Lecture 29: Kadane's Algorithm || Prefix and Suffix Sum || Array into 2 equal Sum Subarray

[@CoderArmy9]

You Guys are Love❤

[@YashSaini007]

Good Morning Bhaiya ❣️

[@Kalamkaar2607]

@CoderArmy9 is love

[@bonkai.movies]

bhaiya your debugging process in blackboard is making you diffrent then others keep it up BestOluck

[@HONESTLYALOK]

Thank you bhaiya

[@VivekMishra-yq5us]

❤❤

[@Updatedknowledge9]

Bhaiya aap ek baat universal truth hai, ki starting me sab easy lagta hai, fir hard iske baaad, fir se sab easy lagne lagta hai ☺☺☺❤❤❤, main to aise hi feel kar rha hu,

[@inspirationalyouthicons4998]

1:15:03 hw maximum diffrence between two elements:- int maximumdiffrencebetweentwoelements(int arr[],int n){ int suffix[n]; suffix[n-1] = arr[n-1]; for(int i=n-2;i>=0;i--){ suffix[i] = max(suffix[i+1],arr[i]); } int ans = INT_MIN; for(int i=0;i

[@Anjali-ty2fk]

Aap ek baar iss solution main mistake bta do please int suffix=nums[n-1]; int ans=INT_MIN; for(int i =N-2;i>=0;i--) { suffix=max(nums[i],suffix); ans=max(ans,suffix-nums[i]); } return ans; } return -1;

[@ronitapatra]

@@Anjali-ty2fk first you have to calcualte the difference then update suffix...otherwise suffix = nums[n-1] will update first...

[@ujjwalgaur5460]

@@Anjali-ty2fk dont know this method i solved through two pointers mthd#include using namespace std; int main() { int arr[8] = {9, 8, 8, 12, 0, 3, 7, 4}; int s = 0, e = 1; // Start and end pointers int maxdiff = 0; // To store the maximum difference int n = 8; // Size of the array while (e < n) { if (arr[e] > arr[s]) { // Calculate the difference int diff = arr[e] - arr[s]; // Update maxdiff if the current difference is greater if (diff > maxdiff) { maxdiff = diff; } } else { // Move the start pointer to the current end pointer s = e; } // Always move the end pointer forward e++; } cout

[@Sajal_Mallick_vlogs]

Day 39/180 done and dusted.... Awesome explanation and exceptional learning.. Thank you bhaiya... #consistency

[@imavinashsingh]

0:00:01 [Prefix and Suffix Sum] 0:09:50 [Divide Array in two Sub arrays with equal sum ] 0:28:15 [Code Part] 0:35:15 [Largest Sum Contiguous Sub array] 0:47:40 [Kadane's Algorithm ] 1:01:15 [Code Part] 1:03:08 [Max. Difference b/w 2 elements]

[@TheSilentObserver30]

Day 39/180..Growing strong each day and learning new everyday. You are making us consistent and confident. Thank you.

[@sarikasaxena5567]

suffixsum array H.W- vectorsuff(n); suff[n-1]=arr[n-1]; for(int i = n-2;i>=0;i--){ suff[i] = suff[i+1]+arr[i]; }

[@rvpool9830]

To print all subarrays: #include using namespace std; int main(){ int n; cin>>n; int arr[n]; for(int i=0; i> arr[i]; for(int i=0; i

[@xavdok]

I run your code and I did not get the ans, your code is wrong.

[@maroofgpm]

5:08 Likh Diya!! // Homework: Sum of suffix #include #include using namespace std; int main() { int vect[] = {1,2,3,4,5}; int n = 5; for (auto i : vect) { cout

[@joydeep-halder]

5:10 H/W Suffix Sum: suffix[n-1] = arr[n-1] for(int i=n-2; i>=0; --i) { suffix[i] = suffix[i+1] + arr[i] }

[@lotifurshabbir7302]

why you are giving answer ☹️

[@joydeep-halder]

@@lotifurshabbir7302 To help others 😊

[@yashisrivastava6196]

Will never forget kadane's algorithm now ✨

[@VivekMishra-yq5us]

Chamk gya hai concept ❤ bhiya

[@PandaSingh-o3j]

At 48:45 when you add 7 and -5 it is 2 but don't replace it with maxi as maxi contains 7.

