Minimize The Total Price Of The Trips - Leetcode Solution

Problem:

You want to take n trips, the i-th trip costs ai dollars, and each trip has some benefit, it will increase your happiness by bi units. You can take these trips in any order. When you take a trip i, you also must take all the trips j where j is chosen such that i and j share a common factor greater than 1 (i.e. there must be no trip k where i and j are chosen but k is not). Return the maximum total benefit you can get.

Example:

Input: a = [2,3,5,7,11,13,17,19], b = [2,3,5,7,11,13,17,19] Output: 108 Explanation: We can take the first 7 trips together as is the limit of the biggest factor that can be shared. So, the total price is 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58 and the total benefit is 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58. Next, we take the trip of cost 19 that has a benefit of 19, which will be the last. The total cost will be 58 + 19 = 77 and the total benefit will be 58 + 19 = 77 + 31 = 108.

Solution:

This problem can be solved with the help of dynamic programming. We can create a list dp with size n and initialize it with all zeros. This list will contain the maximum benefit that can be obtained by taking trips starting from the i-th trip. To solve this problem, we need to fill this list dp in reverse order as we can only choose trips that share a common factor greater than 1 with each other.

We will iterate from n-1 to 0, and for each i, we will iterate from i*2 to n in steps of i. If j is the j-th trip, then we can add the benefit of j to the current benefit of the i-th trip and update the corresponding dp[j] with this value. We will track the maximum value of dp[j] using a variable max_ben and return this value after the loop. The time complexity of this solution is O(n^2) because of the nested loop, which may not be suitable for large values of n.

Code:

Here is the Python code for the Minimize The Total Price Of The Trips problem on leetcode:

def maxHappyGroups(batchSize, groups): cnt = [0] * batchSize for g in groups: cnt[g % batchSize] += 1

memo = {}
def dfs(remain, cnt):
    key = tuple(cnt)
    if key in memo:
        return memo[key]

    if remain == 0:
        memo[key] = 1
        return 1

    ans = 0
    for i in range(batchSize):
        if cnt[i] > 0:
            cnt[i] -= 1
            ans = max(ans, dfs((remain - i) % batchSize, cnt))
            cnt[i] += 1

    memo[key] = ans + (remain == 0)
    return memo[key]

return cnt[0] + dfs(0, cnt[1:]) - 1

Conclusion:

The Minimize The Total Price Of The Trips problem on leetcode can be solved with dynamic programming. We need to create a list dp, fill it in reverse order, and track the maximum value of dp. We need to iterate through all the trips and add their benefit to the current benefic of the i-th trip. The time complexity of this solution is O(n^2), which may not be suitable for large values of n.

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