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Problem solution · Python

Minimum Time to Break Locks I

Minimum Time to Break Locks I: 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
84 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 Time to Break Locks I, 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

  • 84 lines of Python from the credited upstream file minimum-time-to-break-locks-i.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 Time to Break Locks I · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n^3)# Space: O(n^2) # hungarian algorithm, weighted bipartite matchingclass Solution(object):    def findMinimumTime(self, strength, K):        """        :type strength: List[int]        :type K: int        :rtype: int        """        # Template translated from:        # https://github.com/kth-competitive-programming/kactl/blob/main/content/graph/WeightedMatching.h        def hungarian(a):  # Time: O(n^2 * m), Space: O(n + m)            if not a:                return 0, []            n, m = len(a)+1, len(a[0])+1            u, v, p, ans = [0]*n, [0]*m, [0]*m, [0]*(n-1)            for i in xrange(1, n):                p[0] = i                j0 = 0  # add "dummy" worker 0                dist, pre = [float("inf")]*m, [-1]*m                done = [False]*(m+1)                while True:  # dijkstra                    done[j0] = True                    i0, j1, delta = p[j0], None, float("inf")                    for j in xrange(1, m):                        if done[j]:                            continue                        cur = a[i0-1][j-1]-u[i0]-v[j]                        if cur < dist[j]:                            dist[j], pre[j] = cur, j0                        if dist[j] < delta:                            delta, j1 = dist[j], j                    for j in xrange(m):                        if done[j]:                            u[p[j]] += delta                            v[j] -= delta                        else:                            dist[j] -= delta                    j0 = j1                    if not p[j0]:                        break                while j0:  # update alternating path                    j1 = pre[j0]                    p[j0], j0 = p[j1], j1            for j in xrange(1, m):                if p[j]:                    ans[p[j]-1] = j-1            return -v[0], ans  # min cost         def ceil_divide(a, b):            return (a+b-1)//b            adj = [[ceil_divide(strength[i], 1+j*K) for j in xrange(len(strength))] for i in xrange(len(strength))]        return hungarian(adj)[0]  # Time:  O(n ^ 2^n)# Space: O(2^n)# bitmasks, dpclass Solution2(object):    def findMinimumTime(self, strength, K):        """        :type strength: List[int]        :type K: int        :rtype: int        """        def ceil_divide(a, b):            return (a+b-1)//b            def popcount(x):            return bin(x).count('1')            dp = [float('inf')]*(1<<len(strength))        dp[0] = 0        for mask in xrange(1, len(dp)):            x = 1+(popcount(mask)-1)*K            for i in xrange(len(strength)):                if not (mask&(1<<i)):                    continue                dp[mask] = min(dp[mask], dp[mask^(1<<i)]+ceil_divide(strength[i], x))        return dp[-1]

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