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)```
```

Array-DS

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Dynamic Array

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Left Rotation

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Array Manipulation

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