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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`

```
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];
}
```

```
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]
```

```
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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