[@mohdshabaz9675]

Good explanation of previous videos

[@CoderArmy9]

Good morning bhai❤

[@rockrock350]

never saw this much of dedication

[@ajaythedaredevil7220]

Print all Subarray in an array: for i in range(len(res)): for j in range(i+1,len(res)): print(res[i:j])

[@PotatoPcGamer01]

1:15:03 - Find Largest Difference between 2 elements (HOMEWORK) Time Complexity - O(N) Code : int largestDiff(vector& v, int index) { // 1) initialization : int size = v.size()-1; int difference = 0, ans=INT_MIN, findMax=INT_MIN; // 2) handling edge cases : if(index size) return -1; // 3) find max difference : for(int i = size; i>=index; i--){ findMax = max(findMax, v[i]); difference = findMax - v[i]; ans = max(ans, difference); } return ans; } PS - Suggestions / Replies are welcomed :)

[@joydeep-halder]

9:32 H/W Solution: Print all subarrays: for (int size = 1; size

[@joydeep-halder]

​@@ddoy7378bhai compiler me run karke dekho.

[@arjungurjar7702]

1:14:30 int maxDifference(int arr[],int n){ int ans = INT_MIN ; int suffix = 0 ; for(int i = n-1 ; i >=0 ; i--){ suffix = max(suffix , arr[i]); ans = suffix - arr[i]; cout

[@RohitCoder]

at 32:17 why dont we used for(int i=0; i

[@bee.eee.]

n-1 refers to the last element If we take i

[@atifmalik8012]

59:29 //Kadane's algorithm class Solution { public: // Function to find the sum of contiguous subarray with maximum sum. int maxSubarraySum(vector &arr) { int prefix = 0,MaxSum = INT_MIN; for(int i = 0;i

[@AryanShrivastav-n2v]

Gltt ha bhai..prefix vala line sum kr nicche aayga

[@rahulpatil6871]

yaha tk smja ya ni smja ,haan ji bhaiya samj gya😂

[@mahuaranu4435]

HW problem ans 1:15:03 int max_difference(int arr[], int n){ int suffix = arr[n-1]; int maxi = INT32_MIN ; int diff = 0; for(int i = n-2; i>=0; i--){ suffix = max(suffix , arr[i]); diff = suffix - arr[i]; maxi = max(maxi , diff); } return maxi; }

[@Anonymous_EngineerX]

Kadane's algo to be honest outstanding explanation Thanks bhaijaan

[@rohanbarwade578]

best mentor ever

[@GayatriShejule-o1k]

Thank you so much sir 🥰

[@aayushrajpurohit321]

#Chamaka diya bhaiya it was an amazing session.

[@saadrml]

bhaiya, i think there is slight mistake at 48:54 prefix( mein 2 aane pe) maxi(jo pehle se 7 tha ) usko 2 update nhi karenge , balki 7 hi rehne denge , i hope someone finds its correct 🕊

[@_hulk748]

Great Video as always sir🙏🙇‍♂✨💖

[@GoogleSoftwareEnginear]

Bhaiyo itna badiya content bhaiya hme de rhe hai to Subscribe to bnta hai sabhi kre Thank you so much Bhaiya for this amazing DSA series 😊😊👍👍

[@atifmalik8012]

int equalSubarray(vectorA) { int prefix=0,Totalsum=0; for(int i=0;i

[@ankushladani496]

Done Bhaiya... 39/180 ✅

[@OsamaShabih]

Chamak gya bhaiya 😊

[@Sangi-sj2dy]

Simply Awesome

[@nitinpal880]

bhaiya apki vajah se improvement dekh para hu bhot jyada bas revision ka tarika bhi bata do aap

[@abhisheksinghmehra9576]

Amazing explanation bhai as always🔥🔥🔥

[@PinkuModi-c6i]

bhaiya yaar apne pucha tha ki beech me to ni aa rhe h ad, but aa rhe h-av just aaya jab main 33:15 time stamp pe pahucha aur jaise hi ek question complete hua!!

