Question: Given the following algorithms ( Algorithm A i=1 while(i 0 and Aljl> key A[j+1] =Aljl j-j-1 A[j+1] = key j-j+1 Find for both algorithms:

Given the following algorithms ( Algorithm A i=1 while(i 0 and Aljl> key A[j+1] =Aljl j-j-1 A[j+1] = key j-j+1 Find for both algorithms: a) What is the general case T (n)? b) What is the best-case T (n)? What is the worst-case T (n)? Algorithm B int min = 1000; i-1 while(i < arr.length) { if(arr[i]
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a The general case Tn for both algorithms is On2 Algorithm A In the worst case when the array is sor... View full answer
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