Power Set of Distinct Values

1. Step 1: Initialize Solution.subsets

   Replace the empty starter with the first real state owned by Solution.subsets. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Pass the Empty Choice Tree case

   Complete the readable core algorithm for one representative Interview case. Record every recursion node as a valid subset and extend only with later values.

3. Step 3: Harden the Three Values boundary

   Repair the reviewed boundary and pass the complete submission contract. Every subset has a unique increasing sequence of input indices; the recursion visits that sequence once and records its path, including the empty sequence at the root.

• The empty subset is part of every power set and must be recorded before choosing anything.
• recursion and backtracking

• LeetCode 78 — Subsets
• Hello Interview — Backtracking Overview

• Complete One Backtracking Frame(opens in a new tab)

  Complete One Backtracking Frame isolates before expanding each choice, path equals the original caller-owned prefix; every emitted candidate contains exactly one additional element. That focused state discipline is required when implementing subsets as a complete Interview Problem.

Append a copy of the current path at every recursion node, then extend it with each value from a monotone start index.

Why it works: Every subset has a unique increasing sequence of input indices; the recursion visits that sequence once and records its path, including the empty sequence at the root.

class Solution:
    def subsets(self, nums):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        result = []
        path = []
        def backtrack(start):
            result.append(path[:])
            for index in range(start, len(nums)):
                path.append(nums[index])
                backtrack(index + 1)
                path.pop()
        backtrack(0)
        return result

• Permutations of Distinct Values(opens in a new tab)
• Unique Subsets With Duplicate Inputs(opens in a new tab)