House Robber

At each house, you have exactly two choices: rob it and collect its loot plus the best total from two houses back, or skip it and keep whatever was best up to the previous house. You only need to remember the best results from the previous two steps to make the next call.

Full solution

Constant-space rolling variables (the notes' recommended approach): only the best loot from one and two houses back is ever needed, so the O(n) dp array collapses to two scalars.

def rob(nums: list[int]) -> int:
    # Rolling registers: best loot using houses up to i-2 and up to i-1.
    prev2, prev1 = 0, 0
    for n in nums:
        prev2, prev1 = prev1, max(prev1, prev2 + n)
    return prev1

function rob(nums) {
  let prev2 = 0, prev1 = 0;
  for (const n of nums) {
    [prev2, prev1] = [prev1, Math.max(prev1, prev2 + n)];
  }
  return prev1;
}

The theory behind it

Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems

What Dynamic Programming is

Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.

When to reach for it

Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.

How the pattern works

Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.