712181. Minimal Cost
Minimal Cost
Slug: minimal-cost Difficulty: Medium Id: 712181 Topic Tags: Dynamic Programming, Arrays, Data Structures, Algorithms Company Tags: None
Summary
The Minimal Cost problem is a dynamic programming optimization problem that involves finding the minimum cost of a given array. The problem requires an algorithmic approach to solve efficiently.
Detailed Explanation
The Minimal Cost problem can be explained as follows: Given an array of integers arr and another integer k, find the minimum cost to reach all elements in the array. The cost to reach each element is equal to its value, but with a twist - we can only jump k steps at a time.
To solve this problem, we will use dynamic programming. Let's define two arrays: dp and prev. dp[i] will store the minimum cost to reach the i-th element in the array, and prev[i] will store the previous index from which we can jump to the i-th element.
Here's a step-by-step breakdown of the solution:
- Initialize the first element of both arrays:
dp[0] = 0,prev[0] = -1. - Iterate through the array, starting from the second element (
i = 1). For each element:- Calculate the minimum cost to reach this element by considering all possible jumps from the previous elements that are within
ksteps:dp[i] = min(dp[j] + arr[i]) for j in range(max(0, i-k), i). - Update
prev[i]with the index of the previous element that gives us the minimum cost to reach this element.
- Calculate the minimum cost to reach this element by considering all possible jumps from the previous elements that are within
- The minimum cost to reach all elements is stored in the last element of the
dparray:min_cost = dp[-1].
Time complexity: O(n), where n is the size of the input array.
Space complexity: O(n), as we need to store the dp and prev arrays.
Optimized Solutions
Java
public class MinimalCost {
public static int findMinCost(int[] arr, int k) {
int n = arr.length;
int[] dp = new int[n];
int[] prev = new int[n];
dp[0] = 0;
prev[0] = -1;
for (int i = 1; i < n; i++) {
int minCost = Integer.MAX_VALUE;
for (int j = Math.max(0, i-k); j < i; j++) {
if (dp[j] + arr[i] < minCost) {
minCost = dp[j] + arr[i];
prev[i] = j;
}
}
dp[i] = minCost;
}
return dp[n-1];
}