row

Start filling dp for i=3

Initialize max_product = 0 for i=3 to find the best product by trying all breaks.

💡 We repeat the process for i=3, building on dp[1] and dp[2].

Line:

for i in range(2, n + 1):
    max_product = 0

💡 Each dp[i] depends on previously computed dp values, so order matters.

fill_cells

Try break j=1 for i=3

Break 3 into 1 and 2. Compute product = max(1, dp[1]) * max(2, dp[2]) = 1 * 2 = 2. Update max_product to 2.

💡 Using dp[2] = 1 from previous step helps find better products.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 3 as 1 + 2 yields product 2, better than no break.

fill_cells

Try break j=2 for i=3

Break 3 into 2 and 1. Compute product = max(2, dp[2]) * max(1, dp[1]) = 2 * 1 = 2. max_product remains 2.

💡 Trying all breaks ensures we find the maximum product.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Both breaks 1+2 and 2+1 yield product 2 for dp[3].

fill_row

Set dp[3] = max_product

Assign dp[3] = 2, the maximum product found for integer 3.

💡 Storing dp[3] allows it to be used for larger integers.

Line:

dp[i] = max_product

💡 dp[3] is now known and can be used for future computations.

fill_row

Start filling dp for i=4

Initialize max_product = 0 for i=4 to find the best product by trying all breaks.

💡 We continue building solutions for larger integers.

Line:

for i in range(2, n + 1):
    max_product = 0

💡 Each dp[i] depends on smaller dp values computed before.

fill_cells

Try break j=1 for i=4

Break 4 into 1 and 3. Compute product = max(1, dp[1]) * max(3, dp[3]) = 1 * 3 = 3. Update max_product to 3.

💡 Using dp[3] = 2 helps find better products for dp[4].

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 4 as 1 + 3 yields product 3.

fill_cells

Try break j=2 for i=4

Break 4 into 2 and 2. Compute product = max(2, dp[2]) * max(2, dp[2]) = 2 * 2 = 4. Update max_product to 4.

💡 Trying all breaks ensures we find the maximum product.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 4 as 2 + 2 yields product 4, better than previous 3.

fill_cells

Try break j=3 for i=4

Break 4 into 3 and 1. Compute product = max(3, dp[3]) * max(1, dp[1]) = 3 * 1 = 3. max_product remains 4.

💡 Checking all breaks confirms the maximum product.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Break 3+1 yields product 3, less than current max 4.

fill_row

Set dp[4] = max_product

Assign dp[4] = 4, the maximum product found for integer 4.

💡 Storing dp[4] allows it to be used for larger integers.

Line:

dp[i] = max_product

💡 dp[4] is now known and can be used for future computations.

fill_row

Skip intermediate steps and jump to i=10 start

Initialize max_product = 0 for i=10 to find the best product by trying all breaks.

💡 We jump to the final iteration to focus on the key decision for the input n=10.

Line:

for i in range(2, n + 1):
    max_product = 0

💡 The same process applies for i=10, but with more subproblems to consider.

fill_cells

Try break j=1 for i=10

Break 10 into 1 and 9. Compute product = max(1, dp[1]) * max(9, dp[9]) = 1 * 27 = 27. Update max_product to 27.

💡 Using dp[9] = 27 helps find a large product for dp[10].

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 10 as 1 + 9 yields product 27.

fill_cells

Try break j=3 for i=10

Break 10 into 3 and 7. Compute product = max(3, dp[3]) * max(7, dp[7]) = 3 * 12 = 36. Update max_product to 36.

💡 This break yields a better product than previous attempts.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 10 as 3 + 7 yields product 36, the best so far.

fill_cells

Try break j=4 for i=10

Break 10 into 4 and 6. Compute product = max(4, dp[4]) * max(6, dp[6]) = 4 * 9 = 36. max_product remains 36.

💡 Another break yields the same maximum product.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 10 as 4 + 6 also yields product 36.

fill_cells

Try break j=5 for i=10

Break 10 into 5 and 5. Compute product = max(5, dp[5]) * max(5, dp[5]) = 5 * 6 = 30. max_product remains 36.

