Single Element in a Sorted Array | Amazon | Microsoft | Leetcode 540

[@SonamKumari-cm5ef]

Subhanallah 😊😊 kya shaandar samjhaya aapne

[@BiswajitRout-dl5ry]

most underrated channel for dsa keep up doing sir

[@Satya-g5t]

nice explanation with diagrams. 😄

[@souravjoshi2293]

Thanks a lot for making this look too easy

[@thakuradarsh3637]

thanks sir for providig us this types of explanantion

[@priyanshuchaturvedi3706]

after watching this......problem become piece of cake, THANKYOU

[@_Avinash_CSEB]

great work , keep up doing sir.

[@gamersgame43]

this is also a problem on bit manipulation taking xor sum, if the array wouldn't have been sorted

[@sarcaastech]

why ?

[@dayashankarlakhotia4943]

Class solution {
Public int single Non Duplicate (int[]nums){
int s=0 e=nums.length;
While (s

[@vivek.tiwary]

the if else block can be simplified.
public class Solution
{
public int SingleNonDuplicate(int[] nums)
{
/*
#1. elements are in pair, then in ideal case
it will start from even index & end at odd index
#2. for single element, element != nums[prev] && element != nums[next]
*/
int n = nums.Length;
if (n == 1)
return nums[0];
if (nums[0] != nums[1])
return nums[0];
if (nums[^1] != nums[^2])
return nums[^1];
int l = 2;
int r = n - 2;
while (l

[@harsh.jain22]

Similar solution but using indexes : Java Solution
class Solution {
public int singleNonDuplicate(int[] nums) {
int n = nums.length;
int s = 0;
int e = n-1;

while(s < e){
int mid = s + (e-s)/2;
boolean odd = false;
boolean even = false;
if((mid % 2) == 1){
odd = true;
}
if( (mid % 2) == 0){
even = true;
}
boolean isNextEqual = false;
if(nums[mid] == nums[mid+1]){
isNextEqual = true;
}
if( (odd && isNextEqual) || (even && ! isNextEqual) ){
e = mid;
}
else if( (odd && ! isNextEqual) || (even && isNextEqual) ){
s = mid+1;
}
}
return nums[s];
}
}

[@riaa55]

understood

[@dayashankarlakhotia4943]

Class solution {
Public int singlesNon Duplicate (int[]nums){
int s=0 e=nums.length;
While (s

[@Zenetz-le4mq]

class Solution {
public:
int singleNonDuplicate(vector& nums) {
int n = nums.size();
int ans = 0;

for(int i = 0;i

[@doomhead9109]

Thank you soo much

[@codestorywithMIK]

Thank you for watching ❤️❤️❤️

[@dayashankarlakhotia4943]

2nd approach correction s=mid+2

[@codestorywithMIK]

❤️❤️❤️

[@mereRevyoos]

I am watching this solution after attempting Amazon SDE Online Assessment

[@abhinav2161]

Can you guide me for prepation for coding test

[@utkarshnigam1573]

Hii can we do this question with map as well?

[@abhisheknishad2282]

time complexity by using map will become O(n logn) , which will be bigger than O(logn)

[@satyasanjay1339]

My solution is also the same but there is little difference. can you please check and tell me if it does make a difference in the performance of the code?
class Solution {
public int singleNonDuplicate(int[] nums) {
int start = 0;
int end = nums.length-1;
while(start < end){
int mid = start + (end-start)/2;
//Check number of elements on the right side
boolean isEven = (end-mid)%2 == 0 ? true : false;
if(nums[mid] == nums[mid+1]){
if(isEven){
start = mid+2;
}
else{
end = mid-1;
}
}
else{
if(isEven){
if(nums[mid-1] == nums[mid]){
end = mid-2;
}
else{
return nums[mid];
}
}
else{
start = mid+1;
}
}
}
return nums[start];
}
}