3179. Find the N-th Value After K Seconds | Pascal's Triangle | DP | Bottom Up Optimised DP

In this video, I'll talk about how to solve Leetcode 3179. Find the N-th Value After K Seconds | Pascal's Triangle | DP | Bottom Up Optimised DP
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[@deepakchalla495]

Great Video.
cant we just do the same using for loops
vectorarr(n,1);
while(k--)
{
for(int i=1;i

[@srikarvinjamara2003]

heyy @Aryan Mittal this was the code i had written during contest.
int valueAfterKSeconds(int n, int k) {
vector nums(n,1);
for(int i = 0;i

[@jatinjangra5278]

actually we dont need to take two vectors , only one is enough

[@raviprasath.k.j9462]

actually we don't need a vector if we do it math in O(n) complexity.
class Solution:
def valueAfterKSeconds(self, n: int, k: int) -> int:
ans = 1
mod = 1000000007
for i in range(0,k) :
ans = (ans*(n+i))%mod
ans = (ans*pow(i+1,mod-2,mod))%mod
return ans

[@raviprasath.k.j9462]

I came with this solution in the contest in O(n) time complexity aryan if wonder if you could explain about modular inverse concept
cauz the actual logic i thought is this
class Solution:
def valueAfterKSeconds(self, n: int, k: int) -> int:
ans = 1
mod = 1000000007
for i in range(0,k) :
ans = (ans*(n+i))%mod
ans = (ans//(i+1))
return ans
but due to some reasons i got wrong answer I want to know why
and i end up using this solution after hearing about modular inverse concept
class Solution:
def valueAfterKSeconds(self, n: int, k: int) -> int:
ans = 1
mod = 1000000007
for i in range(0,k) :
ans = (ans*(n+i))%mod
ans = (ans*pow(i+1,mod-2,mod))%mod
return ans

[@ayushsrinivas]

aryan you wrote code worng

[@cse100amisha6]

class Solution {
public:
int valueAfterKSeconds(int n, int k) {
int MOD=1e9+7;
vector a(n,1);
for(int i=0;i

[@KunalSingh-ol6zx]

koi mathematical formula?

[@jd3287]

nice!