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Postdef generateBinaryPathCode(n):

6 April 2025
D
Deloitte | COGNIZANT | IBM | CAPGEMINI
def generateBinaryPathCode(n): t = 1 << n l = [] s = set() dp = [0] * n def isValid(prev, curr): d = prev ^ curr if bin(d).count('1') != 1: return False b= d.bit_length() - 1 tb = (curr >> b) & 1 k = 1 if dp[b] % 2 == 0 else 0 return tb == k def backtrack(): if len(l) == t: return bin(l[0] ^ l[-1]).count('1') == 1 prev = l[-1] for i in range(n): b = 1 << i c = prev ^ b if c in s: continue if not isValid(prev, c): continue l.append(c) s.add(c) dp[i] += 1 if backtrack(): return True l.pop() s.remove(c) dp[i] -= 1 return False l.append(0) s.add(0) if backtrack(): return l return [] 7/7 cases passed✅✅ contact : @MLCODER2
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