**A Flight of Stairs - Amazon Top Interview Questions**

### Problem Statement :

There's a staircase with n steps, and you can climb up either 1 or 2 steps at a time. Given an integer n, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters, so each different order of steps counts as a way. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 Example 1 Input n = 4 Output 5 Explanation There are 5 unique ways: 1, 1, 1, 1 2, 1, 1 1, 2, 1 1, 1, 2 2, 2 Example 2 Input n = 1 Output 1 Explanation There's only one way: climb up 1 step. Example 3 Input n = 3 Output 3 Explanation There are three ways: [1, 1, 1], [2, 1], and [1, 2].

### Solution :

` ````
Solution in C++ :
const int mod = 1e9 + 7;
struct mtx {
long long int a, b, c, d;
mtx(long long int w, long long int x, long long int y, long long int z)
: a(w), b(x), c(y), d(z) {
}
mtx operator*(mtx& other) {
mtx A(a * other.a + b * other.c, a * other.b + b * other.d, c * other.a + d * other.c,
c * other.b + d * other.d);
A.a %= mod;
A.b %= mod;
A.c %= mod;
A.d %= mod;
return A;
}
};
mtx power(mtx A, long long int k) {
mtx B(1, 0, 0, 1);
while (k) {
if (k & 1) {
B = B * A;
}
A = A * A;
k /= 2;
}
return B;
}
int solve(int n) {
mtx A(1, 1, 1, 0);
mtx B = power(A, n);
// cout << B.a << " " << B.b << endl << B.c << " " << B.d << endl;
return B.a;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int n) {
int a = 1;
int b = 1;
int c;
for (int i = 2; i <= n; i++) {
c = a;
a = b;
b = (b + c) % 1000000007;
}
return b;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, n):
MOD = 10 ** 9 + 7
def fib():
a = b = 1
yield a
yield b
while True:
yield (a + b) % MOD
a, b = b, (a + b) % MOD
f = fib()
for _ in range(n):
next(f)
return next(f)
```

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