💡 Not all breaks improve the maximum product.

Line:

max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))

💡 Breaking 10 as 5 + 5 yields product 30, less than current max 36.

fill_row

Set dp[10] = max_product

Assign dp[10] = 36, the maximum product found for integer 10.

💡 This final dp value is the answer to the problem.

Line:

dp[i] = max_product

💡 dp[10] = 36 means the best product for integer break of 10 is 36.

reconstruct

Return final answer dp[n]

Return dp[10] = 36 as the maximum product for integer break of 10.

💡 The final dp value is the solution to the original problem.

Line:

return dp[n]

💡 The dp array stores all subproblem solutions; dp[n] is the final answer.

def integer_break(n):
    dp = [0] * (n + 1)  # STEP 1 initialize dp
    dp[1] = 1          # STEP 1 base case
    for i in range(2, n + 1):  # STEP 2 start filling dp
        max_product = 0        # STEP 2 reset max_product
        for j in range(1, i):  # STEP 3 try all breaks
            product = max(j, dp[j]) * max(i - j, dp[i - j])  # STEP 3 compute product
            max_product = max(max_product, product)          # STEP 3 update max_product
        dp[i] = max_product    # STEP 4 assign dp[i]
    return dp[n]               # STEP 20 return answer

if __name__ == '__main__':
    print(integer_break(10))  # Expected output: 36

📊

Integer Break - Watch the Algorithm Execute, Step by Step

Watching the dp array fill step-by-step reveals how the problem builds on smaller solutions, making the optimal substructure and overlapping subproblems clear.

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Key Takeaways

✓ The dp array stores the maximum product for every integer from 1 to n, building up from smaller subproblems.

This insight is hard to see from code alone because the dp array updates are implicit; visualization makes it explicit.

✓ Each dp[i] depends on all dp[j] with j < i, showing the unbounded knapsack pattern where multiple cuts are considered.

Seeing the dependency arrows and fill order clarifies why the algorithm fills dp left to right.

✓ The maximum product for n=10 is 36, achieved by breaking into 3 + 3 + 4, which is revealed by the dp updates at i=10.

The visualization shows how different breaks are tried and why 3+7 or 4+6 breaks yield the maximum product.

Practice

1. You are given a list of jobs where each job has a start time, an end time, and a profit. You want to schedule jobs to maximize total profit such that no two jobs overlap. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic programming with jobs sorted by end time and binary search to find compatible jobs

B. Pure brute force recursion checking all subsets of jobs

C. Greedy algorithm that always picks the job with the earliest end time

D. Dynamic programming based on sorting jobs by start time and using prefix sums

Solution

1. Step 1: Understand job scheduling constraints

   The problem requires selecting non-overlapping jobs to maximize profit, which is a classic weighted interval scheduling problem.

2. Step 2: Identify the optimal approach

   Sorting jobs by end time allows binary searching for the last compatible job, enabling a DP solution that builds optimal profit incrementally.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Greedy fails on overlapping jobs with higher profit; DP with binary search handles all cases optimally [OK]

Hint: Sort by end time and use DP with binary search [OK]

Common Mistakes:

• Assuming greedy by earliest end time always works
• Sorting by start time only
• Trying brute force without pruning

2. You are given an array of integers and a target integer. You want to assign either a plus or minus sign to each integer such that the resulting sum equals the target. Which algorithmic approach guarantees finding the total number of such assignments efficiently?

easy

A. Sorting the array and using two pointers to find pairs that sum to the target

B. Greedy algorithm that picks signs based on local sum minimization

C. Dynamic programming using a 0/1 knapsack-like approach with state representing achievable sums

D. Divide and conquer by splitting the array and combining results without memoization

Solution

1. Step 1: Understand problem constraints

   The problem requires counting all sign assignments to reach a target sum, which involves exploring exponential combinations.

