Java Program to Search ArrayList Element Using Binary Search - GeeksforGeeks

11 Dec, 2020
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.
 Linear Search Binary Search.
 Linear Search
 Binary Search.
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
// Java Program to Search ArrayList Element// Using Binary Search  // Importing generic java librariesimport 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
// Java Program to Search ArrayList Element// Using Binary Search  // Importing generic java librariesimport 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
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