Count Number of ways to reach Nth step using steps of lengths 1,2,3,4,......m.(where m<=n)

Question

I'm given a number n of order 10⁵. I've to find number of ways to reach n^th step from ground using steps of length 1 or 2 or 3 or.....or m, here m<=n.

As the answer can be too large output it modulo 10⁹+7.

#include<iostream.h>
using namespace std;

#define ll long long
#define MOD 1000000007

ll countWays_(ll n, ll m){
    ll res[n];
    res[0] = 1; res[1] = 1;
    for (ll i=2; i<n; i++)
    {
       res[i] = 0;
       for (ll j=1; j<=m && j<=i; j++)
         res[i] =(res[i]%MOD+ res[i-j]%MOD)%MOD;
    }
    return res[n-1];
}

ll countWays(ll s, ll m){
    return countWays_(s+1, m);
}
int main (){
    scanf("%lld%lld",&s,&m);
    printf("%lld\n",countWays(s,m));
    return 0;
}

As the complexity O(m*n), I want to reduce it.

Answer 1

Your inner loop adds res[i-1] + res[i-2] + ... + res[i-m] to the result.

Let s be the sum of the first i elements in res. Then you can simply add s[i-1] - s[i-m-1] to the result.

ll countWays_(ll n, ll m){
    ll res[n];
    res[0] = 1; res[1] = 1;
    s[0] = 1; s[1] = 2;
    for (ll i=2; i<n; i++)
    {
       if (i <= m)
           res[i] = s[i-1] % MOD;
       else
           res[i] = (s[i-1] - s[i - m - 1] + MOD) % MOD; 
       s[i] = (s[i-1] + res[i]) % MOD;
    }
    return res[n-1];
}

The new complexity will be O(n). You can even get rid of s as an array and use a single variable with a little more bookkeeping.

Answer 2

I think use can use a variable Sum to store sum of res form i-m+1 to i like this:

ll mod(ll a, ll b){ 
    return (a%b+b)%b; 
}
ll countWays_(ll n, ll m){
    ll res[n],sum;
    res[0] = 1; res[1] = 1;
    sum = res[0] + res[1];
    int head_sum = 0;
    for (ll i=2; i<n; i++)
    {
       if ((i - head_sum) > m) {
            sum=mod((sum- res[head_sum]),MOD);
            head_sum++;
       }  
       res[i] = sum;
       sum = mod((sum% MOD + res[i]% MOD),MOD);
    }
    return res[n-1];
}