Minimum Time to Remove All Cars Containing Illegal Goods

Hard

#1974Time: O(n^2) due to the nested loops iterating through all possible start and end points of the middle segment. The precomputation of the prefix sum array takes O(n).Space: O(n) to store the prefix sum array for counting ones.

Patterns

Dynamic Programming

Data structures

String

Prompt

You are given a 0-indexed binary string s which represents a sequence of train cars. s[i] = '0' denotes that the i^th car does not contain illegal goods and s[i] = '1' denotes that the i^th car does contain illegal goods.

As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations any number of times:

Remove a train car from the left end (i.e., remove s[0]) which takes 1 unit of time.
Remove a train car from the right end (i.e., remove s[s.length - 1]) which takes 1 unit of time.
Remove a train car from anywhere in the sequence which takes 2 units of time.

Return the minimum time to remove all the cars containing illegal goods.

Note that an empty sequence of cars is considered to have no cars containing illegal goods.

Example 1:

Input: s = "1100101"
Output: 5
Explanation: 
One way to remove all the cars containing illegal goods from the sequence is to
- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
- remove a car from the right end. Time taken is 1.
- remove the car containing illegal goods found in the middle. Time taken is 2.
This obtains a total time of 2 + 1 + 2 = 5. 

An alternative way is to
- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.
- remove a car from the right end 3 times. Time taken is 3 * 1 = 3.
This also obtains a total time of 2 + 3 = 5.

5 is the minimum time taken to remove all the cars containing illegal goods. 
There are no other ways to remove them with less time.

Example 2:

Input: s = "0010"
Output: 2
Explanation:
One way to remove all the cars containing illegal goods from the sequence is to
- remove a car from the left end 3 times. Time taken is 3 * 1 = 3.
This obtains a total time of 3.

Another way to remove all the cars containing illegal goods from the sequence is to
- remove the car containing illegal goods found in the middle. Time taken is 2.
This obtains a total time of 2.

Another way to remove all the cars containing illegal goods from the sequence is to 
- remove a car from the right end 2 times. Time taken is 2 * 1 = 2. 
This obtains a total time of 2.

2 is the minimum time taken to remove all the cars containing illegal goods. 
There are no other ways to remove them with less time.

Constraints:

1 <= s.length <= 2 * 10⁵
s[i] is either '0' or '1'.

Approaches

2 approaches with complexity analysis and trade-offs.

This approach considers all possible ways to partition the removal operations. We can imagine that we remove a prefix, a suffix, and then handle the remaining middle part by removing each car with illegal goods individually. The total cost is the sum of costs for these three parts. We can iterate through all possible middle segments s[i...j], calculate the cost for each, and find the minimum.

Algorithm

To make the calculation of countOnes(s[i...j]) efficient, first precompute a prefix sum array, prefixOnes, where prefixOnes[k] stores the total number of '1's in the prefix s[0...k-1].
Initialize a variable minCost to a very large value.
Iterate through all possible start indices i of the middle segment, from 0 to n.
For each i, iterate through all possible end indices j of the middle segment, from i-1 to n-1. The case j = i-1 represents an empty middle segment.
Inside the loops, calculate the cost for the current partition (i, j):
- Cost of removing the prefix s[0...i-1] is i.
- Cost of removing the suffix s[j+1...n-1] is n - (j+1).
- Cost of removing '1's from the middle s[i...j] is 2 * (prefixOnes[j+1] - prefixOnes[i]).
- The total cost is i + (n - j - 1) + 2 * onesInMiddle.
Update minCost with the minimum cost found so far.
After checking all (i, j) pairs, minCost will hold the result.

