Easy Diffeculty
Climbing Stairs

Math

Dynamic programming

Memoizarion

easy

## Climbing Stairs

The problem of climbing stairs involves finding the number of distinct ways to climb a staircase with n steps, given that you can climb one or two steps at a time. Here's an example of how to solve this problem using JavaScript:

``````function climbStairs(n) {
if (n === 1) {
return 1;
}
let dp = new Array(n + 1).fill(0);
dp = 1;
dp = 2;
for (let i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}``````

The function climbStairs takes a single argument, an integer `n`, and returns the number of distinct ways to climb a staircase with n steps. The function uses a dynamic programming algorithm to compute the solution. Specifically, it initializes an array dp of length n+1 to store the number of distinct ways to climb the staircase with 1, 2, ..., n steps, and fills in the array in a bottom-up manner by computing the value of each element as the sum of the values of the two previous elements.

The time complexity of this algorithm is O(n), where n is the number of steps in the staircase. This is because the algorithm fills in the dp array with n elements using a single for-loop. The space complexity of the algorithm is O(n), because the algorithm uses an array of size n to store the intermediate solutions. However, it's possible to optimize the space complexity by only keeping track of the two previous elements instead of storing the entire dp array. This would reduce the space complexity to O(1).