************ more optimized ************ int findMin(vector& nums){ int left = 0, right = nums.size() - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; }
guys, this is one of the most important question if u r preparing for amazon and adobe.. hell they asked it a lot in previous 6 months according to Leetcode stats..
If hurts the most , when you realize this easy question took you more than 2 hour and failed several time while in the test cases below 5 when you submit in leetcode. I did a course on java in youtube from a famous youtuber an year ago. I took a short break of two months for my semster exam and other activities. I can remind that I solved a similar type of question in an year ago. But now , I am struggling to do a easy problem like this.... Taking a break, even it is small gets your mind out of the programming mode.
00:38 Problem statement 01:21 Brute-force approach 01:38 Optimal approach 06:30 Dry-run 10:35 Pseudocode 12:57 Code 13:20 Optimised version : If you want try to do then 15:04 Code 15:27 Complexity
I felt proud of myself of solving this question in o(logn) time without even looking this video cz obviously I've seen your previous 2 videos and of course able to solve this one 😎.. Thank you Striver.
man what an explanation ! I saw several solutions on leetcode discuss but couldn't understand all of it, some if else conditions were going above my head, but you made it so clear I did code it myself both leetcode 153 and 154 with AC solutions
I was solving this for the first time , was struggling because everytime was giving me runtime error and guess what i lerally watched the first 10 mins only infinite times , because my dumb brain was just not convincing myself that how is this done, at 3pm night i passed this that too in 0ms time without seeing your written code 😭😭😭😭😭. It was giving me stress because i haven't been able to pass a single solution that day so i really needed anything to get accepted just to boost my morale . Thanks striver.
I have gone through some solutions to this problem and able to understand but if I try to recollect and code it later I would be a little confused. But your explanation and approach is so detailed and very easy to understand that I am able to solve it on my own without any confusion (even after many days). Thanks a lot!!
Another approach I thought of - -> If left array is sorted, right is unsorted. Then go to right as pivot point can be current element or in right array -> If right is sorted, left is unsorted. Then go to left as pivot point can be current element or in left array -> If both are sorted then break and just compare the ans with arr[low] as complete search space is sorted. int findMin(vector& arr) { int low = 0; int high = arr.size() - 1; int ans = INT_MAX; while(low
Thanks Striver. Adding my thoughts for (duplicates) . If we apply the previous logic of high-- or low++ as the mid and low/high values are equal , we will end up getting the minimum element but that does not gaurentee the no of times array has been rotated . Examples: array=[1,1,2,1,1] , orignal array = [1,1,1,1,2]. The correct answer is 3 (as the element at 0th index has been moved to 3rd index [1,1,2,1,1]). But if we apply the previous logic, the answer would come as 0 as 0th index is the minimum element. To upsolve this, we can do the following : Keep reducing high, untill [high-1] > [high] (2>1). Once you attain this point, high(3rd index) will be the answer.
This works as well: first check if the right side of mid is sorted then check of left side. int findMin(vector& nums) { int ans=INT_MAX; int low=0; int high=nums.size()-1; while(low
solved without any help with BS, but before solving i have watched the previous two videos of the playlist which helped in developing the thinking skill for solving this problem
I guess this algorithm fails for the below input: arr = [3,1,2] Your explanation is so clear bro❤ This condition should be added: min = Math.min(min,nums[mid]);
We know that if it is sorted in nature then arr[low] would always be the minimum number of all. by considering this we can simplify this as following. low=0 high=len(nums)-1 while lownums[high]: low=mid+1 elif nums[low]>nums[mid]: high=mid else: break return nums[low]
Other way to solve this problem is by pivot search.. Here is the approach, if(nums[nums.size()-1] > nums[0]) return nums[0]; // find pivot int s = 0, e = nums.size()-1, mid = s + (e - s) / 2; while(s < e){ if(nums[mid] >= nums[0]){ s = mid + 1; } else{ e = mid; } mid = s + (e - s) / 2; } return nums[mid];
00:04 Find the minimum in a rotated sorted array using binary search. 02:08 Identify the sorted half in a rotated sorted array 04:19 The left portion of the rotated sorted array contains the minimum. 06:37 Identifying and eliminating the minimum element in a rotated sorted array. 08:40 Finding the smallest element in rotated sorted arrays using binary search. 11:06 Binary search on a rotated sorted array 13:13 Optimization in identifying sorted search space 15:16 Binary search stops in a sorted search space, improving time complexity
How I thought about this solution is, the minimum is always in the left part of the array in a normal sorted array. However in a rotated sorted array the minimum is always in the unsorted half. So we move to the unsorted right half if it exists. Otherwise we always move to the left. while(low
I think this below code is simpler and easier to understand: int findPivot(vector &nums){ int low = 0; int high = nums.size()-1; int pivot =0; // if not rotated, it should return 0 while (low
For duplicates I think the whole code will remain same only the optimized part will be changed as if (a[low]==a[mid] and a[low]==a[high]) {ans=min(ans,arr[low]); Low++; High--; Continue; Please check whether it is correct or not .
