Integer Break LeetCode Solution

Problem – Integer Break LeetCode Solution

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

Integer Break LeetCode Solution in Java

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

Integer Break LeetCode Solution in Python

class Solution:
    def integerBreak(self, n: int) -> int:
        case = [0,0,1,2,4,6,9]
        if n < 7:
            return case[n]
        dp = case + [0] * (n-6)
        for i in range(7, n+1):
            dp[i] = 3*dp[i-3]
        return dp[-1]

Integer Break LeetCode Solution in C++

class Solution {
public:
    int integerBreak(int n) {
        
        if (n <= 2)
            return 1;

        vector<int> maxArr(n+1, 0);
                    
        /** For a number i: write i as a sum of integers, then take the product of those integers.
        maxArr[i] = maximum of all the possible products */
        
        maxArr[1] = 0;
        maxArr[2] = 1; // 2=1+1 so maxArr[2] = 1*1
        
        for (int i=3; i<=n; i++) {
            for (int j=1; j<i; j++) {
                /** Try to write i as: i = j + S where S=i-j corresponds to either one number or a sum of two or more numbers
                
                Assuming that j+S corresponds to the optimal solution for maxArr[i], we have two cases:
                (1) i is the sum of two numbers, i.e. S=i-j is one number, and so maxArr[i]=j*(i-j)
                (2) i is the sum of at least three numbers, i.e. S=i-j is a sum of at least 2 numbers,
                and so the product of the numbers in this sum for S is maxArr[i-j]
                (=maximum product after breaking up i-j into a sum of at least two integers):
                maxArr[i] = j*maxArr[i-j]
                */
                maxArr[i] = max(maxArr[i], max(j*(i-j), j*maxArr[i-j]));
            }
        }
        return maxArr[n];
    }
};
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