Easy
You are climbing a staircase. It takes n
steps to reach the top.
Each time you can either climb 1
or 2
steps. In how many distinct ways can you climb to the top?
Example 1:
Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
Constraints:
1 <= n <= 45
class Solution {
fun climbStairs(n: Int): Int {
if (n < 2) {
return n
}
val cache = IntArray(n)
// creating a cache or DP to store the result
// so that we dont have to iterate multiple times
// for the same values;
// for 0 and 1 the result array i.e cache values would be 1 and 2
// in loop we are just getting ith values i.e 5th step values from
// i-1 and i-2 which are 4th step and 3rd step values.
cache[0] = 1
cache[1] = 2
for (i in 2 until n) {
cache[i] = cache[i - 1] + cache[i - 2]
}
return cache[n - 1]
}
}