CareerCup

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no, after 2-3 days

My approach towards this problem is
Lets say we have matrix of M having m rows and n columns
always search row wise for S[0] or S[x-1] where x is the length of string.
Now suppose currently you are at (i,j) position in matrix.
int a =0; // a global variable
if( M[i,j]== S[0] && i<=(m-x)){

bool flag = true;
for( a = 1 to x-1){
if( S[a] == M[i+a,j])
continue;
else{
flag = false;
break;
}
}
a= 0;
if(flag){
we got the string.
return ;
}
}
if( M[i,j]== S[0] && j<=(n-x)){

bool flag = true;
for( a = 1 to x-1){
if( S[a] == M[i,j+a])
continue;
else{
flag = false;
break;
}
}
if(flag){
we got the string.
return ;
}
}

if( M[i,j]== S[x-1] && i<=(m-x)){

bool flag = true;
for( a = 1 to x-1){
if( S[x-a] == M[i+a,j])
continue;
else{
flag = false;
break;
}
}
a = 0;
if(flag){
we got the string.
return ;
}
}
if( M[i,j]== S[x-1] && j<=(n-x)){

bool flag = true;
for( a = 1 to x-1){
if( S[x-a] == M[i,j+a])
continue;
else{
flag = false;
break;
}
}
if(flag){
we got the string.
return ;
}
}

//If none of the above returns then
//Start searching from S[i,j+ a+1] if a+1 <=n
// else start from S[i+1,0]

Not only coding but all were related to algorithms , design and coding.

In list question. make one scan on list 1 and one scan on list 2. Let sizes be m and n. If m is larger then n, then skip pointer of m from head (m-n) times. now size of both list is same. now skip each pointer side by side, till you reach common node.

in stock question one option in brute force method O(n*2). Another way is to keep track of current max and try to find minimum after that. if you find max again. reset every thing and remember the last max loss.

Can you eloborate more on the stock solution please? What exactly does the question mean and what's the solution in detail.

Thanks