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Climbing_Stairs.cpp
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Climbing_Stairs.cpp
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/*
Problem Description :
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?
Algorithm :
Solving Using Dynamic Programming Approach.
Sub-problem Formula of each trail : steps[i] = steps[i - 1] + steps[i - 2]
Extra Storage Size : No extra storage Required.
*/
#include <iostream>
using namespace std;
long long ClimbStairs(int N)
{
// Base Case : at the first stair (steps[0]) the number of ways to get there is 1
// Base Case : at the second stair (steps[1]) the number of ways to get there is 2
if(N == 1)
return 1;
long long s1 = 1;
long long s2 = 2;
// Calculate the steps at the (i - 1)th stair according to this formula :
// steps[i] = steps[i - 1] + steps[i - 2]
// starting from the third stair, calculate the number of ways to reach stair i according to the previous formula
for(int i = 2; i < N; i++)
{
// calculate s2 and update s1 to be equal the previous s2
long long tmp = s2;
s2 = s1 + s2;
s1 = tmp;
}
//Return the steps at the final stair.
return s2;
}
int main()
{
int n;
cout << "Enter the number of stairs to calculate the number of ways to get to it's top :\t";
cin >> n;
if(n > 0)
cout << "The number of distinct ways can you climb to the top :\t" << ClimbStairs(n) << endl;
else
cout << "The number of stairs must be greater than 0." <<endl;
return 0;
}
/*
Test Cases
1. Input : 5
Output : 8
2. Input : 1
Output : 1
3. Input : 10
Output : 89
4. Input : 1000
Output : 9079565065540428013
Time Complexity
Theta(N), Where N is the input number (the number of stairs).
Space Complexity
O(1), Where N is the input number (the number of stairs).
*/