70. Climbing Stairs

What will be learned

• A basic dynamic programming problem that models how to build solutions incrementally.

Solutions

function climb(start: number, end: number) {
  if(start === end) return 1;
  if(start > end) return 0;
 
  return climb(start + 1, end) + climb(start + 2, end);
}
 
function climbStairs(n: number): number {
  return climb(0, n);
};

Video Solution

...this is gonna be another dynamic programming problem, and this is definitely more on the easy side, definitely compare to some of the really hard dynamic programming problem that we've been solving recently, but I still think this is a good problem to really understand the fundamental, understand the brute force, and taking that brute force using memoization, and that of cource getting the true dynamic programming solution...