Linear Search can be implemented for sorting and non-sorting elements of a Data structure particular Data structure but the average case time complexity is O(n). Whereas as Binary Search can be implemented only when the items are in sorted order and average-case time complexity is O(logn) and both Transversal have best-case Time complexity is O(1). Now, given an Array List containing sorted elements Check whether the element exists in the ArrayList or not.
There are two types of Transversal while searching elements in Linear Data structure.
Illustration:
Input: ArrayList:[1, 2, 3, 4, 6, 7, 8, 9] key:3 Output: true
Case 1: Use Binary Search Because the list is sorted in order and Binary Search has less average time complexity as compared to Linear Search i.e O(logn).
// Java Program to Search ArrayList Element // Using Binary Search // Importing generic java libraries import java.io.*;
import java.util.*;
class GFG {
// Method to search elements in ArrayList
static boolean search( int key, ArrayList<Integer> A)
{
// low pointer
int low = 0 ;
// high pointer
int high = A.size() - 1 ;
while (low <= high) {
// find the mid pointer
int mid = low + (high - low) / 2 ;
if (A.get(mid) == key) {
return true ;
}
else if (A.get(mid) < key) {
// shift the low pointer
low = mid + 1 ;
}
else {
// shift the high pointer
high = mid - 1 ;
}
}
return false ;
}
// Main driver method
public static void main(String[] args)
{
// Creating an ArrayList
ArrayList<Integer> A = new ArrayList<>();
// Adding items in the list
A.add( 1 );
A.add( 2 );
A.add( 3 );
A.add( 4 );
A.add( 6 );
A.add( 7 );
A.add( 8 );
A.add( 9 );
// Random element to be searched
int key = 19 ;
// Binary search
boolean check = search(key, A);
System.out.println(check);
int key1 = 2 ;
// Binary search
boolean check1 = search(key1, A);
System.out.println(check1);
}
} |
Output:
false true
Case 2: Suppose in order to find the maximum index of the greatest element less than the key in sorted repeated elements of ArrayList Using Binary Search.
Input: List: [2, 3, 3, 4, 4, 5, 6, 7] key: 2 key: 4 Output: -1 2
Example: Modify the Binary Search according to the condition
// Java Program to Search ArrayList Element // Using Binary Search // Importing generic java libraries import java.io.*;
import java.util.*;
class GFG {
// Method to search elements in arrayList
static int search( int key, ArrayList<Integer> A)
{
// Low pointer
int low = 0 ;
// High pointer
int high = A.size() - 1 ;
int index = - 1 ;
while (low <= high) {
// Mid pointer
int mid = low + (high - low) / 2 ;
if (A.get(mid) == key) {
// Shift the High Pointer
high = mid - 1 ;
}
else if (A.get(mid) < key) {
// Storing the index of mid index value
index = mid;
// Shift the Low Pointer
low = mid + 1 ;
}
else {
high = mid - 1 ;
}
}
// Return the index
return index;
}
public static void main(String[] args)
{
// Creating an ArrayList
ArrayList<Integer> A = new ArrayList<>();
// Adding elements in ArrayList
A.add( 2 );
A.add( 3 );
A.add( 3 );
A.add( 4 );
A.add( 4 );
A.add( 5 );
A.add( 6 );
A.add( 7 );
// Key
int key = 5 ;
// Index of smallest greater element
int index = search(key, A);
// Print searched element
System.out.println(index);
}
} |
Output:
4
Time complexity: O(logn)
Auxiliary space: O(1) as it is using constant variables