Climbing Stairs
Problem Statement :
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
Solution :
Solution in C :
int climbStairs(int n){
if(n <= 2)
return n;
int a = 1;
int b = 2;
for(int i = 3; i <= n; i++)
{
int sum = a + b;
a = b;
b = sum;
}
return b;
}
Solution in C++ :
class Solution {
public:
int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}
return climbStairs(n-1) + climbStairs(n-2);
}
};
Solution in Java :
class Solution {
public int climbStairs(int n) {
if (n == 0 || n == 1) {
return 1;
}
return climbStairs(n-1) + climbStairs(n-2);
}
}
Solution in Python :
class Solution:
def climbStairs(self, n: int) -> int:
if n == 0 or n == 1:
return 1
return self.climbStairs(n-1) + self.climbStairs(n-2)
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