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Given two sorted list of size m and n respectively. The number of comparisons needed the worst case by the merge sort algorithm will be
- a)m x n
- b)maximum of m and n
- c)minimum of m and n
- d)m + n - 1
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Given two sorted list of size m and n respectively. The number of comp...
To merge two lists of size m and n, we need to do m+n-1 comparisons in worst case. Since we need to merge 2 at a time, the optimal strategy would be to take smallest size lists first. The reason for picking smallest two items is to carry minimum items for repetition in merging. So, option (D) is correct.
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Given two sorted list of size m and n respectively. The number of comp...
Merge Sort Algorithm and Comparison Complexity
Merge Sort Algorithm
Merge sort is an efficient, general-purpose, comparison-based sorting algorithm. It works by dividing an array into two halves, sorting each half separately, and then merging the two halves back together. This process is repeated recursively until the entire array is sorted.
Comparison Complexity of Merge Sort Algorithm
The comparison complexity of an algorithm is the number of comparisons it makes during its execution. In the case of merge sort, the worst-case comparison complexity is determined by the number of comparisons made in the merging step, which is the final step in the algorithm.
Number of Comparisons in Merging Step
The merging step of merge sort involves comparing the elements of the two sorted subarrays and merging them into a new sorted array. The number of comparisons needed in this step depends on the sizes of the subarrays being merged.
In general, if the two subarrays being merged have sizes m and n, then the worst-case number of comparisons needed is m+n-1. This occurs when the smallest element in one subarray is greater than the largest element in the other subarray, and every element in one subarray must be compared with every element in the other subarray.
Answer Explanation
Given two sorted list of size m and n respectively, the worst-case comparison complexity of the merge sort algorithm will be the number of comparisons made in the merging step. From the above discussion, we know that the worst-case number of comparisons needed to merge two subarrays of sizes m and n is m+n-1.
Therefore, for the given two sorted lists of sizes m and n, the worst-case number of comparisons needed to merge them is m+n-1. Hence, option D is the correct answer.
Merge Sort Algorithm
Merge sort is an efficient, general-purpose, comparison-based sorting algorithm. It works by dividing an array into two halves, sorting each half separately, and then merging the two halves back together. This process is repeated recursively until the entire array is sorted.
Comparison Complexity of Merge Sort Algorithm
The comparison complexity of an algorithm is the number of comparisons it makes during its execution. In the case of merge sort, the worst-case comparison complexity is determined by the number of comparisons made in the merging step, which is the final step in the algorithm.
Number of Comparisons in Merging Step
The merging step of merge sort involves comparing the elements of the two sorted subarrays and merging them into a new sorted array. The number of comparisons needed in this step depends on the sizes of the subarrays being merged.
In general, if the two subarrays being merged have sizes m and n, then the worst-case number of comparisons needed is m+n-1. This occurs when the smallest element in one subarray is greater than the largest element in the other subarray, and every element in one subarray must be compared with every element in the other subarray.
Answer Explanation
Given two sorted list of size m and n respectively, the worst-case comparison complexity of the merge sort algorithm will be the number of comparisons made in the merging step. From the above discussion, we know that the worst-case number of comparisons needed to merge two subarrays of sizes m and n is m+n-1.
Therefore, for the given two sorted lists of sizes m and n, the worst-case number of comparisons needed to merge them is m+n-1. Hence, option D is the correct answer.
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Given two sorted list of size m and n respectively. The number of comparisons needed the worst case by the merge sort algorithm will bea)m x nb)maximum of m and nc)minimum of m and nd)m + n - 1Correct answer is option 'D'. Can you explain this answer?
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