Integer Break

📋

Problem

Imagine you have a rope of length n and want to cut it into integer lengths to maximize the product of those lengths. How should you cut it?

Given an integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

2 ≤ n ≤ 10^5

Edge cases: n = 2 → output: 1 (only 1+1 possible)n = 3 → output: 2 (1+2 or 2+1, product 2)n = 4 → output: 4 (2+2, product 4)

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IDE

def integer_break(n: int) -> int:public int integerBreak(int n)int integerBreak(int n)function integerBreak(n)

def integer_break(n: int) -> int:
    # Write your solution here
    pass

class Solution {
    public int integerBreak(int n) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int integerBreak(int n) {
    // Write your solution here
    return 0;
}

function integerBreak(n) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for all inputs due to missing base case or incorrect initialization.✅ Initialize dp array with dp[1] = 1 and handle base cases explicitly.

Wrong: n-1Returning n-1 as max product by splitting into 1 + (n-1) without further splits.✅ Use dp to consider recursive splits and max of dp[j] and j for all j.

Wrong: Incorrect smaller values for n=7 or n=8Using greedy approach picking only 3s or treating problem as 0/1 knapsack.✅ Implement full DP considering all splits and allowing repeated parts.

Wrong: Timeout or no output for large nUsing O(n^2) DP approach for large n causing TLE.✅ Use mathematical formula based on dividing n into 3s and 2s for O(n) or O(1) solution.

✓

Test Cases

Focus on handling smallest inputs and base cases correctly.

Watch out for common algorithmic pitfalls like greedy and 0/1 confusion.

Think about time complexity and how to optimize beyond DP.

t1_01basic

Input{"n":10}

Expected36

⏱ Performance - must finish in 2000ms

10 can be broken into 3 + 3 + 4, and 3*3*4 = 36 is the maximum product.

💡 Think about breaking n into parts to maximize product.

💡 Use dynamic programming to store max product for each length.

💡 For each length i, try all splits j and take max of j * (i-j) or j * dp[i-j].

Why it failed: Incorrect output means recurrence relation is wrong; ensure you consider max(j, dp[j]) and max(i-j, dp[i-j]) when computing products.

✓ Correctly computes max product using unbounded knapsack style DP.

t1_02basic

Input{"n":5}

Expected6

⏱ Performance - must finish in 2000ms

5 can be broken into 2 + 3, and 2*3 = 6 is the maximum product.

💡 Try all splits of 5 into two parts and check products.

💡 Remember to compare splitting further or not for each part.

💡 Use dp[i] = max over j of max(j, dp[j]) * max(i-j, dp[i-j]).

Why it failed: Failing this test usually means not considering recursive splits properly; fix by using dp values for subproblems instead of just raw splits.

✓ Handles smaller inputs correctly with proper DP.

t2_01edge

Input{"n":2}

Expected1

⏱ Performance - must finish in 2000ms

Only split is 1 + 1, product is 1.

💡 Check smallest valid input boundary.

💡 Ensure base cases for n=2 are handled correctly.

💡 Return 1 for n=2 since 1*1=1 is max product.

Why it failed: Incorrect output means base case for n=2 is missing or wrong; add explicit base case dp[2] = 1.

✓ Correctly handles smallest input boundary.

t2_02edge

Input{"n":3}

Expected2

⏱ Performance - must finish in 2000ms

3 can be split into 1 + 2, product 2 is max.

💡 Check base cases for small n.

💡 Remember dp[3] = max(1*2, 1*dp[2], dp[1]*dp[2]) etc.

💡 Ensure dp array is correctly initialized and updated.

Why it failed: Failing this test means dp initialization or transition is incorrect for small n; fix dp updates to consider max(j, dp[j]).

✓ Correctly computes max product for small inputs.

t2_03edge

Input{"n":4}

Expected4

⏱ Performance - must finish in 2000ms

4 can be split into 2 + 2, product 4 is max.

💡 Check exact boundary where splitting into equal parts is optimal.

💡 Try all splits and verify dp updates.

💡 Ensure dp[4] = max of splits including 2*2.

Why it failed: Incorrect output means not checking all splits or missing equal split case; fix by iterating all j from 1 to i-1.

✓ Handles exact boundary cases correctly.

t3_01corner

Input{"n":7}

Expected12

⏱ Performance - must finish in 2000ms

7 can be split into 3 + 4, product 12 is max.

💡 Beware of greedy approach picking largest 3s only.

💡 Try all splits, not just 3-based splits.

💡 Use dp to consider all partitions, not greedy.

Why it failed: Greedy approach fails here by choosing only 3s; fix by using DP to consider all splits.

✓ Correctly avoids greedy trap by full DP.

t3_02corner

Input{"n":8}

Expected18

⏱ Performance - must finish in 2000ms

8 can be split into 3 + 3 + 2, product 18 is max.

💡 Check if solution mistakenly treats problem as 0/1 knapsack.

💡 Remember this is unbounded: parts can repeat.

💡 Allow multiple uses of same split length in DP.

Why it failed: Treating problem as 0/1 knapsack causes undercounting; fix by allowing repeated splits in DP.

✓ Correctly implements unbounded knapsack logic.

t3_03corner

Input{"n":9}

Expected27

⏱ Performance - must finish in 2000ms

9 can be split into 3 + 3 + 3, product 27 is max.

💡 Watch for off-by-one errors in loops.

💡 Ensure dp array indices and ranges are correct.

💡 Check loop boundaries from 1 to n-1 inclusive.

Why it failed: Off-by-one errors cause missing optimal splits; fix loop ranges to include all valid splits.

