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The minimum number of comparisons required to determine if an integer appears more than $\frac{n}{2}$ times in a sorted array of $n$ integers is

1. $\Theta(n)$
2. $\Theta(\log n)$
3. $\Theta(\log^*n)$
4. $\Theta(1)$

If an element occurs more than n/2 times, it WILL be present in the middle of the sorted array.

Now, counting the occurrences in a sorted array can easily be done by binary search.

We need to find boundary indices.

We can intuitively do it by performing a linear search on left and right directions to find the boundary indices. But, this takes linear time in the worst case (all elements are same)

Better way is to delete the equality condition from the binary search algorithm.

ie, delete a[k] = x method.

Now, we're only left with a[k] > x and a[k] < x.

Now, instead of returning -1 on an unsuccessful search, return i

We can do manipulations like these to get a boundary index.

Then, reverse this algo to get the other boundary index.

Subtract both these indices to check if the value is $\geq n/2$
Can anyone plz explain with giving a better example? I'm not getting it .Plz explain it with the better example

whenever there exists an element which is present in the array : more than $\frac{n}{2}$ times, then definitely it will be present at the middle index position; in addition to that it will also be present at anyone of the neighbourhood indices namely $i-1$ and $i+1$
No matter how we push that stream of More than $\frac{n}{2}$ times of elements of same value around the Sorted Array, it is bound to be present at the middle index $+$ atleast anyone of its neighbourhood

once we got the element which should have occurred more that $n/2$ times we count its total occurrences in $\mathcal{O}(\log n)$ time.

Then it should be (B)
Why do we need to  count the numbe rof occurrences of the element?..we just need to find whether i occurs more than n/2 times or not..and that can be done in O(1) time..isnt it? ,by checking the middle and next element..please clarify.
It may be the case that an integer does not appear more than n/2 times but appears at all those locations where you are suggesting the checks.
suppose

1 3 4 4 4 4  4 6 7 9

4 occurs in middle but not  majority by n/2

it is present both in middle as well as the neighbourhood.

The algo checks to find the position of frst occurance of 4.

here for 4 it is,3

>n/2 times means

there should be span of n/2 4 s,since element are in sorted order,nothing between span of 4 s.

Hence,i+n/2 th index must also be 4 .here it is not ,so here 4 is not occurring > n/2

core of the algo is a BS which is logn
@amar
I think your last statement is wrong. O(log n) would be needed to find "an integer" and O(1) would be needed to check if its present more than n/2 times. Recheck.
Read example given by GateMm, it should make you understand.
"It may be the case that an integer does not appear more than n/2 times but appears at all those locations where you are suggesting the checks." Can you give sone examples for such situations and which also agrees with the guven question?

example : 1 3 4 4 4 4 4 6 7 9

See $4$ appears 5 times only. But according to the question to be a majority element it should appear more than $\frac{10}{2}=5$ times i.e. atleast 6 times it should appear in the array. Here $\mathcal{O}(1)$ time solution FAILS.

Thank You
Actually in the sorted array if an element is present more than than n/2 times then it must be present in the middle element so in a sorted array we can find in O(1) time and to locate the element where it ends and where it starts we have to take a binary search which is O(log n) time so total O(log n) time.
edited

Does Divide and Conquer work here ?
I know I have to count occurrences of middle element.
I will divide array into two halves, then i will count occurrance of that element indivisually, then i will combine and check if total occurance if greater than n/2 or not.
Like this

mid=n/2
Count_occurrance(i,j)
{
if(i==j && A[i] == A[mid])
return 1;
if(i==j && A[i] != A[mid])
return 0;
else if (A[(i+j)/2] = A[mid])
return(1+Count_occurrance(i, k)+Count_occurrance(k+1,j))    //here k is (i+j)/2
else
return(Count_occurrance(i, k)+Count_occurrance(k+1,j))
}

this is
T(n) = 2T($\frac{n}{2}$)+c
which gives $logn$.

