Minimal Cost (k jumps)

Medium Dynamic Programming 1D DP Original on GeeksforGeeks

Minimal Cost (k jumps) is a medium dynamic programming problem solved with the 1d dp pattern. The best approach, optimal (tabulation over k previous stones), runs in O(n · k) time and O(n) space. Below are 2 approaches in Java, from recursion up.

Problem

A frog starts on stone 0 and wants to reach the last stone. From stone i it can jump to any of stones i + 1 through i + k, and a jump costs the absolute difference in heights. Return the minimum total cost.

Examples

Example 1

Input: heights = [10, 30, 40, 50, 20], k = 3
Output: 30
Why: 0 → 1 (cost 20) → 4 (cost 10).

Example 2

Input: heights = [10, 20, 10], k = 1
Output: 20

Constraints

1 <= n <= 10^5, 1 <= k <= 100
A jump from i to j costs |heights[i] - heights[j]|.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Recursion	`O(kⁿ)`	`O(n)`
Optimal (tabulation over k previous stones)	`O(n · k)`	`O(n)`

1Recursion

TimeO(kⁿ)

SpaceO(n)

To reach stone i, the last jump came from one of the k previous stones. Try them all.

cost(0) = 0; cost(i) = min over j of cost(j) + |h[i] - h[j]|.

Java

class Solution {
    public int minimizeCost(int k, int[] heights) {
        return cost(heights, k, heights.length - 1);
    }

    private int cost(int[] h, int k, int i) {
        if (i == 0) return 0;
        int best = Integer.MAX_VALUE;
        for (int j = Math.max(0, i - k); j < i; j++)
            best = Math.min(best, cost(h, k, j) + Math.abs(h[i] - h[j]));
        return best;
    }
}

2Optimal (tabulation over k previous stones)

TimeO(n · k)

SpaceO(n)

Fill dp from left to right. Each dp[i] checks the last k entries. This generalises Frog Jump from 2 choices to k choices.

dp[0] = 0.
For i: dp[i] = min over j in [max(0, i - k), i - 1] of dp[j] + |h[i] - h[j]|.

Java

class Solution {
    public int minimizeCost(int k, int[] heights) {
        int n = heights.length;
        int[] dp = new int[n];
        for (int i = 1; i < n; i++) {
            dp[i] = Integer.MAX_VALUE;
            for (int j = Math.max(0, i - k); j < i; j++)
                dp[i] = Math.min(dp[i], dp[j] + Math.abs(heights[i] - heights[j]));
        }
        return dp[n - 1];
    }
}

Edge cases to test

k >= n (any stone reachable directly)
n = 1 (answer 0)

Hints

Hint 1

dp[i] = min over j in [i - k, i - 1] of dp[j] + |h[i] - h[j]|.

FAQ

What is the best time complexity for Minimal Cost (k jumps)?

Optimal (tabulation over k previous stones) runs in O(n · k) time and O(n) extra space.

Which pattern does Minimal Cost (k jumps) use?

It is a dynamic programming problem that uses the 1d dp pattern. Other problems with the same pattern: Fibonacci Number, Climbing Stairs, Min Cost Climbing Stairs (2 jumps).

Is there a brute force solution for Minimal Cost (k jumps)?

Yes. Recursion takes O(kⁿ) time and O(n) space. To reach stone i, the last jump came from one of the k previous stones.

Which edge cases should I test for Minimal Cost (k jumps)?

k = n (any stone reachable directly); n = 1 (answer 0).