LeetCode 1578: Minimum Time to Make Rope Colorful Solution

Master LeetCode problem 1578 (Minimum Time to Make Rope Colorful), 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.

1578. Minimum Time to Make Rope Colorful

Medium

Array String Dynamic Programming Greedy

Problem Explanation

Explanation:

To solve this problem, we can iterate through the given colors and neededTime arrays while keeping track of the total time needed to make the rope colorful. For each balloon color, we can check if the current color is the same as the previous one. If it is, we can compare the time needed to remove the current balloon with the time needed to remove the previous balloon. We should remove the balloon that requires less time to minimize the total time needed.

Algorithmic Idea:

Initialize a variable totalTime to 0.
Iterate through the colors and neededTime arrays.
For each balloon, check if the color is the same as the previous one.
If it is, compare the time needed to remove the current balloon and the previous balloon. Remove the one with less time.
Add the time taken to totalTime.
Return totalTime.

Time Complexity:

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

Space Complexity:

The space complexity of this solution is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int minCost(String colors, int[] neededTime) {
        int n = colors.length();
        int totalTime = 0;
        
        for (int i = 1; i < n; i++) {
            if (colors.charAt(i) == colors.charAt(i - 1)) {
                if (neededTime[i] < neededTime[i - 1]) {
                    totalTime += neededTime[i];
                    neededTime[i] = neededTime[i - 1];
                } else {
                    totalTime += neededTime[i - 1];
                    neededTime[i - 1] = neededTime[i];
                }
            }
        }
        
        return totalTime;
    }
}

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