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An array of n numbers is given, where n is an even number. The maximum as well as the minimum of these n numbers needs to be determined. Which of the following is TRUE about the number of comparisons needed?
- a)At least 2n − c comparisons, for some constant c, are needed.
- b)At most 1.5n − 2 comparisons are needed.
- c)At least nlog2 comparisons are needed.
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
An array of n numbers is given, where n is an even number. The maximum...
B)At least n+log2n−2
c)At least 3n/2−2
d)At least 2n−1
Answer: d) At least 2n-1
Explanation:
To find both maximum and minimum, we can't just compare the numbers one by one and keep track of maximum and minimum separately. That would take 2n comparisons.
We can use the divide and conquer approach to reduce the number of comparisons. We divide the array into two halves and recursively find the maximum and minimum of each half. Then we compare the maximums of the two halves to get the maximum of the entire array, and compare the minimums of the two halves to get the minimum of the entire array.
Let's assume that the number of comparisons needed to find the maximum and minimum of n numbers is T(n). Then we have the following recurrence relation:
T(n) = 2T(n/2) + 2
The first term on the right side is for finding the maximum and minimum of each half recursively, and the second term is for comparing the maximums and minimums of the two halves.
By using the master theorem, we can show that T(n) = O(n). Therefore, the minimum number of comparisons needed is at least 2n-1.
c)At least 3n/2−2
d)At least 2n−1
Answer: d) At least 2n-1
Explanation:
To find both maximum and minimum, we can't just compare the numbers one by one and keep track of maximum and minimum separately. That would take 2n comparisons.
We can use the divide and conquer approach to reduce the number of comparisons. We divide the array into two halves and recursively find the maximum and minimum of each half. Then we compare the maximums of the two halves to get the maximum of the entire array, and compare the minimums of the two halves to get the minimum of the entire array.
Let's assume that the number of comparisons needed to find the maximum and minimum of n numbers is T(n). Then we have the following recurrence relation:
T(n) = 2T(n/2) + 2
The first term on the right side is for finding the maximum and minimum of each half recursively, and the second term is for comparing the maximums and minimums of the two halves.
By using the master theorem, we can show that T(n) = O(n). Therefore, the minimum number of comparisons needed is at least 2n-1.
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Community Answer
An array of n numbers is given, where n is an even number. The maximum...
One possible way to do this is we select first element as max and also min. Then we compare it with all others.
Atmost this would take 2n comparisons during linear search.
But if we use divide and conquer as for merge sort, we have
T(n) = 2T(n/2) + 2 for n>2
Atmost this would take 2n comparisons during linear search.
But if we use divide and conquer as for merge sort, we have
T(n) = 2T(n/2) + 2 for n>2
= 1.5n - 2
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An array of n numbers is given, where n is an even number. The maximum as well as the minimum of these n numbers needs to be determined. Which of the following is TRUE about the number of comparisons needed?a)At least 2n − c comparisons, for some constant c, are needed.b)At most 1.5n − 2 comparisons are needed.c)At least nlog2 comparisons are needed.d)None of the aboveCorrect answer is option 'B'. Can you explain this answer?
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