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1578. Minimum Time to Make Rope Colorful
Description
Alice has n
balloons arranged on a rope. You are given a 0indexed string colors
where colors[i]
is the color of the i^{th}
balloon.
Alice wants the rope to be colorful. She does not want two consecutive balloons to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it colorful. You are given a 0indexed integer array neededTime
where neededTime[i]
is the time (in seconds) that Bob needs to remove the i^{th}
balloon from the rope.
Return the minimum time Bob needs to make the rope colorful.
Example 1:
Input: colors = "abaac", neededTime = [1,2,3,4,5] Output: 3 Explanation: In the above image, 'a' is blue, 'b' is red, and 'c' is green. Bob can remove the blue balloon at index 2. This takes 3 seconds. There are no longer two consecutive balloons of the same color. Total time = 3.
Example 2:
Input: colors = "abc", neededTime = [1,2,3] Output: 0 Explanation: The rope is already colorful. Bob does not need to remove any balloons from the rope.
Example 3:
Input: colors = "aabaa", neededTime = [1,2,3,4,1] Output: 2 Explanation: Bob will remove the balloons at indices 0 and 4. Each balloons takes 1 second to remove. There are no longer two consecutive balloons of the same color. Total time = 1 + 1 = 2.
Constraints:
n == colors.length == neededTime.length
1 <= n <= 10^{5}
1 <= neededTime[i] <= 10^{4}
colors
contains only lowercase English letters.
Solutions

class Solution { public int minCost(String colors, int[] neededTime) { int ans = 0; int n = neededTime.length; for (int i = 0, j = 0; i < n; i = j) { j = i; int s = 0, mx = 0; while (j < n && colors.charAt(j) == colors.charAt(i)) { s += neededTime[j]; mx = Math.max(mx, neededTime[j]); ++j; } if (j  i > 1) { ans += s  mx; } } return ans; } }

class Solution { public: int minCost(string colors, vector<int>& neededTime) { int ans = 0; int n = colors.size(); for (int i = 0, j = 0; i < n; i = j) { j = i; int s = 0, mx = 0; while (j < n && colors[j] == colors[i]) { s += neededTime[j]; mx = max(mx, neededTime[j]); ++j; } if (j  i > 1) { ans += s  mx; } } return ans; } };

class Solution: def minCost(self, colors: str, neededTime: List[int]) > int: ans = i = 0 n = len(colors) while i < n: j = i s = mx = 0 while j < n and colors[j] == colors[i]: s += neededTime[j] if mx < neededTime[j]: mx = neededTime[j] j += 1 if j  i > 1: ans += s  mx i = j return ans

func minCost(colors string, neededTime []int) (ans int) { n := len(colors) for i, j := 0, 0; i < n; i = j { j = i s, mx := 0, 0 for j < n && colors[j] == colors[i] { s += neededTime[j] if mx < neededTime[j] { mx = neededTime[j] } j++ } if ji > 1 { ans += s  mx } } return }