Integer Break in C++

Suppose we have a positive integer n, we have to break it into the sum of at least two positive numbers and maximize the product of those integers. We have to find the maximum product we can get. So if the number is 10, then the answer will be 36, as 10 = 3 + 3 + 4, 3 * 3 * 4 = 36

To solve this, we will follow these steps −

• Define a method solve(), this will take n, array dp and flag

• if n is 0, then return 1

• if dp[n] is not -1, then return dp[n]

• end := n – 1 when flag is set, otherwise n

• ret := 0

• for i in range 1 to end

  • ret := max of ret and i* solve(n – i, dp, false)

• set dp[n] := ret and return

• From the main method, create an array dp of size n + 1 and fill this with – 1

• return solve(n, dp)

Example (C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int solve(int n, vector <int>& dp, bool flag = true){
      if(n == 0) return 1;
      if(dp[n] != -1) return dp[n];
      int end = flag? n - 1: n;
      int ret = 0;
      for(int i = 1; i <= end; i++){
         ret = max(ret, i * solve(n - i, dp, false));
      }
      return dp[n] = ret;
   }
   int integerBreak(int n) {
      vector <int>dp(n + 1, -1);
      return solve(n, dp);
   }
};
main(){
   Solution ob;
   cout << (ob.integerBreak(10));
}

Input

10

Output

36