for(int i = 0; i < Array.length ;i++)
{
...
if(s[i]>s[m]){
m = i;
}
}
Being n
the size of the array, and assuming that the computation done in "..." is constant, so let us label the time taken on that "..." as c1
, then the time taken with if(s[i]>s[m])
as c2
, and m = i
as c3. In the worst case-scenario this code has a time complexity of :
n * (c1 + c2 + c3)
which can be represented by O(n)
.
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