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Problem solution · Python

Minimum Possible Maximum Waiting Time

Minimum Possible Maximum Waiting Time: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
144 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Minimum Possible Maximum Waiting Time, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 144 lines of Python from the credited upstream file minimum-possible-maximum-waiting-time.py.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Possible Maximum Waiting Time · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n * rlogr * (f / 64)) = O(n * rlogr), r = max(demand), f = fuel[0]# Space: O(r * (f / 64)) = O(r) # binary search, dp, bitmasksclass Solution(object):    def minMaxWaitingTime(self, demand, fuel):        """        :type demand: List[int]        :type fuel: List[int]        :rtype: int        """        FULLMASK = ((1<<64)-1)        def binary_search(left, right, check):            while left <= right:                mid = left+(right-left)//2                if check(mid):                    right = mid-1                else:                    left = mid+1            return left         def low(x):            return FULLMASK if x >= 63 else ((1<<(x+1))-1) if x >= 0 else 0         def high(x):            return FULLMASK^low(x-1)         def find_max_served():            mask = 1            total = 0            for i, x in enumerate(demand):                mask = ((mask<<x)&low(fuel[0]))|(mask&high(total+x-fuel[1]))                if not mask:                    return i                total += x            return len(demand)         def check(w):            def update(last, gap, mask):                if last == 0:                    mask = (mask<<demand[i])&low(fuel[0])                else:                    mask &= high(total+demand[i]-fuel[1])                new_dp[last][gap] |= mask             # dp[last][gap] = bit used0 is set if the state (last, gap, used0) is reachable.            # - last: which dispenser serving the previous car            # - gap: remaining busy time of the other dispenser            # - used0: total fuel consumed by dispenser 0            dp = [[0]*(mx+1) for _ in xrange(2)]            dp[0][0] = 1            total = 0            for i in xrange(l):                new_dp = [[0]*(mx+1) for _ in xrange(2)]                for last in xrange(len(dp)):                    for gap in xrange(len(dp[0])):                        if not dp[last][gap]:                            continue                        if (demand[i-1] if i-1 >= 0 else 0) <= w:                            update(last, max(gap-(demand[i-1] if i-1 >= 0 else 0), 0), dp[last][gap])                        if gap <= w:                            update(last^1, max((demand[i-1] if i-1 >= 0 else 0)-gap, 0), dp[last][gap])                dp = new_dp                total += demand[i]            return any(x for row in dp for x in row)         l = find_max_served()        mx = max(demand)        return binary_search(0, mx, check) if l else -1  # Time:  O(n * rlogr * f),  r = max(demand), f = fuel[0]# Space: O(r * f)# binary search, dpclass Solution2(object):    def minMaxWaitingTime(self, demand, fuel):        """        :type demand: List[int]        :type fuel: List[int]        :rtype: int        """        def binary_search(left, right, check):            while left <= right:                mid = left+(right-left)//2                if check(mid):                    right = mid-1                else:                    left = mid+1            return left         def find_max_served():            dp = [False]*(fuel[0]+1)  # dp[used0] = whether the current prefix can be served with state (used0).            dp[0] = True            total = 0            for i, x in enumerate(demand):                new_dp = [False]*(fuel[0]+1)                for used0 in xrange(len(dp)):                    if not dp[used0]:                        continue                    if used0+x <= fuel[0]:                        new_dp[used0+x] = True                    if (total-used0)+x <= fuel[1]:                        new_dp[used0] = True                if not any(new_dp):                    return i                dp = new_dp                total += x            return len(demand)         def check(w):            def update(last, gap, used0):                if last == 0:                    if used0+demand[i] <= fuel[0]:                        new_dp[last][gap][used0+demand[i]] = True                else:                    if (total-used0)+demand[i] <= fuel[1]:                        new_dp[last][gap][used0] = True             # dp[last][gap][used0] = whether the current prefix can be served with state (last, gap, used0).            # - last: which dispenser serving the previous car            # - gap: remaining busy time of the other dispenser            # - used0: total fuel consumed by dispenser 0            dp = [[[False]*(fuel[0]+1) for _ in xrange(mx+1)] for _ in xrange(2)]            dp[0][0][0] = True            total = 0            for i in xrange(l):                new_dp = [[[False]*(fuel[0]+1) for _ in xrange(mx+1)] for _ in xrange(2)]                for last in xrange(len(dp)):                    for gap in xrange(len(dp[0])):                        for used0 in xrange(len(dp[0][0])):                            if not dp[last][gap][used0]:                                continue                            if (demand[i-1] if i-1 >= 0 else 0) <= w:                                update(last, max(gap-(demand[i-1] if i-1 >= 0 else 0), 0), used0)                            if gap <= w:                                update(last^1, max((demand[i-1] if i-1 >= 0 else 0)-gap, 0), used0)                dp = new_dp                total += demand[i]            return any(x for matrix in dp for row in matrix for x in row)         l = find_max_served()        mx = max(demand)        return binary_search(0, mx, check) if l else -1

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