[@heetpatel3037]

Chamka bhaiya thank you 👍🏻

[@rejaulislam2593]

alhamdulillah lec done

[@AnuragSingh-im2md]

Answer for the last question: int n = nums.size(); int suffix = nums[n-1]; int ans = 0; int diff; for(int i = n-2; i>=0; i--){ diff = suffix - nums[i]; ans = max(ans , diff); suffix = max(suffix , nums[i]); } if(ans==0) return -1; else return ans; o(n) T.C

[@arpitkhandelwal8280]

amazing lecture bhaiya

[@Hiisbnxkskkwl]

49:10 time stap pae toh 7 bada ha toh vohi rahega na sir

[@ankushladani496]

Thanks Bhaiya...❤🎉

[@Coder-rohits]

maza aa gaya bhaiya🙌

[@neerajrawat4294]

smallest sum contingous subarray is also a variation of kadane's algorithm;

[@swapnadip]

Good morning Bhaiya 😊🤗🤗

[@vermapowerhub]

Bhaiya aaj aapne day nhi dala video discription m ,ki konsa day h 180 days challenge ka ☺️

[@Updatedknowledge9]

Maza aa gya 😊😊😊😊❤

[@atifmalik8012]

23:08 //i tried it myslef without watching //In this approch i made two verctors one is prefix other one is suffix //if the perticular index in prefix is same as the element present at index+1 in suffix then return true otherwise false(for all elements) class Solution { public: bool canPartition(vector& nums) { int n = nums.size(); vectorprefix(n); vectorsuffix(n); prefix[0] = nums[0]; for(int i = 1;i=0;i--){ suffix[i] = suffix[i+1] + nums[i]; } for(int i = 0;i

[@amritpathak8630]

Great effort

[@csforcommonsense3360]

tq for ur lots of efforts

[@DesiMrBean7]

Good morning negi bhai❤

[@shailjayadav7995]

Good Morning Bhaiya❤

[@GOATSOFFOOTBALL]

completed,samajh aya

[@study-yd6es]

Amazing Content !!!

[@AzeemKhan-oe8rt]

bhai mujhse homework nhi ho paa rha, jbke me solution sochta rehta hoon 1-2 hours but code me convert ho hi nhi paa rha hai?

[@shubhamkumarjha9192]

Good morning Bhaiya..❤️

[@CoderArmy9]

Good morning bhai❤

[@AzeemKhan-oe8rt]

last wale 2-3 lecture nhi dekh paaya yaha se continue kr skata honnn?????????/ @coder army

[@AMITKUMAR-ds4hp]

Good morning boss ❤❤❤

[@Armoured_Traveller]

Day 39 Bhaiya ek motivational video chahiye apka🎉🎉❤

[@sourabhgupta1466]

best lecture bhai

[@Rishi_s.corner]

chamak gaya bhaiya.

[@Codewithkumar-gc1kr]

chamak gaya elder brother😊😊 09-Nov-24

[@NoorulAbdeenSiddiqui]

u are the best

[@Lifeofmoana22]

thankyou bhaiyaaa

[@Saumyasingh0001]

Great explanation but i think at 18 min par jo inner loop par 2n kah rahe hai wo 2n nahi hoga waha par dono inner loop milakar n time he chala tho n hona chahiye 😅

[@divyanshsharma673]

In worst case, first loop will run n-1 times and the second will 1 time. So n((n-1)+1) = n². It could be for both end so this is why it's 2n.

[@akashkewat1219]

Completed 😊.

