Read the following pseudocode, note that A is a sorted array: BINARY-SEARCH(A, v) low := 1...
Read the following pseudocode, note that A is a sorted
array:
BINARY-SEARCH(A, v)
low := 1
high := n
while low <= high
mid = floor((low+ high)/2)
if v == A[mid]
return mid
else if v > A[mid]
low = mid + 1
else
high = mid - 1
return NIL
a) Find the Recurrence Relation.
b) Is there a tight bound? If the answer is yes, find it.
c) Find the best case and worst case running time.
Homework Answers
Add Answer to:
Read the following pseudocode, note that A is a sorted
array:
BINARY-SEARCH(A, v)
low := 1...
Read the following pseudocode, note that A is a sorted array: BINARY-SEARCH(A, v) low := 1...
Read the following pseudocode, note that A is a sorted array: BINARY-SEARCH(A, v) low := 1 high := n while low <= high mid = floor((low+ high)/2) if v == A[mid] return mid else if v > A[mid] low = mid + 1 else high = mid - 1 return NIL a) Find the Recurrence Relation. b) Is there a tight bound? If the answer is yes, find it. c) Find the best case and worst case running time.
Show the steps of the binary search algorithm (pseudocode is given below); low = index of...
Show the steps of the binary search algorithm (pseudocode is given below); low = index of first item in list high = index of last item in list while low is less than or equal to high mid = index halfway between low and high if item is less than list[mid] high = mid - 1 else if item is greater than list[mid] low = mid + 1 else return mid end while return not found For each step of...
Complete the binary search function. def binary Search (array, first, last, key): if last >= first:...
Complete the binary search function. def binary Search (array, first, last, key): if last >= first: mid = int(first + (last - first)/2) if array[mid] return mid key: elif array[mid] > key: ##Statement 1 else: ##Statement 2 else: return -1 Statement 1: return binarySearch(array, last, mid+1, key) Statement 2: return binarySearch(array, mid-1, first, key) Statement 1: return binarySearch(array, first, mid+1, key) Statement 2: return binarySearch(array, mid-1, last, key) Statement 1: return binarySearch(array, mid-1, first, key) Statement 2: return binary Search(array,...
Consider the following: Algorithm 1 Smallest (A,q,r) Precondition: A[ q, ... , r] is an array...
Consider the following: Algorithm 1 Smallest (A,q,r) Precondition: A[ q, ... , r] is an array of integers q ≤ r and q,r ∈ N. Postcondition: Returns the smallest element of A[q, ... , r]. 1: function Smallest (A , q , r) 2: if q = r then 3: return A[q] 4: else 5: mid <--- [q+r/2] 6: return min (Smallest(A, q, mid), Smallest (A, mid + 1, r)) 7: end if 8: end function (a) Write a recurrence...
without coding Give the Big O run-time of the following algorithms. Binary Search: def binary-search (arr,...
without coding
Give the Big O run-time of the following algorithms. Binary Search: def binary-search (arr, low, high, x): # Check base case if low > high : return None else: mid = (high + low) // 2 element arr[mid] == X: if element return mid elif element > X: return binary-search(arr, low, mid 1, x) else: return binary_search(arr, mid + 1, high, x) Selection Sort: def selection_sort (arr): for i in range (len(arr)): smallest index = i smallest value...
Given the binary search function, answer following question: int binarySearch(int a[], int size, int target, int...
Given the binary search function, answer following question: int binarySearch(int a[], int size, int target, int low, int high) { while (low <= high) { int mid = (low + high) / 2; if (a[mid] == target) return mid; else if (a[mid] < target) low = mid + 1; else high = mid 1: } return -1; } 1) If array a[] = {4, 5, 18, 25, 66, 70, 78}, size = 7, target = 71, low = 0, high...
The binary search algorithm is used to search the following array for the value 59. In...
