Max  Min ProblemProblem: 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) Analysis:Method 1: if we apply the general approach to the array of size n, the number of comparisons required are 2n2. Method2: In another approach, we will divide the problem into subproblems 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.
T (n) = 2 T → Eq (i) T (2) = 1, time required to compare two elements/items. (Time is measured in units of the number of comparisons) → Eq (ii) Put eq (ii) in eq (i) Similarly, apply the same procedure recursively on each subproblem or anatomy {Use recursion means, we will use some stopping condition to stop the algorithm} Recursion will stop, when → (Eq. 4) Put the equ.4 into equation3.
Number of comparisons requires applying the divide and conquering algorithm on n elements/items = Number of comparisons requires applying general approach on n elements = (n1) + (n1) = 2n2 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 (2n2), (2x8)2=14 comparisons It is evident; we can reduce the number of comparisons (complexity) by using a proper technique.
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