✓ Handles loop boundaries correctly.

t4_01performance

Input{"n":100000}

Expectednull

⏱ Performance - must finish in 2000ms

Input n=100000 tests O(n^2) DP solutions which must time out; efficient O(n) math solution required.

💡 O(n^2) DP will TLE for n=100000.

💡 Use mathematical formula based on dividing n into 3s and 2s.

💡 Implement O(n) or O(1) solution using known integer break math.

Why it failed: TLE occurs due to O(n^2) DP; fix by using math-based O(n) or O(1) approach.

✓ Solution runs within time using optimized approach.

Practice

1. You are given a set of stones with positive integer weights. You want to split them into two groups such that the difference between the sum of weights in the two groups is minimized. Which algorithmic approach guarantees finding the minimal possible difference?

easy

A. A dynamic programming approach similar to 0/1 knapsack that finds achievable sums up to half the total weight

B. A breadth-first search exploring all subsets of stones without pruning

C. A greedy algorithm that repeatedly smashes the two heaviest stones until one or none remain

D. A divide-and-conquer approach that recursively splits stones into halves and merges results

Solution

1. Step 1: Understand the problem as partitioning stones into two subsets with minimal difference

   This is a classic subset-sum variation where we want to find a subset with sum as close as possible to half the total.

2. Step 2: Recognize that 0/1 knapsack DP can find achievable sums up to half total weight

   By using DP to track which sums are possible, we can find the closest sum to half, minimizing the difference.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Greedy fails on some inputs; DP guarantees minimal difference [OK]

Hint: Minimal difference -> subset sum DP, not greedy [OK]

Common Mistakes:

• Believing greedy smashing always yields minimal difference
• Thinking divide-and-conquer without DP suffices
• Assuming BFS is efficient enough here

2. You need to find the number of distinct ways to make a target amount using unlimited supply of given coin denominations. Which algorithmic approach guarantees counting all unique combinations without overcounting permutations?

easy

A. Greedy algorithm that picks the largest coin first until the amount is reached

B. Dynamic programming using a 1D array iterating coins first, then amounts

C. Backtracking exploring all permutations of coins to sum to the amount

D. Sorting coins and using two pointers to find pairs that sum to the amount

Solution

1. Step 1: Understand problem constraints

   The problem requires counting unique combinations, not permutations, of coins to reach the target amount.

2. Step 2: Identify algorithm that counts combinations without duplicates

   Dynamic programming iterating coins first ensures each coin is considered in order, avoiding counting permutations multiple times. The 1D DP approach efficiently accumulates counts for each amount.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP with coin-first iteration counts combinations correctly [OK]

Hint: Coin-first DP avoids permutation overcounting [OK]

Common Mistakes:

• Using greedy misses some combinations
• Counting permutations instead of combinations
• Using two pointers only works for pair sums

3. Consider the following Python function implementing the space-optimized subset sum algorithm. Given nums = [2, 3, 7] and S = 5, what is the value of dp[5] after processing all elements?

easy

A. True

B. False

C. IndexError due to out-of-range access

D. None (function does not return a boolean)

Solution

1. Step 1: Trace dp array after processing num=2

   Initially dp=[True, False, False, False, False, False]. After num=2, dp[2]=True because dp[0] was True.

2. Step 2: Trace dp array after processing num=3

   From w=5 down to 3, dp[5]=dp[5] or dp[2]=False or True -> True, dp[3]=dp[3] or dp[0]=False or True -> True.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Subset {2,3} sums to 5 -> dp[5] is True [OK]

Hint: Trace dp updates backward to avoid double counting [OK]

Common Mistakes:

• Iterating dp forward causing overcounting
• Off-by-one errors in dp indices

4. The following code attempts to implement the space-optimized 0/1 Knapsack solution. Identify the line that contains a subtle bug that can cause incorrect results.

def knapsack_space_optimized(weights, values, W):
    n = len(weights)
    dp = [0] * (W + 1)
    for i in range(n):
        for w in range(weights[i], W + 1):  # Note: iterating forwards
            dp[w] = max(dp[w], dp[w - weights[i]] + values[i])
    return dp[W]

medium

A. Line 5: Inner loop iterating forwards over weights.

B. Line 4: Outer loop iterating over items.

C. Line 2: Initializing dp array with zeros.

D. Line 6: Updating dp[w] with max of current and new value.

Solution

1. Step 1: Analyze iteration order in inner loop

   The inner loop iterates forwards from weights[i] to W, which causes the current item to be counted multiple times in the same iteration, violating 0/1 knapsack constraints.

2. Step 2: Correct iteration direction

   Iterating backwards (from W down to weights[i]) ensures each item is only counted once per capacity, preserving correctness.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Forward iteration causes overcounting items [OK]

Hint: Inner loop must iterate backwards to avoid overcounting [OK]

Common Mistakes:

• Iterating forwards in dp array updates for 0/1 knapsack.

5. What is the time complexity of the space-optimized bottom-up subset sum algorithm given an input list of size n and target sum S?

medium

A. O(2^n)

B. O(n + S)

C. O(n * S)

D. O(n * log S)

Solution

1. Step 1: Identify loops in the algorithm

   The algorithm has an outer loop over n elements and an inner loop over sums from S down to each num.

2. Step 2: Calculate total operations

   Each element causes up to S iterations, so total time is O(n * S). Recursive brute force is exponential, and linear or log factors are incorrect.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested loops over n and S -> O(n * S) [OK]

Hint: Nested loops over n and S -> multiply complexities [OK]

Common Mistakes:

• Confusing recursion stack space with time
• Assuming linear or exponential time incorrectly