This is one of the most standard DP problems. It can be easily seen that the person can come to a step i either from step i-1 or step i-2. Define dp[i] as number of ways to end up at i-th stair. Clearly, dp[i] = dp[i-1] + dp[i-2].
int climbStairs(int n) {
int dp[n+1] = {0};
if (n==0)
return 1;
if (n>=1)
dp[1] = 1;
if (n>=2)
dp[2] = 2;
for (int i=3; i<n+1; i++) {
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}
Time complexity: O(n)
Space complexity: O(n)
We can further reduce the time complexity by observing that we only require dp[i-1] and dp[i-2] to determine dp[i]. Also observe that this is a classical Fibonacci series.
public int climbStairs(int n) {
// base cases
if(n <= 0) return 0;
if(n == 1) return 1;
if(n == 2) return 2;
int one_step_before = 2;
int two_steps_before = 1;
int all_ways = 0;
for(int i=2; i<n; i++){
all_ways = one_step_before + two_steps_before;
two_steps_before = one_step_before;
one_step_before = all_ways;
}
return all_ways;
}