Search in Rotated Sorted Array | LeetCode problem 33

[@adesoyeshubomi4010]

Best Explanation I've seen !!! Thank you 🙏🙏🙏🙏🙏🙏 keep up the great work!

[@sidharthsharma2147]

Great logic building!!⭐

[@mayappapujari9259]

very easy solution.... make more videoes on leetcode problems

[@imtiyazali5894]

perfect code for pass all case of leetcode:=>
public int search(int[] nums, int target) {
int start = 0;
int end = nums.length-1;
while(start nums[mid] && target

[@sanjaymurugasen3386]

please continue the top interview 150 leetcode question series 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@rajukomaturi4341]

Great explanation always , your solution wont work for this array int[] nums = {7, 8, 9, 0, 1, 2, 3, 4, 5, 6}; , i think you have to correct left portion of array checking with if(nums[start]

[@navalprakash3727]

class Solution {
public int search(int[] nums, int target) {
for(int i=0; i

[@amanchand01]

you saw it too 🍻

[@lv8705]

hello mam . in else part we should write target

[@PramodK-y4m]

Can you share git repository?

[@cheerlachinnikrishnasai1947]

Thanks

[@SandhyaraniTorase]

Thank u mam

[@KapilSharma-qh4zi]

But it is not working on [5,1,3] array

[@shubham_shubham]

you can do a small change here .. change second if condition inside while loop as if(nums[st]

[@navalprakash3727]

class Solution {
public int search(int[] nums, int target) {
for(int i=0; i

[@ChinmayPatil-22]

bro it's time complexity is O(n) we have to do the question by O(log n)✌

[@garun4950]

@@ChinmayPatil-22 and also they ask to print -1 if we dont find element, we cant run another loop as it increases time complexity

[@allasandeep1466]

class Solution {
public int search(int[] nums, int target) {
if(nums.length == 0 || nums == null){
return -1;
}
int n= nums.length;
int i;
for( i=0;i