Count Subarray sum Equals K | Brute - Better -Optimal
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- čas přidán 26. 07. 2024
- Problem Link: bit.ly/3Kn10eZ
Please watch this video before watching this one: • Longest Subarray with ...
Notes/C++/Java/Python codes: takeuforward.org/arrays/count...
We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.
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00:00 Intro
01:03 Problem Statement
01:13 Sub-array definition
01:52 Example
03:13 Brute force solution
06:07 Complexity
06:29 Better solution
08:07 Complexity
08:21 Optimal Solution
08:41 Prefix sum concept
09:35 Thought process
13:55 Dry run
21:22 Code
22:38 Complexity
When I first started DSA, I was just so scared of the code and knowing very little about the logic and the intuition behind but Striver you are the one who has ever showed the completely difference - Extreme natural approach and observation and crystal clear logic. You've made the coding fun and interesting toward people like me. Hats off for your dedication! Understood!
i am doing his dsa sheet , the array part and almost every question has a completely different logic and approach and its becoming quite frustrating as im not able to come up with the solutions so do i just keep solving more problems or am i doing something wrong
keep going! I was in the same boat, you'll slowly be able to form logic@@utkarshverma3314
@@utkarshverma3314 same problem i am also facing
ashishbhattacharyya yup, it works for me.
could you please tell if i should make notes ?? how do i make them so that i dont put hours in it
I saw this video at around 10 am in college, was only able to think of o(n²) solution, I took the pause at your hint of using prefix sum array, kept thinking for some time while in college lectures but was able to get the solution at around 6pm 😝, I know it took time but it felt good to actually think of the solution, please keep up the consistency
Just when I think the video is getting long enough for a simple problem,
I go search the web and find not a single tutor with such a powerful explanation!
This man is my hero, I struggled so much with this question to visualize it and subsequent problems like it. I've done ove 270 LC questions, but this man made it so simple
understood bhaiya . You can not even imagine how much people love u and praise you and speak about your videos all the time. I am sure u get hiccups every now and then . Live long bhaiya u are our real teacher.
Its really amazing that you are taking your time in explaining via the dry run. It helps me a lot. Thanks :)
This explanation is so 100x better than Neetcode's. Thank you !!!
Thank you so much. I saw many other youtuber's video but this video is the best . I was able to understand after watching this video for 2 times
Understood! Super fantastic explanation as always, thank you very very much for your effort!!
I saw many times but I couldn't understand...But now i finally understood and realised that ur explaination is too good...!
Amazingly Understood, Thank you for such a in depth thought process
Struggled to understand this solution, now understood, Keep up the great work Striver
16:29 if we write if(sum==k) count++
We don't need to include 0 at beginning
good one!!
THAT IS WHAT I WAS DOING
Instead of adding 0 in the hashmap, we can just use the statement if(sum==k) ans++; That works fine.
See the Below Code
int subarraySum(vector& nums, int k)
{
int ans=0;
unordered_map x;
int sum=0;
for(int i=0;i
very good .
` if(x.find(sum-k)!=x.end()) ans+=x[sum-k]; `
And why this check before incrementing and.
class Solution {
public:
int subarraySum(vector& nums, int k)
{
int ans=0,sum=0;
unordered_map x;
for(int i=0;i
brilliant
So here, if in the question it is stated that it contains only positive as in codestudio then we can apply 2-pointer(Sliding window) as done in longest subarray with sum k. TC: O(N) SC: O(1)
Else, if it contains both positive and negative HashMap is optimal
Excellent explanation, understood ! ✨
That's really amazing 🤩Striver! code is too sort but logic is superb for the optimal. from brute (TC-> O(N^2), SC -> O(1)) to optimal (TC -> O(N) when using unordered_map, O(N * log N) when using ordered map, SC -> O(N) for using map data structure.
thanks striver you have really explained this so well
god bless you
you are a wonderful teacher for many 🙏🙏
Understood. Thank you so much!
Understood Bhaiya! This was a tough one to understand for me...will soon revisit it
Thank you🙏
Excellent explanation yaar! Great to see such kind of explanations that guide from brute force to optimised solution in such a detail! Thank you so much and keep making such great videos :)
Really good explanation & kudos to efforts, understood 👍
You have given more than anyone can ask for. Thank you for all your support and such amazing videos. They are really helpful. Just a small thing can we have a dark mode setting for "take you forward" pages.
You can build one chrome extension may be ;)
appreciates striver by telling "u have given more than anyone can ask" and proceeds to place demand .
@@takeUforward lol would be a good project for the resume?
striver actually give the dark mode option
Previous explanation and current explanation is good .understood
God for beginner❤hats of u bhaiya
Understood Striver ❤. You are a living legend sir. 🤩🤩
Understood💖amazing explanation
Striver you are great 🙌🙌 .I just want to thank you ❤️.
Understood, thanks!
understood sir ,great explanation
tks for explanation, you are savior
Understood, thank you.
The sliding window approach actually performs better with O(n) time complexity.
int subarraySum(vector& nums, int sum) {
int count = 0;
int currSum = 0;
unordered_map sumFreq;
for (int i = 0; i < nums.size(); ++i) {
currSum += nums[i];
if (currSum == sum)
count++;
if (sumFreq.find(currSum - sum) != sumFreq.end())
count += sumFreq[currSum - sum];
sumFreq[currSum]++;
// If currSum becomes greater than sum,
// subtract elements from the start
while (currSum > sum) {
currSum -= nums[i - count + 1];
sumFreq[currSum]--;
if (sumFreq[currSum] == 0)
sumFreq.erase(currSum);
count--;
}
}
return count;
}
understood captain...thanks
understood ,thnx for amazing video ❤❤❤❤❤❤
Thanks bhaiya, understood
thanks really good explination
you are awesome bhaiya...
