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Consider the following code that generates all permutations of 3 elements using bitmask DP with iterative tabulation. What is the content of dp[7] after the loops finish?

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Bit Manipulation - Generate All Permutations Using Bitmask DP
Consider the following code that generates all permutations of 3 elements using bitmask DP with iterative tabulation. What is the content of dp[7] after the loops finish?

def generate_permutations_iterative(n):
    dp = [[] for _ in range(1 << n)]
    dp[0] = [[]]
    for mask in range(1 << n):
        for i in range(n):
            if not (mask & (1 << i)):
                for perm in dp[mask]:
                    dp[mask | (1 << i)].append(perm + [i])
    return dp[(1 << n) - 1]

perms = generate_permutations_iterative(3)
print(perms)
A[[0,1,2], [0,2,1], [1,0,2], [1,2,0], [2,0,1], [2,1,0]]
B[[0,1,2], [1,0,2], [2,0,1]]
C[[0,1,2], [0,2,1], [1,0,2], [2,1,0]]
D[[0,1,2], [1,2,0], [2,1,0], [2,0,1]]
Step-by-Step Solution
  1. Step 1: Trace dp[0] initialization

    dp[0] starts with [[]], representing empty permutation.
  2. Step 2: Iteratively build dp for all masks

    For each mask, append unused elements to existing permutations, eventually filling dp[7] with all permutations of [0,1,2].
  3. Final Answer:

    Option A -> Option A
  4. Quick Check:

    dp[7] contains all 3! = 6 permutations of [0,1,2] [OK]
Quick Trick: dp[(1<
Common Mistakes:
MISTAKES
  • Confusing mask indexing leading to missing permutations
  • Off-by-one errors in bitmask checks
Trap Explanation:
PITFALL
  • Partial dp states may look correct but only dp[(1<
Interviewer Note:
CONTEXT
  • Tests candidate's ability to mentally execute bitmask DP code and understand state buildup.
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