yea you are right so basically the duplicates case arises when there is like mid low high all three are same so basically this case is same as search in a rotated sorted array 2 right? there also this case has rose.
slightly lengthier, but core logic is , we need to handle 2 cases where we are uncertain where to move, that was the crux for me, a) mid is less than lo and hi, b) mid greater than lo and hi, if we exclude that , then searching is regular BS, attaching java code for the same int ans = Integer.MAX_VALUE; while(lo
@@harishms5330 So consider an example arr=11111111111111 size=14 it will take about O(7), if arr[low]==arr[mid]==arr[high] low++; high--; till low crosses high, it will follow the condition , at one time, low increases to 1 and high decreases to 1, so arr size reduced by 2 similarly it will go till O(n/2).
I think This is more intutive moving towards unsorted and if sorted move left while(low arr[mid] ) high = mid - 1; else if( arr[mid] > arr[high] ) low = mid + 1; else { ans = min(ans , arr[low] ); break; } } return ans;
Solution based on applying binary search on sorted side and storing start index element as possible solution : tops 100% in leetcode int findMin(vector& nums) { int ans=INT_MAX; int start=0; int end=nums.size()-1; // logic is to find sorted side and store start index value into min each time while(start
You don't know how relived I am to know that you won't change format, I was afraid that it is some advance technique. I always hated binary search, hope I wouldn't after this your Binary Search Playlist
Can also be done by finding pivot. class Solution { public: int findMin(vector& nums) { int n=nums.size(); int left=0,right=n-1; while(right >= left){ int mid=(left+right)>>1; int val=nums[mid]; int prev=1e9; int next=1e9; if(mid-1>=0)prev=nums[mid-1]; if(mid+1 val && next > val)return val; if(val >= nums[left] && val >= nums[right]){ left=mid+1; } else right=mid-1; } return -1; } };
@@floatingpoint7629 In duplicates we cant return index because it would have no meaning consider [3,3,3,3,3] this as example, minimum here is 3 but index is not defined because index can be anything here. Please correct me if i am wrong
To find minimum element in rotated sorted aray with duplicates,just add the condition used before while finding an element in rotated sorted array with duplicates and the solution works fine.
i think below code is easier class Solution { public: int findMin(vector& nums) { int n = nums.size(); int low=0, high=n-1; while(low nums[high]) low=mid+1; } return -1; } };
Please comment understood and give us a like if you got everything :)
mass
Understood
************ more optimized ************
int findMin(vector& nums){
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return nums[left];
}
East or west. Striver is the best
I came only for Minimum in Rotated Sorted Array so why I would seen previews video #striver ?
the feeling of writing the exact same code by your own without even looking at the lecture is insane, ALL thanks to your level of explanation
Abe easy hai ye upper ke 3 phele se kiye hue the esiliyee btw good
@@ravikr3463
@@ravikr3463 bhai merse wo karne ke baad bhi nahi hua yrr
guys, this is one of the most important question if u r preparing for amazon and adobe.. hell they asked it a lot in previous 6 months according to Leetcode stats..
Thanks brother 🙏🏾
The fact that you acknowledged the problem of having different formats in binary search deserve a like. keep up the good work!
