predict the complexity of the code:
vector<vector > v(100,vector (100,-1) );
int func(int n,int m)
{
if((n==0)||(m==0))
{
return 0;
}
else
{
if(dp[m][n]!=-1)
{
return dp[m][n];
}
else
{
return func(n-1,m-2)+func(n-2,m-1);
}
}
}
O(n^2)
O(n logn)
O(2^n)
O(n^3)
Can you explain how 2^n is the answer
the given code is :-
vector<vector<int > > v(100,vector <int> (100,-1) );
int func(int n,int m)
{
if((n==0)||(m==0))
{
return 0;
}
else
{
if(dp[m][n]!=-1)
{
return dp[m][n];
}
else
{
return func(n-1,m-2)+func(n-2,m-1);
}
}
}
As you can see we are not storing result in dp matrix due to which dp[i][j] is always -1 . so both recursive call will take place even for overlapping cases
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