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  1. Recursion: Davis' Staircase
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Recursion: Davis' Staircase

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411 Discussions

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  • Starfreck
     + 0 comments
    def stepPerms(n):
    memo = [-1] * (n+1)
    res = countSteps(n, memo) % 10000000007
    return res

def countSteps(n, memo):
    if n == 0:
        return 1 # Stay where you are
    if n == 1:
        return 1 # There is only 1 way to reach on top by takine 1 step
    if n == 2: 
        return 2 # There are 2 ways 1+1 or 2
    if n == 3:
        return 4 # There are 4 ways 1+1+1 or 2+1 or 1+2 or 3
    if memo[n] != -1:
        return memo[n]
    memo[n] = countSteps(n-1, memo) + countSteps(n-2, memo) + countSteps(n-3, memo)
    return memo[n]

  • 1337er
     + 0 comments
    memo = {1: 1, 2: 2, 3: 4}

def r(n):
    # Write your code here
    if n <= 0:
        return 0
    if n in memo:
        return memo[n]
    result = r(n-1) + r(n-2) + r(n-3)
    memo[n] = result
    return result
    
def stepPerms(n):
    return r(n) % (10**10 + 7)

  • ygorsansone
     + 0 comments

    Python with dynamic programming

    def stepPerms(n, memo=None):
    if memo is None:
        memo = [None] * (n + 1)
    if n == 0:
        return 1
    if n < 0:
        return 0
    if memo[n] is None:
        memo[n] = stepPerms(n-1, memo) + stepPerms(n-2, memo) + stepPerms(n-3, memo)
    return memo[n]

  • sonali9696
     + 0 comments

    Why is n = 0 (base case), the ans is 1 not 0?

  • gauravgargdsb
     + 0 comments

    Recursion function to avoid TLE :

    int solve(int n, int index, vector &dp){

    if(index == n){
    return 1;
}
if(index > n){
    return 0;
}
if(dp[index]!=-1){
    return dp[index];
}
return dp[index] = solve(n,index+1, dp)+solve(n,index+2, dp)+solve(n,index+3, dp);

    }

    int stepPerms(int n) { vectordp(n+1,-1); int ans = solve(n, 0, dp) % mod;

    return ans;

    }