Understood!! Hi I'm a Pakistani and I study at US. I really like your DSA playlist and recommend it to everyone. Thank you so much for your effort.
Pucha kisine?
Ab ji*@d recruitment mai bhi dsa poochne lage kya? 😂😂
6 minutes into the video, I was able to deduce the whole logic as soon as I understood the approach !!!!!! Amazing content !!!!!!
The way you cover each and every edge case is amazing. Understood everything 👍
just watched like 7 min and understood the solution and what i am missing out
thanks man you are awesome
If hurts the most , when you realize this easy question took you more than 2 hour and failed several time while in the test cases below 5 when you submit in leetcode.
I did a course on java in youtube from a famous youtuber an year ago.
I took a short break of two months for my semster exam and other activities.
I can remind that I solved a similar type of question in an year ago.
But now , I am struggling to do a easy problem like this....
Taking a break, even it is small gets your mind out of the programming mode.
bro where did u do dsa in java from?
00:38 Problem statement
01:21 Brute-force approach
01:38 Optimal approach
06:30 Dry-run
10:35 Pseudocode
12:57 Code
13:20 Optimised version : If you want try to do then
15:04 Code
15:27 Complexity
I felt proud of myself of solving this question in o(logn) time without even looking this video cz obviously I've seen your previous 2 videos and of course able to solve this one 😎.. Thank you Striver.
the same exact code i was able to write down without even watching your lecture first , kudos to u🙌❣
Striver sir plz start a series for strings to simplify strings for us just like you did with the other topics🙏
I'm really blessed to watch and learn the concept form you striver and i'm solving the problems now without referring the solutions. Thanks a lot.
man what an explanation ! I saw several solutions on leetcode discuss but couldn't understand all of it, some if else conditions were going above my head, but you made it so clear I did code it myself both leetcode 153 and 154 with AC solutions
I was solving this for the first time , was struggling because everytime was giving me runtime error and guess what i lerally watched the first 10 mins only infinite times , because my dumb brain was just not convincing myself that how is this done, at 3pm night i passed this that too in 0ms time without seeing your written code 😭😭😭😭😭. It was giving me stress because i haven't been able to pass a single solution that day so i really needed anything to get accepted just to boost my morale . Thanks striver.
I have gone through some solutions to this problem and able to understand but if I try to recollect and code it later I would be a little confused. But your explanation and approach is so detailed and very easy to understand that I am able to solve it on my own without any confusion (even after many days). Thanks a lot!!
Understood, the last optimisation was great !!
Another approach I thought of -
-> If left array is sorted, right is unsorted. Then go to right as pivot point can be current element or in right array
-> If right is sorted, left is unsorted. Then go to left as pivot point can be current element or in left array
-> If both are sorted then break and just compare the ans with arr[low] as complete search space is sorted.
int findMin(vector& arr)
{
int low = 0;
int high = arr.size() - 1;
int ans = INT_MAX;
while(low
Was wondering, What if I search for target = INT_MIN and keep the code same as the previous video wont the arr[low] be my answer?
you are binary search playlist is goated bro
I used to think that neetcode gives the best explanation, but after watching this playlist i changed my thoughts.
Thanks Striver. Adding my thoughts for (duplicates) .
If we apply the previous logic of high-- or low++ as the mid and low/high values are equal , we will end up getting the minimum element but that does not gaurentee the no of times array has been rotated .
Examples:
array=[1,1,2,1,1] ,
orignal array = [1,1,1,1,2].
The correct answer is 3 (as the element at 0th index has been moved to 3rd index [1,1,2,1,1]).
But if we apply the previous logic, the answer would come as 0 as 0th index is the minimum element.
To upsolve this, we can do the following :
Keep reducing high, untill [high-1] > [high] (2>1).
Once you attain this point, high(3rd index) will be the answer.
this does not cover all the cases
I think if we remove the condition of checking if(a[low]
u unfolded the mysterious binary search....thank you!!!
YOU ARE AN AMAZING TEACHER! THANKYOU FOR EXISTING!
Understood! Awesome explanation as always, thank you soooo much for your effort!!