The binary search algorithm is used to search the following array for the value 59. In the table below, list the values for the subscript variables low, high, and mid for each pass through the algorithm's while loop. (There may be fewer than five passes.) 10 19 24 37 42 48 52 59 65 78 Pass 1: low = high = mid = Pass 2: low = high = mid = Pass 3: low = high = mid = Pass 4: low = high = mid = Pass...
plz solve Modify binary search so that it always returns the element with the smallest index...
plz solve Modify binary search so that it always returns the element with the smallest index * that matches the search element (and still guarantees logarithmic running time). import java.util.Arrays; public class BinarySearch2 { public static int rank(int key, int[] a) { // Array must be sorted. int lo = 0; int hi = a.length - 1; while (lo <= hi) { // Key is in a[lo..hi] or not present. int mid = lo + (hi - lo) / 2;...
PLEASE write in C Language, ( you can use array, recursive function, Pointers, Dynamic call by re...
PLEASE write in C Language, ( you can use array, recursive
function, Pointers, Dynamic call by refernce, Structure)
a) 70 marks] Write a program that Sorts a given integer array via Selection Sort obtained by rotation around 3. Searches for a key point in the rotated array in O(logn) time, where the rotation Rotates the sorted array around a random point, e.g., 1 2 345->34512 is point is known in advance. ts above by first finding the unknown rotation point...
When binary searching a sorted array that contains more than one key equal to the search...
When binary searching a sorted array that contains more than one
key equal to the search key, the client may want to know the index
of either the first or the last such key. Accordingly, implement
the following API: public class BinarySearchDeluxe { /* Returns the
index of the first key in a[] that equals the search key, or -1 if
no such key. */ public static int firstIndexOf(Key[] a, Key key,
Comparator comparator) /* Returns the index of the...
Complete the binary search function. def binary Search (array, first, last, key): if last >= first: mid = int(first + (last - first)/2) if array[mid] return mid key: elif array[mid] > key: ##Statement 1 else: ##Statement 2 else: return -1 Statement 1: return binarySearch(array, last, mid+1, key) Statement 2: return binarySearch(array, mid-1, first, key) Statement 1: return binarySearch(array, first, mid+1, key) Statement 2: return binarySearch(array, mid-1, last, key) Statement 1: return binarySearch(array, mid-1, first, key) Statement 2: return binary Search(array,...
without coding
Give the Big O run-time of the following algorithms. Binary Search: def binary-search (arr, low, high, x): # Check base case if low > high : return None else: mid = (high + low) // 2 element arr[mid] == X: if element return mid elif element > X: return binary-search(arr, low, mid 1, x) else: return binary_search(arr, mid + 1, high, x) Selection Sort: def selection_sort (arr): for i in range (len(arr)): smallest index = i smallest value...
Given the binary search function, answer following question: int binarySearch(int a[], int size, int target, int low, int high) { while (low <= high) { int mid = (low + high) / 2; if (a[mid] == target) return mid; else if (a[mid] < target) low = mid + 1; else high = mid 1: } return -1; } 1) If array a[] = {4, 5, 18, 25, 66, 70, 78}, size = 7, target = 71, low = 0, high...
PLEASE write in C Language, ( you can use array, recursive
function, Pointers, Dynamic call by refernce, Structure)
a) 70 marks] Write a program that Sorts a given integer array via Selection Sort obtained by rotation around 3. Searches for a key point in the rotated array in O(logn) time, where the rotation Rotates the sorted array around a random point, e.g., 1 2 345->34512 is point is known in advance. ts above by first finding the unknown rotation point...
When binary searching a sorted array that contains more than one
key equal to the search key, the client may want to know the index
of either the first or the last such key. Accordingly, implement
the following API: public class BinarySearchDeluxe { /* Returns the
index of the first key in a[] that equals the search key, or -1 if
no such key. */ public static int firstIndexOf(Key[] a, Key key,
Comparator comparator) /* Returns the index of the...
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