Leetcode Climbing Stairs problem solution

// Climbing stairs: Distinct ways can you climb to the top

int climbStairs(int n){
    // Base cases n = 0, 1 or 2, return n
    if((n == 0) || (n == 1) || (n == 2))
        return n;
    
    int *result = (int *)calloc(n, sizeof(int));
    int i = 0;
    
    result[i++] = 1; // For the stair 1, it can be climbed in only one way
    result[i++] = 2; // For the stair 2, it can be climed 1-1 or 2, so 2 ways
    
    // Calculate for the n steps by calculating till n-1 steps and n-2 steps
    for(i = 2; i < n; i++) {
        result[i] = result[i-1] + result[i-2];
    }
    
    return result[n - 1];
}

class Solution {
public:
    int helper(int i, int n) {
        if(i == n)
            return 1;
        if(i > n)
            return 0;
        
        return helper(i + 1, n) + helper(i + 2, n); 
    }
    int climbStairs(int n) {
        return helper(0, n);
    }
};

class Solution {
    public int climbStairs(int n) {
    if(n==1)
        return 1;
    else if(n==2)
        return 2;
    else
    {
        int a=1,b=2,c=0,i;
        for(i=3;i<=n;i++)
        {
            c=a+b;
            a=b;
            b=c;
        }
        return c;
    }
}
}

class Solution:
    def climbStairs(self, n):
        def dynprog(m, memo={}):
            if m in memo:
                return memo[m]
            if m <= 2:
                return m
            memo[m] = dynprog(m-1, memo) + dynprog(m-2, memo)
            return memo[m]
        res = dynprog(n, {})
        return res

public class Solution {
    Dictionary<int, int> map = new Dictionary<int, int>();
    public int ClimbStairs(int n) {
        if(n <= 2) return n;
        if(map.ContainsKey(n))
            return map[n];
        int cal = ClimbStairs(n - 2) + ClimbStairs(n-1);
        map.Add(n, cal);
        return cal;
    }
}