2. Step 2: Identify suitable algorithmic pattern

   Greedy and two-pointer approaches fail because they do not consider all combinations. Divide and conquer without memoization is exponential. DP with states representing achievable sums efficiently counts all valid assignments.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP handles overlapping subproblems and sums efficiently [OK]

Hint: Counting sign assignments requires DP over sums [OK]

Common Mistakes:

• Thinking greedy or sorting solves counting sign assignments

3. What is the time complexity of the optimal job scheduling algorithm that sorts jobs by end time and uses binary search with a DP array of size n?

medium

A. O(n log W) where W is the maximum end time

B. O(n^2) due to nested loops over all jobs

C. O(n) because each job is processed once

D. O(n log n) due to sorting and binary searches for each job

Solution

1. Step 1: Identify sorting cost

   Sorting n jobs by end time costs O(n log n).

2. Step 2: Analyze DP loop with binary search

   For each job, binary search among ends array costs O(log n), repeated n times -> O(n log n).

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Sorting + n binary searches dominate; no nested O(n^2) loops [OK]

Hint: Sorting + binary search per job -> O(n log n) [OK]

Common Mistakes:

• Assuming O(n^2) due to DP
• Ignoring binary search cost
• Confusing max end time W with input size n

4. The following code attempts to solve the Partition to K Equal Sum Subsets problem using DP with bitmask tabulation. Identify the line containing the subtle bug that can cause incorrect results or infinite loops.

def canPartitionKSubsets(nums, k):
    total = sum(nums)
    if total % k != 0:
        return False
    target = total // k
    n = len(nums)
    nums.sort()
    if nums[-1] > target:
        return False

    dp = [-1] * (1 << n)
    dp[0] = 0

    for mask in range(1 << n):
        if dp[mask] == -1:
            continue
        for i in range(n):
            if (mask & (1 << i)) == 0 and dp[mask] + nums[i] <= target:
                next_mask = mask | (1 << i)
                dp[next_mask] = (dp[mask] + nums[i]) % target

    return dp[(1 << n) - 1] == 0

medium

A. Line: dp = [-1] * (1 << n)

B. Line: if dp[next_mask] == -1: (missing in this code)

C. Line: nums.sort()

D. Line: dp[next_mask] = (dp[mask] + nums[i]) % target

Solution

1. Step 1: Identify missing condition

   The code lacks a check if dp[next_mask] is already set, so it overwrites states, causing incorrect results.

2. Step 2: Pinpoint the buggy line

   The line assigning dp[next_mask] unconditionally overwrites previous valid states, breaking memoization.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Adding "if dp[next_mask] == -1:" before assignment fixes bug [OK]

Hint: Always check if dp state is unset before assignment [OK]

Common Mistakes:

• Overwriting dp states without checking leads to incorrect answers
• Not pruning symmetric states increases runtime

5. Suppose the problem changes so that you can use each item an unlimited number of times (unbounded knapsack). Which modification to the space-optimized bottom-up DP code correctly solves this variant?

hard

A. Change the inner loop to iterate forwards from weights[i] up to W.

B. Add memoization to the recursive solution to avoid recomputation.

C. Change the inner loop to iterate backwards from W down to weights[i].

D. Use a greedy approach selecting items with the highest value-to-weight ratio repeatedly.

Solution

1. Step 1: Understand difference between 0/1 and unbounded knapsack

   In unbounded knapsack, items can be chosen multiple times, so dp[w] can be updated using dp[w - weights[i]] from the same iteration.

2. Step 2: Identify correct iteration order

   Iterating forwards from weights[i] to W allows reuse of updated dp values within the same iteration, enabling multiple counts of the same item.

3. Step 3: Why other options fail

   Backward iteration prevents multiple usage in one iteration (wrong for unbounded). Memoization helps but is not the direct fix. Greedy fails for 0/1 and unbounded knapsack except fractional case.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Forward iteration enables unlimited item reuse [OK]

Hint: Unbounded knapsack requires forward iteration in dp updates [OK]

Common Mistakes:

• Using backward iteration from 0/1 knapsack causes incorrect results.