LeetCode: Single Element in a Sorted Array

Problem Statement

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

Example 1:

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

Example 2:

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

Note: Your solution should run in O(log n) time and O(1) space.

Solution

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int L=0, U=nums.size()-1;
        int result;
        while(L<=U)
        {
            if(L==U)
            {
                result = nums[L];
                break;
            }
            int M = L+(U-L)/2;
            int consideredLength = M-L+1;
            if(consideredLength%2 == 1)
            {
                if(M-1>=0 && nums[M] == nums[M-1]) {
                    U = M-2;
                }
                else if(M+1<= nums.size()-1 && nums[M] == nums[M+1]) {
                    L = M+2;
                }
                else {
                    result = nums[M];
                    break;
                }
            }
            else {
                if(M+1<= nums.size()-1 && nums[M] == nums[M+1]) {
                    U = M-1;
                }
                else if(M-1>=0 && nums[M] == nums[M-1]) {
                    L = M+1;
                }
                else {
                    result = nums[M];
                    break;
                }
            }
        }
        return result;

    }
};

Solution Thought Process

I solved this problem using binary search. First, we get the middle element of the range using low and high.

We get the length considered by calculating:

consideredLength = M-L+1

1. Let's consider if this length is odd.

nums  1    1    2    2    3
idx   0    1    2    3    4

L = 0, U = 4

M = 0 + (4-0)/2 = 2

So[L, M] is odd in length. If this is the case, if nums[M] == nums[M+1], it means that the element can be found from element index M+2. So we make L=M+2

Let's see another case,

nums  1    2    2    3    3
idx   0    1    2    3    4

L = 0, U = 4

M = 0 + (4-0)/2 = 2

So [L, M] is odd in length. If nums[M] == nums[M-1] , it means that the element can be found before element index M-2 , included.

If those are not the case, then nums[M] is the result and we break the loop.

1. Let's consider if the considered length is even.

nums  1    1    2    3    3    5    5  
idx   0    1    2    3    4    5    6

L = 0, U = 6

M = 0 + (6-0)/2 = 3

So [L, M] is even in length. If nums[M] == nums[M+1], it means that the element can be found before element index M-1, included . So we make U = M-1

nums  1    1    2    2    3    5    5  
idx   0    1    2    3    4    5    6

L = 0, U = 6

M = 0 + (6-0)/2 = 3

So [L, M] is even in length. If nums[M]==nums[M-1], it means that the element can be found from element index M+1. So we make L=M+1.

If we have got L and U as the same element, we return the element as the result.

Complexity

Time Complexity: O(logn), we are making the consideration space half in every iteration.

Space Complexity: O(1)