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The dynamic programming approach involves creating an array dp where dp[i] represents the maximum product obtainable for the integer i. The idea is to iterate through numbers from 2 to n and for each number, explore all possible ways to split it into two positive integers j and i-j. The solution for each i is derived by taking the maximum of j * (i-j) or j * dp[i-j] for all j from 1 to i-1.
Time Complexity: O(n^2) because of the nested loop iterating through i and j.
Space Complexity: O(n) for the dp array.
1#include <stdio.h>
2
3int integerBreak(int n) {
4 if (n <= 3) return n - 1;
5 int dp[n + 1];
6 dp[0] = dp[1] = 0;
7 dp[2] = 1;
8 dp[3] = 2;
9 for (int i = 4; i <= n; ++i) {
10 dp[i] = 0;
11 for (int j = 1; j < i; ++j) {
12 dp[i] = dp[i] > j * (i - j) ? dp[i] : j * (i - j);
13 dp[i] = dp[i] > j * dp[i - j] ? dp[i] : j * dp[i - j];
14 }
15 }
16 return dp[n];
17}
18
19int main() {
20 int n = 10;
21 printf("Maximum product for %d is %d\n", n, integerBreak(n));
22 return 0;
23}This C code defines a dynamic programming approach to solving the integer break problem. We initialize an array dp, where dp[i] is the maximum product obtainable for integer i. A nested loop calculates the maximum product for each integer by checking the product of all possible splits (j * (i-j) and j * dp[i-j]).
The mathematical approach relies on the observation that numbers are best broken down into 2s and 3s to maximize their product. In particular, the number 3 plays a critical role. For any number greater than 4, it is beneficial to use as many 3s as possible, because any larger piece can be expressed as 3k + r, where adding more 3s generally yields a larger product (e.g., breaking 4 into 2 + 2 is optimal, compared to using one 3 and a 1). The approach is to keep factoring out 3 from n as long as the remaining part is not less than 4, then multiply the results together to get the maximum product.
Time Complexity: O(log n), as we repeatedly subtract 3.
Space Complexity: O(1) since no extra space is used besides variables.
This C code snippet uses a mathematical approach to solve the integer break problem. Instead of solving it iteratively, we repeatedly factor out 3 from the integer until the remaining integer is less or equal to 4, at which point the product is simply multiplied by the final remainder.