Is this ok ?

Only binary search, will it work good?
Say if 1233346
will this not violate assumtion?
u have 7  elements means (n=7) here 1233346...and "3" is present only 3 times but in question it says that  it should be present more (n/2) times which is must be greater than 3.5..means at least 4 times it should present..according to me..
@Chhotu See first of all how to get the element that must have occurred more than n/2 times?
Tell me that.
Sol:: It takes O(lgn) time
Thanks for valuable reference and help. I was not thinking properly.
But sir we know that if a number has to present and occur more than n/2 it must have to present at middle position. And the given array is sorted so why we can't just directly go to the middle position and check for the number there and it's neighbor. In this way it should be O(1)

Let us take an ex -  1 1 2 2 3 3 3. So looking at 2 can we it is present n/2 times.  Look top comments you will get idea.

Chhotu  @Vs

Why are we searching for x in the middle position ?

As per question,If an integer appears more than n/2 times,so does this mean our function will take integer and check if it is occurring more than n/2. Or does this mean we need to check if there exists such integer?I interpreted as the former way but the answer assume as per later way.Please clear this

edited
here binary search will be apply on both side of the middle element to count the >=(n/2) occurence

so time will be = 0(1) + 2(logn)           {0(1) for finding middle element}

=0(1) + 0(logn)

=0(logn)
Question is about minimum number of comparisons right? so why are we not considering 1st element it self for getting best results....means by checking occurrences of 1st element it will take only O(1) time...so ans should be O(1) ....
this give O(n) not O(logn)
In question they haven't asked the count of the number...so i think answer is constamt time...

What is the meaning of option (c) log* n ?

Sir doesn't minimum number of comparisons mean that we have to consider the best case.In best case  we have to check the middle element,both the first and last element.Suppose  the element is present in last index then we simply need to check the element before the middle element to see if the element is occurring more than n/2 or not.So the best case number of comparisons should be constant.so,omega(1).correct me if I am wrong.

@ sir,
I think Divide and Conquer would take O(n). and not O(logn). see by Masters theo.

@shubham0109

It's iterated logarithm.

Also present in Cormen.

@Sachin Mittal 1 sir,

else if (A[(i+j)/2] = A[mid])
return(1+Count_occurrance(i, k)+Count_occurrance(k+1,j))    //here k is (i+j)/2
else
return(Count_occurrance(i, k)+Count_occurrance(k+1,j))



here in else we can simply return 0 because if that element is not present in the mid position then it is obviously not having more than n/2 occurrance’s .

The starting position of required element can be anywhere between

1 to n/2-1(inclusive).

We have to search that elements starting index to do that Modified Binary Search.

Step1--ELE=A[n/2]

Step2:--Search for starting index of ELE with in range of 1 to n/2-1 which is sorted array. O(logn) time cost of modified binary search.

Now after find such index say i Now see if A[i] ==A[ i+ n/2 ] then it is the majority Ele (Cz Sorted array) else no majority Ele present. Above done in O(1).

So overall time complexity O(logn)+O(1)=O(logn)

Hence Option B is Ans.

Why are we determining position of x.?Isn't x input to the algorithm?
very thorough and easily interpreted....nice one

To check whether a given number is repeated n/2 times in the array can be done in O(log n) time.

Algo

1. find the first occurrence (index i) of x(given number) in the array which can be done in O(log n) time (a variant of binary search).

2. check if A[i] == A[n/2+i]

return true

3. else return false

its An integer not The integer, so we should determine first what is the Integer $x$
edited
thanks

Fool proof way to find is, to perform 2 binary searches, one for n-1 and other for n+1.

Even if n-1 and n+1 are not present, we can modify BST to return the index where min,max are equal.

Then check the difference between two resuls, if greater than n/2.

Hence 2logn comparisons required and answer is B.

### 1 comment

But what if the element (x-1) and (x+1) are also present multiple times/ O(n) times?