LeetCode 343: Integer Break Solution

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will iterate through all possible ways to break the integer n into k positive integers, calculating the maximum product at each step. The key idea is to keep track of the maximum product we can achieve for each integer i from 2 to n.

Create an array dp of size n+1 to store the maximum products for each integer i from 2 to n.
Initialize dp[2] = 1 and dp[3] = 2 as the base cases.
For each integer i from 4 to n, calculate dp[i] by iterating j from 2 to i-1, and updating dp[i] using the formula: max(dp[i], max(j * (i-j), j * dp[i-j])).
Return dp[n] as the maximum product.

Time complexity: O(n^2)
Space complexity: O(n)

Solution Code

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n + 1];
        dp[2] = 1;
        dp[3] = 2;
        
        for (int i = 4; i <= n; i++) {
            for (int j = 2; j < i; j++) {
                dp[i] = Math.max(dp[i], Math.max(j * (i - j), j * dp[i - j]));
            }
        }
        
        return dp[n];
    }
}

Frequently Asked Questions

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The time complexity for LeetCode 343 (Integer Break) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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343. Integer Break