LeetCode //C - 343. Integer Break

343. Integer Break

Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.

Return the maximum product you can get.
 

Example 1:

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

Example 2:

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

Constraints:

• 2 <= n <= 58

From: LeetCode
Link: 343. Integer Break

Solution:

Ideas:

1. Initialization:

• Create a dp array where dp[i] represents the maximum product obtainable by breaking the integer i.

2. Dynamic Programming Fill:

• For each integer i from 2 to n, compute the maximum product by trying every possible break of i into two parts j and i - j.
• Update dp[i] with the maximum of the current value, j * (i - j), and j * dp[i - j].

3. Result:

• The result for the integer n is stored in dp[n].

Code:

int integerBreak(int n) {
    // Base case for n = 2
    if (n == 2) return 1;
    
    // Create an array to store the maximum product for each number up to n
    int dp[n + 1];
    
    // Initialize the dp array with zeroes
    for (int i = 0; i <= n; i++) {
        dp[i] = 0;
    }
    
    // Fill the dp array with the maximum products
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j < i; j++) {
            // Calculate the maximum product by either breaking or not breaking the number
            dp[i] = (dp[i] > j * (i - j)) ? dp[i] : j * (i - j);
            dp[i] = (dp[i] > j * dp[i - j]) ? dp[i] : j * dp[i - j];
        }
    }
    
    return dp[n];
}