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Q. Calculate that how many key comparisons and assignments an insertion sort makes in its worst case?
Ans:
The worst case performance occurs in insertion sort occurs when the elements of the input array are in descending order. In that case, the first pass requires one comparison, the second pass requires two comparisons, third pass three comparisons till the kth pass requires (k-1), and in the end the last pass requires (n-1) comparisons. Therefore, we obtain the total numbers of comparisons as follows : f(n) = 1+2+3+.........+(n-k)+.....+(n-2)+(n-1) = n(n-1)/2 = O(n2)
The worst case performance occurs in insertion sort occurs when the elements of the input array are in descending order. In that case, the first pass requires one comparison, the second pass requires two comparisons, third pass three comparisons till the kth pass requires (k-1), and in the end the last pass requires (n-1) comparisons. Therefore, we obtain the total numbers of comparisons as follows :
f(n) = 1+2+3+.........+(n-k)+.....+(n-2)+(n-1)
= n(n-1)/2
= O(n2)
Explain Division Method Division Method: - In this method, key K to be mapped into single of the m states in the hash table is divided by m and the remainder of this division
Q. Explain the insertion sort with a proper algorithm. What is the complication of insertion sort in the worst case?
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