Climbing Stairs

Arriving at step n requires a final leap from either step n - 1 with a single stride or step n - 2 with a double stride. Because those two previous landings are mutually exclusive, the total paths to step n is the sum of ways to reach those two predecessors, reducing the entire climb to a Fibonacci recurrence.

Full solution

Two rolling variables (the 'two variable progression' approach): each step's answer is the sum of the previous two, so there is no need to keep a full dp array.

def climb_stairs(n: int) -> int:
    # Two rolling registers: ways to reach the previous two steps.
    prev2, prev1 = 1, 1  # ways to reach step 0, step 1
    for _ in range(2, n + 1):
        prev2, prev1 = prev1, prev2 + prev1
    return prev1

function climbStairs(n) {
  let prev2 = 1, prev1 = 1;
  for (let i = 2; i <= n; i++) {
    [prev2, prev1] = [prev1, prev2 + prev1];
  }
  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.