This works as well: first check if the right side of mid is sorted then check of left side.
int findMin(vector& nums) {
int ans=INT_MAX;
int low=0;
int high=nums.size()-1;
while(low
woow bro, this is crazy, legendary explanation, hats off man
Best DSA course on you tube.
solved without any help with BS, but before solving i have watched the previous two videos of the playlist which helped in developing the thinking skill for solving this problem
Understood sir thank you very much, sir. Your teaching style is really amazing. I hope I will crack my interview.
Understood...Awesome Binary search Playlist.. 👏
I guess this algorithm fails for the below input:
arr = [3,1,2]
Your explanation is so clear bro❤
This condition should be added:
min = Math.min(min,nums[mid]);
ya you have to save mid as a probable answer and compare it while moving on left
change it to nums[low] , you might get your answer, striver's code passes all the testcases
Great explanation! Thank you so much.
Thanks a lot for making such videos.
Your content is amazing.
Keep on doing the great work
We know that if it is sorted in nature then arr[low] would always be the minimum number of all. by considering this we can simplify this as following.
low=0
high=len(nums)-1
while lownums[high]:
low=mid+1
elif nums[low]>nums[mid]:
high=mid
else:
break
return nums[low]
Understood. Very Good Explanation
Other way to solve this problem is by pivot search..
Here is the approach,
if(nums[nums.size()-1] > nums[0]) return nums[0];
// find pivot
int s = 0, e = nums.size()-1, mid = s + (e - s) / 2;
while(s < e){
if(nums[mid] >= nums[0]){
s = mid + 1;
}
else{
e = mid;
}
mid = s + (e - s) / 2;
}
return nums[mid];
00:04 Find the minimum in a rotated sorted array using binary search.
02:08 Identify the sorted half in a rotated sorted array
04:19 The left portion of the rotated sorted array contains the minimum.
06:37 Identifying and eliminating the minimum element in a rotated sorted array.
08:40 Finding the smallest element in rotated sorted arrays using binary search.
11:06 Binary search on a rotated sorted array
13:13 Optimization in identifying sorted search space
15:16 Binary search stops in a sorted search space, improving time complexity
How I thought about this solution is, the minimum is always in the left part of the array in a normal sorted array. However in a rotated sorted array the minimum is always in the unsorted half. So we move to the unsorted right half if it exists. Otherwise we always move to the left.
while(low
I think this below code is simpler and easier to understand:
int findPivot(vector &nums){
int low = 0;
int high = nums.size()-1;
int pivot =0; // if not rotated, it should return 0
while (low
Bhaiya your explanation is relay fabulous. It makes hard concepts easily understandable to us. Thank you bhaiya for helping us in dsa.
Thanks for the incredible knowledge u give us...understood!!
For duplicates
I think the whole code will remain same only the optimized part will be changed as if (a[low]==a[mid] and a[low]==a[high])
{ans=min(ans,arr[low]);
Low++;
High--;
Continue;
Please check whether it is correct or not .
yeah its correct dude !! thanx i was stuck there 🤖
yea you are right so basically the duplicates case arises when there is like mid low high all three are same so basically this case is same as search in a rotated sorted array 2 right? there also this case has rose.
slightly lengthier, but core logic is , we need to handle 2 cases where we are uncertain where to move, that was the crux for me, a) mid is less than lo and hi, b) mid greater than lo and hi, if we exclude that , then searching is regular BS,
attaching java code for the same
int ans = Integer.MAX_VALUE;
while(lo
understood ! the best explanation bhaiya , you are hero and a real gem❤
You beat neetcode on this🎉. Superb dude.
Understood.. Awesome Binary search Playlist..😍
understood, for duplicate elements worst case time complexity will be O(n/2).
how
@@harishms5330 So consider an example arr=11111111111111 size=14 it will take about O(7),
if arr[low]==arr[mid]==arr[high] low++; high--; till low crosses high, it will follow the condition , at one time, low increases to 1 and high decreases to 1, so arr size reduced by 2 similarly it will go till O(n/2).
