Find a Pair with a Given Difference | Love Babbar DSA Sheet | GFG | FAANG🔥 | Placement

#sorting and searching #competitiveprogramming #coding #dsa
Hey Guys in this video I have explained with code how we can solve the problem 'Find a Pair with a Given Difference'.
Array question Playlist = • Love Babbar DSA 450 Qu...
String question Playlist = • Love Babbar DSA 450 Qu...
Searching and Sorting question Playlist = • Love Babbar DSA 450 Qu...
Binary Tree question Playlist = • Love Babbar DSA 450 Qu...
Dynamic Programming question Playlist = • Love Babbar DSA 450 Qu...
Roadmap for Dynamic Programming = • Complete Roadmap for D...
Great Strategy to solve DSA = • Great Strategy to solv...
My Journey to 5 star at codechef = • My Journey to 5 Star a...
Love Babbar DSA Sheet : drive.google.com/file/d/1FMdN...
Follow us on Instagram:
Shailesh Yogendra : / shaileshyogendra
Yogesh Yogendra : / i_am_yogesh_here
Follow us on Linkedin:
Shailesh Yogendra : / shailesh-yogendra-8b13...
Yogesh Yogendra : / yogesh-yogendra-26bbb518a
Hope you like it. Comment if you have any doubt
LIKE | SHARE | SUBSCRIBE

[@aniketshankar2924]

Example Array - [1,3,2] and x=0; then as per your logic (1+0) that is 1 exists in the map .

[@deepakajay4204]

Handling all the test cases😀😀
//we can use hashing
HashMap map = new HashMap();
for(int i = 0; i < size; i++){
map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
}
for(int i = 0; i < size; i++){
int curr = arr[i];
int pair = n + arr[i];
if(n > 0 && map.containsKey(pair)){
return true;
}
if(n == 0){
//check for -curr
if(map.containsKey(-1 * pair)){
return true;
}
//one more edge case
//check for same curr and count
if(map.containsKey(pair) && map.get(pair) > 1){
return true;
}
}
}
return false;

[@prateeksomani5803]

You haven't considered corner case for x=0

[@anujsoni6089]

Bhot shi kaam kr rhe ho bhiyaa appp🔥🔥🔥

[@ridhamgodha4569]

binary search method is not getting submitted on gfg

[@yashyadav7927]

dude ur approach is good but we in constrain there is mentioned that we don't need to use extra space

[@tombrady7390]

edge case when all elements of array are same is not handled by method 2

[@sajaldewangan4343]

int Solution::solve(vector &A, int B) {
unordered_map mp;
for(auto i : A) mp[i] = 1;
for(auto i : A){
mp[i] = 0;
if(mp[B+i] > 0) return 1;
mp[i] = 1;
}
return 0;
}

[@princemathur2451]

You are solving other question, the question given in love babbar sheet is present on coding ninjas

[@stunnerhash]

map me insert karne ki time complexity logn hoti h shayad toh , second case me nlogn hogi time complexity , im not sure though

[@CodeLibrary]

unordered map has O(1) complexity and Map has O(LogN) complexity...and I am using unordered map

[@ekkranti4896]

@@CodeLibrary yeah you are right

[@SajanKumar-ec2us]

Uper bound kya hota hai.expliain clearly last method

[@rohandevaki4349]

at 9:00 , upper bound etc is not there in java, kindly use methods which are generic, and not standard libraries for c or c++

[@CodeLibrary]

You can create your own upper bound fn in java right ?

[@sanjogbhalla1720]

bhaiya apan -a kyu kar rahe hai upperbound function ke baad?

[@hritikchaudhary5470]

3rd method 7:25

[@googlePowerful452]

Not working for nlogn

[@faizanahmad-oc4fw]

Bro give the source code in description

[@trialaccount2244]

bhai + q ke rahe ho map[arr[i]]-x q nhi?

[@EpicClipsShort]

same question

[@user-jk7nb5ip3w]

tqqqqqqqqqqqqqqqqqqq

[@navendraagrawal]

this code is giving wrong answer you make fool in every video

[@vaibhavchhabra800]

It is pronounced as sorted not shorted

[@saurabhsingh1608]

Failing for testcase :5 0
1 2 2 6 5
with method3 i tried on gfg.
bool findPair(int arr[], int n, int k){

bool flag=false;
sort(arr,arr+n);
if (k==0) return flag;
for(int i=0;i