CPMagical CP Library

Problem solution · Python

Distinct Gate Paths to Lca

Distinct Gate Paths to Lca: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
80 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

Depth-first search

For Distinct Gate Paths to Lca, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 80 lines of Python from the credited upstream file distinct-gate-paths-to-lca.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeDistinct Gate Paths to Lca · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlog n + qlogn)# Spece: O(nlogn) # dfs, binary liftingclass Solution(object):    def distinctPaths(self, n, parent, gates, queries):        """        :type n: int        :type parent: List[int]        :type gates: List[List[int]]        :type queries: List[List[int]]        :rtype: int        """        MOD = 10**9+7        BB, BR, RB, RR = range(4)        def ceil_log2(x):            return (x-1).bit_length()         def mult(x, y):            return (                (x[BB]*y[BB]+x[BR]*y[RB])%MOD,                (x[BB]*y[BR]+x[BR]*y[RR])%MOD,                (x[RB]*y[BB]+x[RR]*y[RB])%MOD,                (x[RB]*y[BR]+x[RR]*y[RR])%MOD,            )         def lca(a, b):            if depth[a] < depth[b]:                a, b = b, a            d = depth[a]-depth[b]            for k in xrange(len(par)):                if d&(1<<k):                    a = par[k][a]            if a == b:                return a            for k in reversed(xrange(len(par))):                if par[k][a] != par[k][b]:                    a, b = par[k][a], par[k][b]            return par[0][a]         def count(u, card, t):            if u == l:                return 1            d = depth[u]-depth[t]            b, r = 1 if card == 0 else 0, 1 if card == 1 else 0            for k in xrange(len(par)):                if not d&(1<<k):                    continue                b, r = (b*cnt[k][u][BB]+r*cnt[k][u][RB])%MOD, (b*cnt[k][u][BR]+r*cnt[k][u][RR])%MOD                u = par[k][u]            return (b+r)%MOD         adj = [[] for _ in xrange(n)]        for u in xrange(n):            if parent[u] != -1:                adj[parent[u]].append(u)        depth = [0]*n        stk = [0]        while stk:            u = stk.pop()            for v in reversed(adj[u]):                depth[v] = depth[u]+1                stk.append(v)        par = [[0]*n for _ in xrange(ceil_log2(n-1)+1)]        cnt = [[(0,)*4 for _ in xrange(n)] for _ in xrange(ceil_log2(n-1)+1)]        for u in xrange(n):            par[0][u] = parent[u]            cnt[0][u] = (gates[u][1], gates[u][2], gates[u][2], gates[u][0])        for k in xrange(1, len(par)):            for u in xrange(n):                if par[k-1][u] == -1:                    continue                par[k][u] = par[k-1][par[k-1][u]]                cnt[k][u] = mult(cnt[k-1][u], cnt[k-1][par[k-1][u]])        result = 0        for q in queries:            l = lca(q[0], q[2])            result ^= (count(q[0], q[1], l)*count(q[2], q[3], l))%MOD        return result

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