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Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?
- a)a1 < a2
- b)a1 > a2
- c)a1 = a2
- d)Depends on the input
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the problem of computing min-max in an unsorted array where m...
When Divide and Conquer is used to find the minimum-maximum element in an array, Recurrence relation for the number of comparisons is
T(n) = 2T(n/2) + 2 where 2 is for comparing the minimums as well the maximums of the left and right subarrays
On solving, T(n) = 1.5n - 2.
While doing linear scan, it would take 2*(n-1) comparisons in the worst case to find both minimum as well maximum in one pass.
T(n) = 2T(n/2) + 2 where 2 is for comparing the minimums as well the maximums of the left and right subarrays
On solving, T(n) = 1.5n - 2.
While doing linear scan, it would take 2*(n-1) comparisons in the worst case to find both minimum as well maximum in one pass.
Most Upvoted Answer
Consider the problem of computing min-max in an unsorted array where m...
When Divide and Conquer is used to find the minimum-maximum element in an array, Recurrence relation for the number of comparisons is
T(n) = 2T(n/2) + 2 where 2 is for comparing the minimums as well the maximums of the left and right subarrays
On solving, T(n) = 1.5n - 2.
While doing linear scan, it would take 2*(n-1) comparisons in the worst case to find both minimum as well maximum in one pass.
T(n) = 2T(n/2) + 2 where 2 is for comparing the minimums as well the maximums of the left and right subarrays
On solving, T(n) = 1.5n - 2.
While doing linear scan, it would take 2*(n-1) comparisons in the worst case to find both minimum as well maximum in one pass.
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Community Answer
Consider the problem of computing min-max in an unsorted array where m...
Explanation:
Algorithm A1:
- In the worst-case scenario, Algorithm A1 will compare every pair of elements in the array to find the min and max.
- Since there are n elements in the array, the total number of comparisons made by Algorithm A1 will be n*(n-1)/2.
- Therefore, a1 = n*(n-1)/2.
Algorithm A2:
- In the worst-case scenario, Algorithm A2 will scan the array linearly to find the min and max elements.
- It will make 2 comparisons for each pair of elements it encounters while scanning the array.
- Therefore, a2 = 2*(n-1).
Relation between a1 and a2:
- Comparing the two formulas for a1 and a2, we get:
a1 = n*(n-1)/2
a2 = 2*(n-1)
- As n*(n-1)/2 is a quadratic function of n and 2*(n-1) is a linear function of n, in general, the quadratic function will grow faster than the linear function.
- Therefore, in worst-case scenarios, a1 will be greater than a2.
- Hence, the relation between a1 and a2 is a1 > a2.
Algorithm A1:
- In the worst-case scenario, Algorithm A1 will compare every pair of elements in the array to find the min and max.
- Since there are n elements in the array, the total number of comparisons made by Algorithm A1 will be n*(n-1)/2.
- Therefore, a1 = n*(n-1)/2.
Algorithm A2:
- In the worst-case scenario, Algorithm A2 will scan the array linearly to find the min and max elements.
- It will make 2 comparisons for each pair of elements it encounters while scanning the array.
- Therefore, a2 = 2*(n-1).
Relation between a1 and a2:
- Comparing the two formulas for a1 and a2, we get:
a1 = n*(n-1)/2
a2 = 2*(n-1)
- As n*(n-1)/2 is a quadratic function of n and 2*(n-1) is a linear function of n, in general, the quadratic function will grow faster than the linear function.
- Therefore, in worst-case scenarios, a1 will be greater than a2.
- Hence, the relation between a1 and a2 is a1 > a2.
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Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer?.
Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?a)a1 < a2b)a1 > a2c)a1 = a2d)Depends on the inputCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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