2022-10-03 Daily-Challenge

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Posted modified:
In: DailyChallenge

Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 3

Description

Minimum Time to Make Rope Colorful

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the ith 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 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the ith balloon from the rope.

Return the minimum time Bob needs to make the rope colorful.

Example 1:

img

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:

img

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:

img

Input: colors = "aabaa", neededTime = [1,2,3,4,1]
Output: 2
Explanation: Bob will remove the ballons at indices 0 and 4. Each ballon 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.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int minCost(string colors, vector<int>& neededTime) {
    char current = -1;
    int curSum = 0;
    int curMax = 0;
    int answer = 0;
    for(int i = 0; i < colors.length(); ++i) {
      if(colors[i] != current) {
        answer += curSum - curMax;
        current = colors[i];
        curSum = neededTime[i];
        curMax = neededTime[i];
      } else {
        curSum += neededTime[i];
        curMax = max(curMax, neededTime[i]);
      }
    }
    answer += curSum - curMax;
    return answer;
  }
};

// Accepted
// 113/113 cases passed (214 ms)
// Your runtime beats 71.93 % of cpp submissions
// Your memory usage beats 18.68 % of cpp submissions (95.8 MB)
Algorithm