Array With Single Entry LeetCode (Single Element in a Sorted Array)

You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.

Return the single element that appears only once.

Your solution must run in O(log n) time and O(1) space.

 Example 1:

Input: nums = [1,1,2,3,3,4,4,8,8]
Output: 2

Example 2:

Input: nums = [3,3,7,7,10,11,11]
Output: 10

 

NOTE:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

 

This problem is popular in LeetCode, LeetCode 540

Solution:

 

	import java.text.DateFormat;
	import java.text.SimpleDateFormat;
	import java.util.ArrayList;
	import java.util.Arrays;
	import java.util.Date;
	import java.util.HashMap;
	import java.util.List;
	/**
	*
	* @author rbaral
	*/
	public class ArrayWithSingleEntry {
	public static void main(String args[]){
	Integer [] arr = {723, 256, 668, 723, 140, 360, 597, 233, 128, 845, 737, 804, 986, 701, 906, 512, 845, 510, 510, 227, 430, 701, 366, 946, 464, 619, 946, 627, 209, 771, 424, 555, 959, 711, 530, 937, 716, 261, 505, 658, 706, 140, 511, 277, 396, 233, 819, 196, 475, 906, 583, 261, 147, 658, 517, 197, 196, 702, 944, 711, 128, 555, 149, 483, 530, 291, 716, 258, 430, 464, 601, 749, 149, 415, 802, 573, 627, 771, 660, 601, 360, 986, 291, 51, 415, 51, 227, 258, 937, 366, 923, 669, 33, 517, 417, 702, 475, 706, 110, 417, 275, 804, 500, 473, 746, 973, 669, 275, 973, 147, 817, 657, 277, 923, 144, 660, 197, 511, 793, 893, 944, 505, 322, 817, 586, 512, 322, 668, 33, 424, 962, 597, 144, 746, 345, 753, 345, 269, 819, 483, 368, 802, 573, 962, 583, 615, 208, 209, 269, 749, 256, 657, 619, 893, 959, 473, 753, 299, 396, 299, 500, 368, 586, 110, 793, 737, 615};
	List<Integer> myList = Arrays.asList(arr);
	DateFormat df = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss:S");
	Date d1 = new Date();
	System.out.println(df.format(d1));
	System.out.println(singleNumber(myList));
	d1 = new Date();
	System.out.println(df.format(d1));
	d1 = new Date();
	System.out.println(df.format(d1));
	System.out.println(singleNumberBitWise(myList));
	d1 = new Date();
	System.out.println(df.format(d1));
	}
	/**
	* check for without additional space and less time complexity
	*
	* @param a
	* @return
	*/
	public static int singleNumber(final List<Integer> a) {
	HashMap<Integer, Integer> numbers = new HashMap<Integer, Integer>();
	for(int i=0;i<a.size(); i++){
	if(numbers.containsKey(a.get(i))){
	numbers.remove(a.get(i));
	}else{
	numbers.put(a.get(i), 1);
	}
	}
	//whatever left is the one with single entry
	return (int)numbers.keySet().toArray()[0];
	}
	public static int singleNumberBitWise(final List<Integer> A) {
	int num = 0;
	for (int val : A) {
	num ^= val;
	}
	return num;
	}
	}

view raw ArrayWithSingleEntry.java hosted with ❤ by GitHub