[@SnatanGaming]

I AM UNABLE TO SOLVE LAST 2 Questions OF THE HOMEWORK SHEET I FOUND THIS LECTURE A BIT TOUGH CONSIDERING ALL THE PREVIOUS LECTURES,BECAUSE IN ALL PREVIOUS HOMEWORK SHEET I ATLEAST THINK OF AN APPROACH TO SOLVE BUT IN THIS LECTURE I AM UNABLE TO DO THAT ALSO.I HAVE TO PRACTICE HARD:(

[@joydeep-halder]

H/W Solutoin: : Maximum prefix sum for a given range vector ans; for (int i = 0; i < Q; ++i) { int start = L[i]; int end = R[i]; int maxPrefix = INT_MIN; int prefix = 0; for (int j = start; j

[@utkarshawasthi4791]

❤

[@jeetusingh6182]

i < Q what does it mean in third line

[@TayyabaYasmine]

Any one know in which lactures he covers the operator overloading and prefis and post fix operator overloading function

[@joydeep-halder]

Good morning bhaiya ❤❤

[@CoderArmy9]

Good morning bhai❤

[@joydeep-halder]

@@CoderArmy9 😊😊

[@parthh1112]

48:55 error 7>2 so no update needed

[@karankumar3983]

Logic behind kadanes algorithm 55:00 to 59:08 kadanes ke pichhe ka kala such 😅

[@GaziZai]

Thanks

[@nishikantDalal]

Maximum Difference between 2 element code: int ans=0; int suffix_max=arr[n-1]; for(int i = n - 2 ; i >= 0 ; i - -){ ans=max(ans , suffix_max - arr[i]); if(arr[i] > suffix_max) suffix_max = arr[i] ; } return ans ;

[@jiraiyagaming6491]

Good morning 🌞

[@CoderArmy9]

Good morning bhai❤

[@santanudebnath1283]

MAX DIFFERENCE #include using namespace std; int main() { int n,max_diff=0,ans; int arr[10]; coutn; cout=0;i--) { if(arr[i]>prefix) prefix=arr[i]; else { ans=prefix-arr[i]; max_diff=max(ans,max_diff); } } cout

[@techmcu]

if I'm not wrong bhaiya n-1 indicating last element position

[@sarikasaxena5567]

max difference between 2 element solution: #include using namespace std; int main(){ int n; cin>>n; vector v(n); for(int i =0;i>v[i]; } int ans=0; int suffix=v[n-1]; for(int i = n-2;i>=0;i--){ if(suffix

[@diyaseth9688]

Thank you bhaiya

[@bsufeotxe]

bhaiya dutch national flag problem bhi samjha do pointer ki help se

[@lotifurshabbir7302]

1:00:59 vaia here if i wrote that if condition before updating the maxi line, will it work correctly?

[@AakashParvani]

Day 39/180 👍✅ Chamak Gaya

[@drlovephukn]

Day's 39/180 full power full power ❤❤❤

[@asad_iliyas]

9:29 howework

[@prashantsingh3474]

1:15:03 hw maximum diffrence between two elements:- int maximumDifference(vector& n) { int su=INT_MIN , maxi=INT_MIN, sub=; for(int i=n.size()-1;i>=1;i--){ su=max(su, n[i]); sub=su-n[i-1]; maxi=max(sub, maxi); } return maxi; }

[@PravilaKumari-uv5be]

Good morning sir

[@OnlyPc-c9n]

function me 2nd for lop n-1 tak chalege na

[@Hermitsays007]

homework done suffix[n-1] = arr[n-1]; for(int i = n-2; i>=0; i--){ suffix[i] = suffix[i+1] + arr[i]; }

[@SurajKumar-vc4du]

radhe radhe

[@OmPrakashKewtiya]

Day 29✅

[@GopalKumar-xu3iw]

Dya 39/180 complete # hard

[@punitpubggaming7531]

Attendance marked ✅

[@rajatjana2513]

Day1🙏🙏🙏

[@arunrajput8153]

Love u bhai

[@sahilCSE999]

Bhaia 51:58 pe 13 kaise hoga 7+6+2 = 15 hoga n

[@vermapowerhub]

Good morning coders ❤

[@Megha3570]

Bhaiya can't we just sort the array first in max difference between 2 elements

[@bee.eee.]

Noo, because the order of element matters, it says the we can only subtract an element from an element which is to it's right side (doesn't matter how far) So yes, if we sort it the order will be neglected and we will basically end up getting Max element - min element which is not the required answer.

[@Megha3570]

@@bee.eee. thanks ☺️

[@anshikathakur3567]

lecture 29✅

[@tanishqdixit6313]

Day 39/180 completed