Array interview problem | Java Code | Find Minimum in Rotated Sorted Array :- Bind 75 : Leetcode 153

#array #interview #problem #binary #search #array #sorting

Learning Array problem for interview with java code, and optimal solution, with Big O time and space complexity. This will also be a walkthrough the problem, understanding core concept of array traversal, hashing, hashtable, sorting, sliding window and array traversal, using And and Or operator at binary level.

Leetcode Medium: array traversal + Binary search.

Not stopping till we get employed in FAANG.

Problem asked recently in following companies in coding interview:
Facebook interview,
Amazon interview,
Microsoft interview,
Bloomberg interview,
Goldman Sachs interview,
Uber interview,
Adobe interview,
Oracle interview,

List of Most Important Questions and Solutions:

GitHub Repository for Questions solved so far:

My dream IT companies:
FAANG,
Facebook,
Google,
Amazon,
Netflix,
Microsoft,
apple,
doordash,
roblox,
stripe,
instacart,
uber,
lyft,
twitter,
linked In,
pinetrest,
bloomberg,
robinhood,
box,
two sigma,
byte dance,
tik tok,
air bnb,
nuro,
ui path,
oracle,
twitch,
data bricks,
waymo,
dropbox,
coinbase,
snowflake,
snap,
nvdia,
broadcom,
slack,
intel,
cisco,
indeed,
salesforce,
reddit,
wayfair,
okta,
splunk,
service now,
coursera,
square,
upgrade,
mozilla,
yelp,
unity,
ebay,
affirm,
stach adapt,
amplitude,
github,
wish,
brex,
etsy,
chime,
shopify,
amd,
GoDaddy,
IBM,

[DestinationFAANG]

Coding

class Solution {
public int findMin(int[] nums) {

int left = 0;
int right = nums.length - 1;
int result = nums[0];
if(right == 0){
return nums[0];
}
while (left <= right){

if(nums[left] < nums[right]){
result= Math.min(result, nums[left]);
}


int mid = (left + right)/2;

result = Math.min(result, nums[mid]);

if(nums[left] <= nums[mid]){
//result = Math.min(result, nums[left]);
left = mid+1;
}
else right = mid-1;
}

return result;
}
}