Walkthrough

The strategy is to define a segment s[i...j] of the train. All cars to the left of this segment, i.e., the prefix s[0...i-1], are removed from the left end, costing i units of time. All cars to the right of this segment, i.e., the suffix s[j+1...n-1], are removed from the right end, costing n - (j+1) units of time. All cars with illegal goods ('1's) within the segment s[i...j] are removed individually from the middle, costing 2 for each such car. The total cost for a chosen i and j is i + (n - j - 1) + 2 * countOnes(s[i...j]). We can iterate through all possible values of i (from 0 to n) and j (from i-1 to n-1) to find the minimum possible total cost. To efficiently calculate countOnes(s[i...j]), we can precompute a prefix sum array that stores the cumulative count of '1's.

1class Solution {2    public int minimumTime(String s) {3        int n = s.length();4        if (n == 0) {5            return 0;6        }7 8        int[] prefixOnes = new int[n + 1];9        for (int i = 0; i < n; i++) {10            prefixOnes[i + 1] = prefixOnes[i] + (s.charAt(i) - '0');11        }12 13        int minCost = Integer.MAX_VALUE;14 15        // i: start index of the middle segment16        // j: end index of the middle segment17        for (int i = 0; i <= n; i++) {18            for (int j = i - 1; j < n; j++) {19                int onesInMiddle = (j < i) ? 0 : prefixOnes[j + 1] - prefixOnes[i];20                // Cost = (remove prefix s[0..i-1]) + (remove suffix s[j+1..n-1]) + (remove middle '1's)21                int cost = i + (n - 1 - j) + 2 * onesInMiddle;22                minCost = Math.min(minCost, cost);23            }24        }25        return minCost;26    }27}

Complexity

Time

O(n^2) due to the nested loops iterating through all possible start and end points of the middle segment. The precomputation of the prefix sum array takes O(n).

Space

O(n) to store the prefix sum array for counting ones.

Trade-offs

Pros

The logic is relatively straightforward to understand as it directly models the partitioning of the string.
It correctly covers all possible scenarios of removals.

Cons

The time complexity of O(n^2) is too slow for the given constraints on the input size s.length, which can be up to 2 * 10^5. This will result in a 'Time Limit Exceeded' error on most platforms.

This approach breaks the problem down by considering each position i as a potential split point. We calculate the minimum cost to clear all '1's to the left of i and the minimum cost to clear all '1's to the right of i (inclusive). The sum of these two costs gives a possible total minimum time. By checking every possible split point, we can find the global minimum.

Algorithm

Create an array left_costs of size n+1. left_costs[i] will store the minimum cost to remove all '1's from the prefix s[0...i-1].
Initialize left_costs[0] = 0.
Iterate i from 1 to n to fill left_costs. If s[i-1] == '0', left_costs[i] = left_costs[i-1]. If s[i-1] == '1', left_costs[i] = min(left_costs[i-1] + 2, i). The two options correspond to removing s[i-1] from the middle or removing the entire prefix s[0...i-1].
Create an array right_costs of size n+1. right_costs[i] will store the minimum cost to remove all '1's from the suffix s[i...n-1].
Initialize right_costs[n] = 0.
Iterate i from n-1 down to 0 to fill right_costs. If s[i] == '0', right_costs[i] = right_costs[i+1]. If s[i] == '1', right_costs[i] = min(right_costs[i+1] + 2, n-i).
Initialize minCost to a large value.
The final answer is the minimum of left_costs[i] + right_costs[i] over all possible split points i from 0 to n.
Iterate i from 0 to n, updating minCost = min(minCost, left_costs[i] + right_costs[i]).
Return minCost.

Walkthrough

We can solve this problem efficiently using dynamic programming by splitting the problem at each possible index i. Let left_costs[i] be the minimum cost to remove all '1's in the prefix s[0...i-1]. This can be calculated by iterating from left to right. For each position k < i, if s[k] is a '1', we can either remove the whole prefix s[0...k] (cost k+1) or remove s[k] individually (cost 2) in addition to the cost of clearing s[0...k-1]. This gives a simple recurrence. Similarly, let right_costs[i] be the minimum cost to remove all '1's in the suffix s[i...n-1], calculated by iterating from right to left.