@@RishabhKumar0094 thank u🫡
Good explanation. Explanation in depth
thanks striver understood everything
got it , understood everything
I think This is more intutive moving towards unsorted and if sorted move left
while(low arr[mid] )
high = mid - 1;
else if( arr[mid] > arr[high] )
low = mid + 1;
else
{
ans = min(ans , arr[low] );
break;
}
}
return ans;
Very good explanation
Thank you
Sir completed the problem which you gave as homework 😇😇
amazing explanation . thanks a lot
a random tip : if you dont understand this (because at first I didn't get it ) take your pen and paper and dry run the example of 45123
@striver take case a = [5[L] 1 2[M] 3 4[H]] mid is lesser in both arrays 2
Super Explanation 🎉🎉
Solution based on applying binary search on sorted side and storing start index element as possible solution : tops 100% in leetcode
int findMin(vector& nums) {
int ans=INT_MAX;
int start=0;
int end=nums.size()-1;
// logic is to find sorted side and store start index value into min each time
while(start
Amazing explanation!!!!!
understood sir. Huge kudos to you :)
You don't know how relived I am to know that you won't change format, I was afraid that it is some advance technique. I always hated binary search, hope I wouldn't after this your Binary Search Playlist
i am learn from you , one day i will cross you 🙇
My thought process:
int low=0;
int high=arr.length-1;
int min=arr[high];
while(lowmin)
{
low=mid+1;
}
else{
min=arr[mid];
high=mid-1;
}
}
return min;
Can also be done by finding pivot.
class Solution {
public:
int findMin(vector& nums) {
int n=nums.size();
int left=0,right=n-1;
while(right >= left){
int mid=(left+right)>>1;
int val=nums[mid];
int prev=1e9;
int next=1e9;
if(mid-1>=0)prev=nums[mid-1];
if(mid+1 val && next > val)return val;
if(val >= nums[left] && val >= nums[right]){
left=mid+1;
}
else right=mid-1;
}
return -1;
}
};
Hometask(LC-154):-
class Solution {
public:
int findMin(vector& nums) {
int s=0;
int e=nums.size()-1;int ans=INT_MAX;
while(s
if interviewers ask you to modify it so it returns the index, this won't work
@@floatingpoint7629 In duplicates we cant return index because it would have no meaning
consider [3,3,3,3,3] this as example, minimum here is 3 but index is not defined because index can be anything here. Please correct me if i am wrong
@@6mahine_mein_google if all are duplicates then array is not a rotated one. it is guaranteed that array is at least rotated once
understood good job bro
int findMin(vector& nums) {
int n = nums.size();
int low = 0 , high=n-1;
while(low
Bahut ache se smgh aaya sir ❤
Understood Sir. Thank You Sir 👌👌
Understood perfect bro
understood, thank you bhaiya
The mid will point to the minimum element at least once
To find minimum element in rotated sorted aray with duplicates,just add the condition used before while finding an element in rotated sorted array with duplicates and the solution works fine.
can you share the code
I actually thought the same idea by myself 😁
Understood Very Well!
understood, thanks a lot!
did this on my own in one go.
Understood👍 thank you sir
I first like💓💓 it then watch it because I know it is the best.😎😎
while(low
understood all clear ❤
Understood, thank you.
Understood, Thanks a lot
UNDERSTOOD❤
Great question!
@takeUForward , can you also add find pivot element in Sorted Roatated array .i am unable to find it out in your playlist
UNDERSTOOD
Understood Bhaiya 😃
Understood everything
very well explained
Understood Bhaiya!
amazing!
Hey god bless you bro❤
Absolutely understand ❤
nice explanation
Super level understood
What is the elements in the array are not unique let array ={5,4,1,2,2,2,2,2} then what should be the approach
question is wrong if the array is rotated 4 cannot come after 5 as it is smaller than 5
i think below code is easier
class Solution {
public:
int findMin(vector& nums) {
int n = nums.size();
int low=0, high=n-1;
while(low nums[high])
low=mid+1;
}
return -1;
}
};
if my array is [2, 4, 1, 2], In this case wouldn't the optimised code would give the answer 2 (because of the if condition (ar[low]
your array has a duplicate element, and the code is for the unique element.