Was able to think O(N^2) approach and totally underatood optimal solution too 😊
Thanks for the help raj bhaiya
solution blew my mind
Thank you Striver. I owe you❤
Understood 💯💯💯
Understood ❤️
Understood Sir!
Thanks bro. Understood
Understood✅🔥🔥
Thank you!!
Understood !! 😍😍
you are topG of DSA
Nice explation ❤
thank you so much
very nice. .thankyou
Hi bhaiya it would be really helpful if you can give similar question links for practise after every question explanation!!
Timestamps
01:03 - Problem Statement
01:13 - Sub-array definition
01:52 - Example
03:13 - Brute force solution
06:07 - Complexity
06:29 - Better solution
08:07 - Complexity
08:21 - Optimal Solution
08:41 - Prefix sum concept
09:35 - Example
13:55 - Dry run
21:22 - Code
22:38 - Complexity
understood striver thanks
Understood 🎉
Understood!
Understood ❤
Awesome Awesome Sir...............
Understood❤❤
Thanks for the great video. I don't see how I can come up with the optimal solution during the interview without hint from the interviewer.
The Best♥
nice ..........explanation
Understood😁😉
understood!!!
UNDERSTOOD
for me optimal is little hard but after see 2-3 time video , got easly understand
thank you
Amazing
Understood.
Thanku sir
though I am still figuring out on the map[0]=1 part but surely the rest of the part was indeed clear. thanks for these videos.
if you select no element in sub array then , map[0]=1
underrstood
undestood ✌
Understood
Please also make video for count pairs with given sum and divisible by k
Regarding putting (0,1) int the Hashmap initally :
Let consider somehow the preSum got exactly equal to our k, means now the sub-array till that presum element gives us the k (~ preSum-k = 0).
means whenever we find our preSum touching the k value, we need to increment count but we will not be able to find it in hashMap as it is yet to be inserted that why
either increment count based on if preSum-k = 0 or put a (0,1) in hashmap to automatically finding preSum-k = 0 in the hashMap.
int preSum = 0, count = 0, start = 0;
Map mp = new HashMap();
for(int i=0;i
Storing (0,1) in the hashmap is bit confusing, instead we can just increase a check i.e. if(currSum == k) count++. The more readable code(JAVA) will look like this:
public int subarraySum(int[] nums, int k) {
int currSum = 0, count = 0;
int n = nums.length;
HashMap map = new HashMap();
for(int i=0; i
mpp[0] = 1 helps us to take the count when (sum == k ). Because when sum will be k, then remove will be 0. And then mpp[1] will give us 0, which will be added to the count. If we don't store 0 to our map. then we need to handle the case in an extra if condtion. which is if(sum == k) count++;
Awesome explanation, one approach of sliding window will not work in this problem if array contains negative elements also.
if instead of specifying map[0]=1 in the beginning if we give a condition that if(preSum == k) { count ++;}
Will it be correct Striver bhaiya??
we can also use sliding window 2 pointer approach for this one code->
int findAllSubarraysWithGivenSum(vector < int > & arr, int k) {
int n =arr.size(),sum = 0,cnt = 0,left = 0;
for(int right = 0;right k) sum-=arr[left++];
if(sum == k) cnt++;
}
return cnt;
}
It will fail .arr[]=[1] k=0
@@amitranjan6998 yes we just need to add a condition of left
Sir, May I know time complexity of the optimal approach in java
Striver you can make these playlists a bit fun if you allow your viewers to suggest you problems they think discussable !!
btw loving the playlist
Plz don't ruin it for us 😂😂😂
plz no
understood
Can we do this first calculate the prefix sum and than suffix sum and than we subtract each and sum the prefix and suffix
It seems like you have edited your voice in the video and made it soft, I am missing that bold voice which is there in the DP playlist. Please don't edit your voice, you don't have to adjust yourself for us. We like the way you are, the original STRIVER🔥🔥
But we need to look at the future, we want to go global, so audio quality has to be maintained, I hope you get that :)
@@takeUforwardand we love you for that too,
It's sort of unconditional 😂
We can also solve this problem in O(1) space complexity but the T.C will be 2N using two pointer approach.
Without adding to map, you need to count occurrences still not in map
public int subarraySum(int[] nums, int k) {
int sum=0;
int count =0;
Map map = new HashMap();
for(int i=0;i
22:54 *please explain what lines*
int remove = presum - k;
count += mp[remove];
mp[presum]++;
do ?
Understand
Can you pls pls plsssss do strings before binary search next plsss🙏 ? From 2023 batch🥺. Wish we could have the entire series in our 1st 2nd 3rd yrs
nice
tq
understood , but took tom much time to understood how to update the frequency if there exist a prefix sum in the HashMap
Bhaiya ye Question ka video dekhne se phale hoo gaya 😁😁Only Minor change
int findAllSubarraysWithGivenSum(vector < int > & nums, int k) {
int right = 0;
int left = 0;
int count = 0;
long long sum = nums[0];
int n = nums.size();
while(right < n){
while(left k){
sum = sum - nums[left];
left++;
}
if(sum == k){
count++;
}
right++;
if(right < n){
sum+=nums[right];
}
}
return count;
}
Hey Strivers Good explanation ❤
Can you just change the gfg link of this question it redirect to same page.
Once again thank you 👍
we can also do this problem by sliding window technique
What if the array contains 0s?