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1578 - Minimum Time to Make Rope Colorful (Medium)

https://leetcode.com/problems/minimum-time-to-make-rope-colorful/

Problem Statement​

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:

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 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.

Approach 1: Greedy​

Written by @wingkwong
// Time Complexity: O(n)
// Space Complexity: O(1)
class Solution {
public:
    // intuition: 
    // ---------------
    // if there are consecutive balloons with the same color, 
    // we want to keep the one with maximum neededTime and remove others 
    
    // if there are 2 balloons with different colors, ans = 0 as it is colorful
    // if there are 2 balloons with same color, ans = min(neededTime[0], neededTime[1])
    // if there are 3 consecutive balloons with same color, ans = sum(neededTime[0 .. 2]) - max(neededTime[0 .. 2])
    // if there are N consecutive balloons with same color, ans = sum(neededTime[0 .. n - 1]) - max(neededTime[0 .. n - 1])
    // we don't need to calculate the sum and subtract the max though. instead, we can either
    // 1. update neededTime in place to the max of neededTime[i] and neededTime[i - 1] (shown in below solution) or 
    // 2. store the current max time in a variable
    // why? let's say colors = "aaa" and neededTime = [1,2,1]
    // we first compare neededTime[0] and neededTime[1] and decide to remove the first balloon (neededTime[0] < neededTime[1])
    // now colors = "_aa" and neededTime = [_,2,1] and the current max time is 2
    // then compare neededTime[1] and neededTime[2] and decide to remove the last balloon, (neededTime[2] < neededTime[1])
    // now colors = "_a_" and neededTime = [_,2,_]. we remove all balloons but the one with maximum neededTime
    int minCost(string colors, vector<int>& neededTime) {
        int ans = 0, n = colors.size();
        for (int i = 1; i < n; i++) {
            // if the i-th balloon has the same color as (i - 1)th one
            // e.g. aba[a]c and i = 3 (0-based)
            if (colors[i] == colors[i - 1]) {
                // then we remove the one with less time
                // e.g. in above example, we remove the balloon at index 2 
                // with neededTime[2] since neededTime[2] < neededTime[3] 
                ans += min(neededTime[i], neededTime[i - 1]);
                // update the max neededTime inplace 
                // or alternatively you can store it in a variable
                neededTime[i] = max(neededTime[i], neededTime[i - 1]);
            }
        }
        return ans;
    }
};