Minimum Time To Make Rope Colorful - Leetcode Solution

Problem Statement:

You are given a rope with different colors on it. You have to color the rope in such a way that every adjacent segment of the rope is of different color. You can only color one segment in one operation. Only one segment can be colored at a time and you can only use the given list of colors to color the rope. You have to find the minimum time required to color the whole rope.

Example:

Input: rope = "RRGGBBYY" colors = ["R", "G", "B", "Y"]

Output: 5

Explanation:

We can color the rope in the following times:

• RGBGBYRR (coloring first 3 characters takes 3 units of time)
• RGBGYBRR (coloring 4th and 5th characters takes 2 units of time)
• RGBGYBRY (coloring last 2 characters takes 2 units of time)

Total time = 3 + 2 + 2 = 7.

Therefore, the minimum time required is 5.

Solution: The given problem can be solved using dynamic programming techniques. We can define DP state as dp[i][j], where i denotes the index of the last colored segment and j denotes the last color used.

We can initialize the DP state dp[0][j] = 0 for all colors j in the given list of colors as at the beginning, there are no colored segments.

Now, we can define the recurrence relation as follows:

dp[i][j] = min(dp[i - 1][k] + (j != k)) for all colors k in the given list of colors.

Here, (j != k) denotes that it will take one unit of time to color the segment with the new color k if it's different from the last used color j, otherwise, there will be no time required.

Finally, we can return the minimum time required to color the last segment dp[n][j] for all colors j in the given list of colors, where n is the length of the rope.

The overall time complexity of this solution is O(n x k^2), where n is the length of the rope and k is the number of colors in the given list.

Code:

Here's the Python code for the above solution:

def minTimeToColorRope(rope: str, colors: List[str]) -> int: n = len(rope) k = len(colors) dp = [[float('inf')] * k for _ in range(n + 1)] for j in range(k): dp[0][j] = 0 for i in range(1, n + 1): for j in range(k): for l in range(k): dp[i][j] = min(dp[i][j], dp[i - 1][l] + (colors[j] != rope[i - 1]) * (j != l)) return min(dp[n])

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