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For merging two sorted lists of sizes $m$ and $n$ into a sorted list of size $m+n$, we require comparisons of

1. $O(m)$

2. $O(n)$

3. $O(m+n)$

4. $O(\log m + \log n)$

edited | 4k views

The number of moves are however always $m+n$ so that we can term it as $\Theta(m+n)$. But the number of comparisons vary as per the input. In the best case the comparisons are $Min(m,n)$ and in worst case they are $m+n-1$.

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@Arjun Sir I think its not rght to say best case as Min(m,n). It depnds like if m>n and whichever side will contain smaller values will be finished first. It may be m or n not always minimum.. ryt sir?
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not always- thats why it is only a best case.
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ohh.. ok sir. Thnx :)
A : 10  20  30  90

B. : 40  50  60  70

Suppose A.length =m ,B.length=n

Here m+n-1 comparisons are required
So O(m+n)//this is one of  the worst case comparison
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Thanks for the example

If there are 2 arrays like this

A: 10   20    60   90

B:  30   50    70  100

And store the resultent array in C[ ]

while((a[]!=NULL) && (b[]!=NULL))
{
if((a[i]<b[j])||(b[j]==NULL))
{
c[k++]=a[i++];
}
else
{
c[k++]=b[j++];
}
}
@srestha, kindly notice that in question its written two sorted list, not array.

in this case,
best case:- all elements of 2nd list is greater than that of 1st list then minimum comparison is Min(m,n) as it will take only one iteration.for comparison rest will be inserted as it is.
and worst case:- let 2 sorted list be 1,2,3,4 and 1,2,3,4 so here total comparison for merging should be =7(last element need not be compared) so total comparison= m+n-1= O(m+n).
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How are you knowing that you have reached the last element of one of the sorted lists? Doesn't that require another comparison?

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