Latest DATA STRUCTURE Aptitude Question SOLUTION: How many minimum number of comparisons required to find minimum and maximum of 100 numbers

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(#M40033766) DATA STRUCTURE Programming question Comparisions Required Keep an EYE

How many minimum number of comparisons required to find minimum and maximum of 100 numbers

Asked In DATA STRUCTURE yellapanthula pranav (10 years ago)

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yellapanthula pranav  (10 years ago)

It requires 148 comparisons.
To find maximum or minimum it take 3n/2 -2 comparissions
So 150-2=148

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gaurav  (10 years ago)

minimum comparisions will be 99
min=arr[0];
max=arr[0];
for(i=1;i

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B Gurunath  (10 years ago)

98 comparisons only

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imtiyaz ahmad khan  (6 years ago)

Ans: minimum number of comparison require to find minimum and maximum is: Approach is divide and conquer ....

T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2
T(2) = 1 // if two element then compare both and return max and min
T(1) = 0 // if one element then return both max and min same
If n is a power of 2, then we can write T(n) as:

T(n) = 2T(n/2) + 2
After solving above recursion, we get

T(n) = 3/2n -2
Thus, the approach does 3/2n -2 comparisons if n is a power of 2. And it does more than 3/2n -2 comparisons if n is not a power of 2.

So, here in this case put n=100 and we will get (3/2)(100) - 2 = 148 comparison .....

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Anchal   (9 years ago)

How 98??
Please explain.

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ATUL BHASKAR  (9 years ago)

Using Merge sort or Heap sort, we can find min. and max. 100 nos. as worst case efficiency for both algorithms is o(nlog n)... i.e. 100*log 100 = 200.

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purvi jain  (9 years ago)

98 comparisions

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Kavitha Madhavaraj  (9 years ago)

n-1 => 99 comparisons given the data is unsorted.

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sailaxmi gajjam  (9 years ago)

1 + 2(n-2) in worst case and 1 + n – 2 in best case
here consider best case.
1+n-2
=1+100-2
=99

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