A binary search begins with the A/ largest B/ middle C/ last D/ first element of an array.
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Do not use any
![A binary search begins with the
A/ largest
B/ middle
C/ last
D/ first
E/ None of these
element of an array.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa730c64c-39bc-466f-b793-cad77ce9c376%2Fa02ab598-47d5-4f31-9adf-173443e3e62f%2Fwl9eq8j_processed.jpeg&w=3840&q=75)
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- Q1: Apply linear and binary search on given array. Q2: apply quick, merge, and bubble sort on given array. index: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 value: 1 8 10 14 15 11 13 18 2 4 6 5 19 3 20 Array: 182. For the following array, you are to perform a binary search: Int nums[ ]= {1, 13, 78, 89, 96, 100, 112, 125, 230, 310, 423, 578, 1000); Search for the number 126 by showing the low, midpoint and high values at each step before it reports that 126 is NOT found in the array. Low Mid HighJava- The Insertion sort is efficient for large arrays. True or False?
- Answer in java script Create a function that returns true if the first array can be nested inside the second. arr1 can be nested inside arr2 if: 1. arr1's min is greater than arr2 's min. 2. arr1's max is less than arr2 's max. Examples canNest([1, 2, 3, 4], [0, 6]) true canNest ([3, 1], [4, 0]) → true canNest ([9, 9, 8], [8, 9]) → false canNest([1, 2, 3, 4], [2, 3]) - falseFollowing array is being sorted by Radix sort. Two iterations are already completed, What will be the new sequence after running the third iteration. 1, 7, 10, 9420, 3221, 5622, 4127, 2030, 3138, 743, 577, 9680, 82, 4793, 2599DO NOT POST EXISTING ONE: Write a java method called SearchKey that return the indices of the first and the last occurences of an element k in an array.The function should Return [-1,1] in case the element was not found
- js code const find a5 = n.findIndex(num => n var a = 5 * i + '\t'');console.log(finda5); correct this code ,find divisible by 5 in an array a5,and save the returned value is to be stored in variable v5What is the upper-bound index of an array whose size is 52?"7. How do you implement binary search to find an element in a sorted array?.
- T/F: Binary Search can correctly determine whether an element is in an array if it is unsorted.// function to generate all subsequences of given arrayfunction generate_subseq(arr): // length of arr n = len(arr) // total subsequences for array of length n m = 2**n // to store all subsequences seqs = [] // running a loop for m times for i in range(1, m): // creating an array of zeros of length n a = [0]*n num = i // to use as an index for 'a' j = n-1 // run this loop till num > 0 while num > 0: if num is odd: a[j] = 1 // divide num by 2 num = num/2 // subtract 1 from 'j' j -= 1 // to store current subsequence seq = [] // iterating for n times for i in range(n): // add ith index value to 'seq' if a[i] == 1: seq.append(arr[i]) // add 'seq' to 'seqs' seqs.append(seq) // return seqs return seqs // given listsS1 = ['B','C','D','A','A','C','D']S2 = ['A','C','D','B','A','C']…What is the tradeoff between using an unordered array versus an ordered array ?
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