703990. Maximum Product Cutting
Maximum Product Cutting
Slug: maximum-product-cutting4522
Difficulty: Easy
Id: 703990
Topic Tags: Dynamic Programming, Algorithms
Company Tags: None
Summary
The Maximum Product Cutting problem is an optimization challenge that requires finding the maximum product of a set of products obtained by cutting a given string into smaller segments. This problem involves dynamic programming concepts and involves maximizing the product of segment lengths.
Detailed Explanation
To solve this problem, we will use dynamic programming to build up a solution from the bottom up. We will initialize an array dp where dp[i] represents the maximum product that can be obtained by cutting the string into segments of length i.
The base case is when the length of the segment is 1. In this case, the maximum product is simply the value of the character at that position.
For lengths greater than 1, we consider all possible cuts and calculate the maximum product for each cut. We take the maximum of these products as our current dp[i] value.
Finally, we return the maximum dp value which represents the maximum product that can be obtained by cutting the string into segments.
Here's a step-by-step breakdown:
- Initialize an array
dpof sizen+1wherenis the length of the input string. - Set
dp[0] = 1as the base case for the empty string. - For each
ifrom 1 ton,- Calculate
max_product = 0. - For each
jfrom 1 toi,- Calculate the product
product = dp[j-1] * (string[i-1]).value. - Update
max_productifproduct > max_product.
- Calculate the product
- Set
dp[i] = max_product.
- Calculate
Time complexity: O(n^2) where n is the length of the input string. Space complexity: O(n) for the dynamic programming array.
Optimized Solutions
Java
public class Solution {
public int maxProduct(String s, int n) {
int[] dp = new int[n+1];
dp[0] = 1;
for(int i=1; i<=n; i++) {
int max_product = 0;
for(int j=1; j<=i; j++) {
int product = dp[j-1] * s.charAt(i-j).getValue();
if(product > max_product) {
max_product = product;
}
}
dp[i] = max_product;
}
return dp[n];
}