Leetcode 343: Integer Break

Input: The input is an integer n.

Example: n = 3

Constraints:

• 2 <= n <= 58

Output: The output is an integer representing the maximum product that can be achieved by breaking n into the sum of positive integers.

Example: 2

Constraints:

• The result is the maximum product of integers summing to n.

Goal: The goal is to find an optimal way to break down n such that the product of the parts is maximized.

Steps:

• If n is 2 or 3, return the corresponding product values directly.

• For larger values of n, repeatedly subtract 3 from n and multiply by 3 until n becomes less than or equal to 4.

• Finally, multiply the remaining value of n with the accumulated product.

Goal: The solution must handle values of n between 2 and 58.

Steps:

• The solution must handle values efficiently with constant space.

Assumptions:

• The input n is a valid positive integer greater than or equal to 2.

• Input: n = 3

• Explanation: Breaking 3 into 2 + 1 yields a product of 2.

• Input: n = 12

• Explanation: Breaking 12 into 3 + 3 + 3 + 3 yields a product of 81.

Approach: To solve the problem, we aim to break down the integer into parts where the product is maximized, using a greedy approach.

Observations:

• The maximum product can be achieved by breaking n into as many 3's as possible.

• The product of 3's is optimal, so we should subtract 3 from n repeatedly until n becomes smaller than or equal to 4.

Steps:

• Check if n is 2 or 3 and return the result.

• For n greater than 4, subtract 3 from n as many times as possible.

• Multiply the accumulated 3's with the remaining value of n.

Empty Inputs:

• The input will always be a valid positive integer, so no need to handle empty inputs.

Large Inputs:

• The solution must handle n values up to 58 efficiently.

Special Values:

• When n equals 2 or 3, the result is directly 1 and 2 respectively.

Constraints:

• The solution must be optimized for n values within the given range.

int integerBreak(int n) {
    if(n == 2) return 1;
    if(n == 3) return 2;
    int product = 1;
    while(n > 4) {
        product *= 3;
        n -= 3;
    }
    product *= n;
    return product;
}

1 : Function Definition

int integerBreak(int n) {

Define the function `integerBreak` which takes an integer `n` and returns the maximum product of integers that sum up to `n`.

2 : Base Case

    if(n == 2) return 1;

If `n` is 2, the only possible integer break is 1 + 1, and the maximum product is 1.

3 : Base Case

    if(n == 3) return 2;

If `n` is 3, the only possible integer break is 2 + 1, and the maximum product is 2.

4 : Variable Initialization

    int product = 1;

Initialize a variable `product` to 1, which will store the product of the integers as the function progresses.

5 : Looping

    while(n > 4) {

Start a loop that continues as long as `n` is greater than 4, ensuring that the remaining number can be divided into parts of 3.

6 : Multiplication

        product *= 3;

Multiply the current `product` by 3, as breaking the number into parts of 3 maximizes the product.

7 : Subtraction

        n -= 3;

Subtract 3 from `n` since we are using 3s to maximize the product.

8 : Final Multiplication

    product *= n;

If `n` is now less than or equal to 4, multiply the remaining value of `n` with `product` to complete the integer break.

9 : Return Statement

    return product;

Return the final computed `product` which is the maximum product for the integer break of `n`.

Best Case: O(1)

Average Case: O(1)

Worst Case: O(1)

Description: The time complexity is constant since the solution involves simple arithmetic operations.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant as we only use a few variables.

Leetcode 343: Integer Break

Solution to LeetCode 343: Integer Break Problem

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