LeetCode 2162: Minimum Cost to Set Cooking Time Solution

Master LeetCode problem 2162 (Minimum Cost to Set Cooking Time), a medium 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.

2162. Minimum Cost to Set Cooking Time

Problem Explanation

Explanation

To solve this problem, we can simulate the process of setting the cooking time by considering all possible combinations of digit pushes and finger movements. We can use dynamic programming to keep track of the minimum cost to reach each state. At each step, we can either move the finger to a different digit or push a digit below the finger. We need to consider all possible transitions and calculate the cost accordingly.

Algorithmic Idea

  1. Initialize a 2D DP array to keep track of the minimum cost to reach each state.
  2. Iterate over all possible target seconds and update the DP array based on the minimum cost to reach that state.
  3. Calculate the cost for each transition (moving the finger or pushing a digit) and update the DP array accordingly.
  4. Finally, return the minimum cost to reach the target seconds.

Time Complexity

The time complexity of this solution is O(n), where n is the target number of seconds.

Space Complexity

The space complexity of this solution is O(n) to store the DP array.

Solution Code

class Solution {
    public int minCost(int startAt, int moveCost, int pushCost, int targetSeconds) {
        int[][] dp = new int[100][100];
        
        for (int seconds = 0; seconds <= 6039; seconds++) {
            int minutes = seconds / 60;
            int secs = seconds % 60;
            for (int i = 0; i <= 9; i++) {
                dp[seconds][i] = Integer.MAX_VALUE;
            }
            for (int i = 0; i <= 9; i++) {
                for (int j = 0; j <= 9; j++) {
                    int newSeconds = minutes * 10 + i * 10 + j;
                    if (newSeconds <= 6039) {
                        dp[newSeconds][j] = Math.min(dp[newSeconds][j], dp[seconds][i] + (i == startAt ? 0 : moveCost) + pushCost);
                    }
                }
            }
        }
        
        int result = Integer.MAX_VALUE;
        for (int i = 0; i <= 9; i++) {
            result = Math.min(result, dp[targetSeconds][i]);
        }
        
        return result;
    }
}

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