How many times does a binary search need to execute to find its value? Recall from our lesson that the number of iterations is roughly a log base 2 relationship. In this exercise, you are going to calculate the maximum iterations and the actual iterations needed to find a random value in arrays of size 100, 1000, 10k, and 100k. You are given helper methods to calculate the maximum iterations, generate the random array, and do the binary search. You are also given a counter variable that increments up each time your recursive binary method is called. You will need to come up with the remainder of the code. Same output is provided below. Sample Output Array Size: 100 Max iterations: 7 Actual iterations: 5 Array Size: 1000 Max iterations: 10 Actual iterations: 5 import java.util.*; public class BinarySearchTest { static int count; public static void main(String[] args) { // Use the helper code to generate arrays, calculate the max // iterations, and then find the actual iterations for a randomly // selected value. } public static int binaryRec(int[] array, int target, int begin, int end) { if (begin <= end) { int mid = (begin + end) / 2; count ++; // Base Case if (target == array[mid]) { return mid; } if (target < array[mid]) { return binaryRec(array, target, begin, mid - 1); } if (target > array[mid]) { return binaryRec(array, target, mid + 1, end); } } return -1; //Alternate Base Case - not found } public static int[] generateArrayOfLength(int length) { int[] arr = new int[length]; for(int i = 0; i < length; i++) { arr[i] = (int)(Math.random() * 100); } Arrays.sort(arr); return arr; } private static int binaryMax(int length) { return (int) (Math.log(length) / Math.log(2)) + 1; } }
How many times does a binary search need to execute to find its value? Recall from our lesson that the number of iterations is roughly a log base 2 relationship.
In this exercise, you are going to calculate the maximum iterations and the actual iterations needed to find a random value in arrays of size 100, 1000, 10k, and 100k.
You are given helper methods to calculate the maximum iterations, generate the random array, and do the binary search. You are also given a counter variable that increments up each time your recursive binary method is called.
You will need to come up with the remainder of the code. Same output is provided below.
Sample Output
Array Size: 100
Max iterations: 7
Actual iterations: 5
Array Size: 1000
Max iterations: 10
Actual iterations: 5
import java.util.*;
public class BinarySearchTest {
static int count;
public static void main(String[] args) {
// Use the helper code to generate arrays, calculate the max
// iterations, and then find the actual iterations for a randomly
// selected value.
}
public static int binaryRec(int[] array, int target, int begin, int end) {
if (begin <= end)
{
int mid = (begin + end) / 2;
count ++;
// Base Case
if (target == array[mid]) {
return mid;
}
if (target < array[mid]) {
return binaryRec(array, target, begin, mid - 1);
}
if (target > array[mid]) {
return binaryRec(array, target, mid + 1, end);
}
}
return -1; //Alternate Base Case - not found
}
public static int[] generateArrayOfLength(int length)
{
int[] arr = new int[length];
for(int i = 0; i < length; i++)
{
arr[i] = (int)(Math.random() * 100);
}
Arrays.sort(arr);
return arr;
}
private static int binaryMax(int length) {
return (int) (Math.log(length) / Math.log(2)) + 1;
}
}
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