Problem regarding time complexity

[naved momin]

I was reading a book which tellls that the outer loops time complexity is O(n-m) whereas for inner loop the books gives explanation as

" The inner while
loop goes around at most m times, and potentially far less when the pattern match
fails. This, plus two other statements, lies within the outer for loop. The outer loop
goes around at most n−m times, since no complete alignment is possible once we
get too far to the right of the text. The time complexity of nested loops multiplies,
so this gives a worst-case running time of O((n − m)(m + 2)). "

i didn't understand for what reason the time complexity of inner loop is O(m+2) instead of O(m) ?please help thanks

int findmatch(char *p, char *t)
{
int i,j; /* counters */
int m, n; /* string lengths */
m = strlen(p);
n = strlen(t);
for (i=0; i<=(n-m); i=i+1) {
j=0;
while ((j<m) && (t[i+j]==p[j]))
j = j+1;
if (j == m) return(i);
}
return(-1);
}

[Jason Reaves]

Algorithm books like to be confusing. The for loop in your example will run n-m+1 times so that becomes the complexity and it'll be multiplied by the complexity of the inner loop. I'll try to demonstrate.

I'll use an example of two while loops because basically a while loop comes out to be the same thing as a for loop in time analysis.

Basically the time analysis on a while loop amounts to

twhile = (tbody + tcond) * r + tcond

twhile being the time of the while loop
tbody being the time of the body of the while loop
tcond being the time of the conditional of the while loop
"r" being how many times the body is to be run

The reason tcond is there at the end is because it will be ran again and become false at the end of loop.

So if you have loops inside of loops for instance:

i=0
while(i<n)
{
    j=0
    while(j<k)
    {
         A[i][j] += 1
     }
}

Then we'll call the first while loop T1while and the second while loop T2while so T1 become

T1while = (T2while + t1cond) * r + t1cond   --now we know that the cond is constant so we can just fill in a c for constant
T1while = (T2while + c) * n + c                      --we also know the r will be n some outside number passed in to be our problem size
T1while = (c * k + c  + c) * n + c                   --now we just simplify like we did in T2while via combining constants and such
T1while = (c * k + c) * n + c
T1while = c*k*n + c*n + c                            --hey look at this our problem size is k*n and this now looks like a quadratic equation of sorts... so this function asymptotically would belong to the set O(k*n)


T2while = (t2body + t2cond) * r + t2cond    --now the cond is a constant like T1while and the r will be k some passed in value ----maybe we're traversing a matrix or something?
T2while = (t2body + c) * k + c                      --the body is a simple statement so it's also constant
T2while = (c + c) * k + c                                -- constant + a constant = a constant
T2while = c * k + c                                        --now we can fill this into T1while