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Leetcode 343. Integer Break

Problem Statement

The problem from Leetcode 343, “Integer Break”, reads:

Given a positive 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.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note:

Clarifying Questions

  1. Q: Can n include values as low as 2?
    • A: Yes, as per the problem statement n should be greater than 1.
  2. Q: Is there an upper constraint on n that we should consider?
    • A: Yes, the constraint typically implies n can be reasonably large, but exact limits can usually be found in the problem constraints section of Leetcode.

Strategy

The problem requires breaking n into at least two smaller positive integers such that their product is maximized. Here are the steps to tackle this:

  1. Mathematical Insight:
    • For any integer n, breaking it down into parts of values 1, 2, and 3 is considered optimal as 3s maximize the product in larger chunks.
    • We generally try to get as many 3s as possible because growing numbers exponentially provide higher products.
  2. Dynamic Programming Approach:
    • We initialize an array dp where dp[i] will store the maximum product obtainable for integer i.
    • Iterate from 2 to n and calculate the possible maximum products.
    • For each integer i, iterate through all j from 1 to i/2, and calculate which combination provides the highest product using previously computed dp arrays.
  3. Greedy Approach Insight:
    • The largest product comes when breaking n into multiple 3s. Handle the remainder cases (if n % 3 == 1 then adjust one 3 to a 4).

Code Implementation

Here is a C++ implementation using the dynamic programming approach:

#include <vector>
#include <algorithm>
#include <iostream>

int integerBreak(int n) {
    if (n == 2) return 1;
    if (n == 3) return 2;
    
    std::vector<int> dp(n+1, 0);
    dp[1] = 1;
    dp[2] = 2;
    dp[3] = 3;
    
    int maxProduct = 0;
    
    for (int i = 4; i <= n; ++i) {
        for (int j = 1; j <= i / 2; ++j) {
            dp[i] = std::max(dp[i], dp[j] * dp[i - j]);
        }
    }
    
    return dp[n];
}

int main() {
    int n = 10; // Example input
    std::cout << "Maximum product of breaking " << n << " is: " << integerBreak(n) << std::endl;
    return 0;
}

Time Complexity

This approach should be efficient enough for reasonable values of n as typically required by such Leetcode problems.

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