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Max - Min Problem

Problem: Analyze the algorithm to find the maximum and minimum element from an array.

Algorithm: Max ?Min Element (a [])
Max:  a [i]
Min:   a [i]
For i= 2 to n do
If a[i]> max then
max = a[i]
if a[i] < min then
min: a[i]
return (max, min)


Method 1: if we apply the general approach to the array of size n, the number of comparisons required are 2n-2.

Method-2: In another approach, we will divide the problem into sub-problems and find the max and min of each group, now max. Of each group will compare with the only max of another group and min with min.

Let n = is the size of items in an array

Let T (n) = time required to apply the algorithm on an array of size n. Here we divide the terms as T(n/2).

2 here tends to the comparison of the minimum with minimum and maximum with maximum as in above example.

Max - Min Problem

T (n) = 2 T Max - Min Problem → Eq (i)

T (2) = 1, time required to compare two elements/items. (Time is measured in units of the number of comparisons)

Max - Min Problem→ Eq (ii)

Put eq (ii) in eq (i)

Max - Min Problem

Similarly, apply the same procedure recursively on each subproblem or anatomy

{Use recursion means, we will use some stopping condition to stop the algorithm}

Max - Min Problem

Recursion will stop, when Max - Min Problem → (Eq. 4)

Put the equ.4 into equation3.

Max - Min Problem

Number of comparisons requires applying the divide and conquering algorithm on n elements/items = Max - Min Problem

Number of comparisons requires applying general approach on n elements = (n-1) + (n-1) = 2n-2

From this example, we can analyze, that how to reduce the number of comparisons by using this technique.

Analysis: suppose we have the array of size 8 elements.

Method1: requires (2n-2), (2x8)-2=14 comparisons
Method2: requires Max - Min Problem

It is evident; we can reduce the number of comparisons (complexity) by using a proper technique.

Next TopicBinary Search

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