The key insight is that for any split point i, the operations to clear the left part s[0...i-1] and the right part s[i...n-1] are independent. A left-end removal doesn't affect the right part, a right-end removal doesn't affect the left part, and middle removals are local. Therefore, the total minimum cost is the minimum of left_costs[i] + right_costs[i] over all possible split points i.

1class Solution {2    public int minimumTime(String s) {3        int n = s.length();4        if (n == 0) {5            return 0;6        }7 8        // left_costs[i] = min cost to clear '1's in s[0...i-1]9        int[] left_costs = new int[n + 1];10        left_costs[0] = 0;11        for (int i = 1; i <= n; i++) {12            if (s.charAt(i - 1) == '0') {13                left_costs[i] = left_costs[i - 1];14            } else {15                left_costs[i] = Math.min(left_costs[i - 1] + 2, i);16            }17        }18 19        // right_costs[i] = min cost to clear '1's in s[i...n-1]20        int[] right_costs = new int[n + 1];21        right_costs[n] = 0;22        for (int i = n - 1; i >= 0; i--) {23            if (s.charAt(i) == '0') {24                right_costs[i] = right_costs[i + 1];25            } else {26                right_costs[i] = Math.min(right_costs[i + 1] + 2, n - i);27            }28        }29 30        int minCost = Integer.MAX_VALUE;31        // Combine costs at each split point i32        for (int i = 0; i <= n; i++) {33            minCost = Math.min(minCost, left_costs[i] + right_costs[i]);34        }35 36        return minCost;37    }38}

Complexity

Time

O(n), as it involves three separate linear passes: one to compute `left_costs`, one for `right_costs`, and one to combine them.

Space

O(n) to store the `left_costs` and `right_costs` arrays.

Trade-offs

Pros

Highly efficient with a linear time complexity, making it suitable for large inputs.
The DP state transitions are simple and easy to implement once the subproblem structure is identified.

Cons

Requires O(n) extra space for the DP arrays, which might be a consideration for extremely large n under strict memory limits.
The logic, while efficient, might be less intuitive to come up with compared to a direct brute-force approach.

Solutions

Solution

1class Solution {2public3  int minimumTime(String s) {4    int n = s.length();5    int[] pre = new int[n + 1];6    int[] suf = new int[n + 1];7    for (int i = 0; i < n; ++i) {8      pre[i + 1] = s.charAt(i) == '0' ? pre[i] : Math.min(pre[i] + 2, i + 1);9    }10    for (int i = n - 1; i >= 0; --i) {11      suf[i] =12          s.charAt(i) == '0' ? suf[i + 1] : Math.min(suf[i + 1] + 2, n - i);13    }14    int ans = Integer.MAX_VALUE;15    for (int i = 1; i <= n; ++i) {16      ans = Math.min(ans, pre[i] + suf[i]);17    }18    return ans;19  }20}

1class Solution:2    def minimumTime(self, s: str) -> int: n = len(s) pre = [0] * (n + 1) suf = [0] * (n + 1) for i, c in enumerate(s): pre[i + 1] = pre[i] if c == '0' else min(pre[i] + 2, i + 1) for i in range(n - 1, - 1, - 1): suf[i] = suf[i + 1] if s[i] == '0' else min(suf[i + 1] + 2, n - i) return min(a + b for a, b in zip(pre[1:], suf[1:]))

1class Solution {2public:3  int minimumTime(string s) {4    int n = s.size();5    vector<int> pre(n + 1);6    vector<int> suf(n + 1);7    for (int i = 0; i < n; ++i)8      pre[i + 1] = s[i] == '0' ? pre[i] : min(pre[i] + 2, i + 1);9    for (int i = n - 1; ~i; --i)10      suf[i] = s[i] == '0' ? suf[i + 1] : min(suf[i + 1] + 2, n - i);11    int ans = INT_MAX;12    for (int i = 1; i <= n; ++i)13      ans = min(ans, pre[i] + suf[i]);14    return ans;15  }16};

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