LeetCode 3383: Minimum Runes to Add to Cast Spell Solution

Master LeetCode problem 3383 (Minimum Runes to Add to Cast Spell), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3383. Minimum Runes to Add to Cast Spell

Hard

Array Depth-First Search Breadth-First Search Union Find Graph Topological Sort

Problem Explanation

Explanation:

To solve this problem, we can use a dynamic programming approach. We can create a 2D array to keep track of the minimum number of runes needed to cast a spell for a given range of indices. We can iterate through the array and update the minimum runes needed based on the current values and previous calculations.

Initialize a 2D array dp of size n x n, where n is the length of the input array.
Iterate through the array with a nested loop, considering all possible subarrays.
For each subarray, calculate the minimum runes needed to cast a spell by considering all possible splits and merging the results from previous calculations.
Update the dp array with the minimum runes needed for each subarray.
Return the value at dp[0][n-1], which represents the minimum runes needed to cast a spell for the entire array.

Time Complexity: O(n^2) where n is the length of the input array. Space Complexity: O(n^2) for the 2D array dp.

Solution Code

class Solution {
    public int minRuneToCastSpell(int[] runes) {
        int n = runes.length;
        int[][] dp = new int[n][n];

        for (int len = 2; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                int j = i + len - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k < j; k++) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k+1][j]);
                }
                dp[i][j] += runes[j] - (i > 0 ? runes[i-1] : 0);
            }
        }

        return dp[0][n-1];
    }
}

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