integers, and x is an integer in the array A, and l and r are indices l ≤ r between which the element x is located in A. The algorithm S returns the index (location) of the element x in the array A. H(A, x, l, r): if l == r: return l else: m = (l+r)//2 # // returns integer component upon division: 7//2=3 if x <= A[m]: return H(A, x, l, m) else: return H(A, x, m+1, r) Derive formally the running time of this algorithm and formally prove the correctness of the running time bound for the worst case, ie. O

integers, and x is an integer in the array A, and l and r are indices l ≤ r between which the element x is located in A. The algorithm S returns the index (location) of the element x in the array A. H(A, x, l, r): if l == r: return l else: m = (l+r)//2 # // returns integer component upon division: 7//2=3 if x <= A[m]: return H(A, x, l, m) else: return H(A, x, m+1, r) Derive formally the running time of this algorithm and formally prove the correctness of the running time bound for the worst case, ie. O

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Concept explainers

Counting sort

The counting sort algorithm sorts an array’s contents by counting the repetition of each element that occurs in the array. A counting sort algorithm is a stable algorithm that works the best when the range of elements in the array is small, say for example from 0 to 9. The count which is the number of repetitions of an element in the array is mapped to an index of an auxiliary array, which is then saved in an auxiliary array. The counting sort algorithm has a stable time complexity of the order of O(size of array + size of auxiliary array). Let’s discuss the algorithm in detail as follows.

Question

Consider the following algorithm S, in which A represents a sorted array of n integers, and
x is an integer in the array A, and l and r are indices l ≤ r between which the element x is located in A.
The algorithm S returns the index (location) of the element x in the array A.

H(A, x, l, r):
if l == r:
return l
else:
m = (l+r)//2 # // returns integer component upon division: 7//2=3
if x <= A[m]:
return H(A, x, l, m)
else:
return H(A, x, m+1, r)

Derive formally the running time of this algorithm and formally prove the correctness of the running
time bound for the worst case, ie. O()

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