How do you solve Integer Break on LeetCode?

Integer Break (LeetCode #343) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Breaking numbers into 2s and 3s generally yields the highest product.

What is the brute force solution for Integer Break?

The brute force approach for Integer Break is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible ways to break the integer n into sums of positive integers and calculating the product for each combination. This method guarantees that we will find the maximum product, but it is inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Integer Break?

Integer Break uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Integer Break?

Always start with a brute force solution to demonstrate understanding, then optimize. Explain your thought process clearly while coding; it shows your problem-solving skills. Be aware of edge cases, such as the smallest values of n.

What are common mistakes when solving Integer Break?

Forgetting to consider all possible splits when calculating products. Not initializing the dp array correctly or using incorrect indices.

#343

Integer Break

Medium

MathDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal solution leverages the mathematical properties of numbers, specifically that breaking numbers into 2s and 3s yields the highest product. We can use dynamic programming to build up the maximum product for each integer up to n, using previously computed results.

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Algorithm

3 steps

1Step 1: Create an array dp of size n+1 to store maximum products for each integer up to n.
2Step 2: Initialize dp[1] = 1 (not used but for consistency) and dp[2] = 1.
3Step 3: For each integer i from 3 to n, compute dp[i] as the maximum of dp[j] * (i - j) for j from 1 to i-1.

solution.py12 lines

1# Full working Python code
2
3def integer_break(n):
4    dp = [0] * (n + 1)
5    dp[1] = 1
6    for i in range(2, n + 1):
7        for j in range(1, i):
8            dp[i] = max(dp[i], j * (i - j), j * dp[i - j])
9    return dp[n]
10
11# Example usage
12print(integer_break(10))  # Output: 36

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops iterating through the integers, while the space complexity is O(n) because we store the maximum products in an array of size n.

1Breaking numbers into 2s and 3s generally yields the highest product.
2Dynamic programming can efficiently build solutions for larger problems based on smaller subproblems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.