3 Sum - Target Sum Unique Triplets | Leetcode 15 Solution in Hindi

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NADOS also enables doubt support, career opportunities and contests besides free of charge content for learning. 1. Given an integer array 'nums', and a 'target', return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k.
2. Another thing is nums[i] + nums[j] + nums[k] == target.
3. Notice that the solution set must not contain duplicate triplets.
Topic: #twoPointerApproach, #twoSum, #Sorting
Used #DataStructure: #Arrays
#TimeComplexity: O(n^2)
#SpaceComplexity: O(1)
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Linked Questions:
1. Two Sum - Target Sum Unique Pair: • 2 Sum - Target Sum Uni...
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[@RishabhRJ04]

Bhai bhai bhai bhai
0:24

[@Pepcoding]

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[@nishantagarwal5274]

😂😂

[@ArvindSrawat-fz4qc]

threesum

[@lakshrustagi507]

F

[@anuragdogra9204]

i think time complexity analysis is also more important while writing the code, so please add tc.

[@pabitrasarkar203]

sir in line number 44 why in the for loop you use i

[@mrprime557]

Great explanation. Please add time complexity Analysis to videos. Thank you

[@Pepcoding]

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[@raj14318]

great explaination brother🔥🔥🔥

[@Pepcoding]

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[@CarelessMni]

Thanks Sir.

[@aniketgahiware6592]

Amazing.......

[@Pepcoding]

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[@darshankalathiya8667]

for(int i = 0 ; i

[@LegitGamer2345]

question not available on portal

[@SumitSingh-ui4do]

Amazing❤️❤️

[@Pepcoding]

Glad you liked it.
Keep learning
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[@Saurabh-fe2bg]

what if the target is not given and we are asked to find the triplets in an array

[@raj14318]

please one suggetion every program u should explain about TC...

[@Pepcoding]

Okay

[@mickyman753]

ye ques shayad rajnish sir ne bhi ek baar karay tha

[@kirtikedia6274]

Your solution will fail for following testcase - [0,0,0,0,0]
Your code will return - [[0,0,0],[0,0,0]]
Where else expected output would be - [[0,0,0]]
To fix,
if(sum == find){
ans.push_back({arr[i],arr[start],arr[end]});
while(start

[@darshankalathiya8667]

hey kirtiji can you provaid full solution in c++ pls

[@kanareikaTv]

Что ж, матан научил нас китайскому, программирование научило меня хинди

[@nikhilkumarjamwal5322]

Code is not showing the desired output.
class Solution {

public List twoSum(int[] nums, int s, int e, int t) {
// 1 1 2 3 4 5 6 7 7
int start=s, end=e, target=t, sum=0;
List l = new ArrayList();

while(start < end){ // log(n)
if(start != s && nums[start] == nums[start-1]){
start++;
continue;
}
sum = nums[start] + nums[end];
if(sum == target){//element found
List l1 = new ArrayList();
l1.add(nums[start]);
l1.add(nums[end]);
l.add(l1);
start++;
end--;
}
else if(sum > target){
end--;
}
else if(sum < target){
start++;
}
}
return l;
}

public List threeSum(int[] nums) {
List l = new ArrayList();
int target = 0, oldTarget = 0, value = 0;
if(nums.length < 3){
return l;
}

Arrays.sort(nums);

for(int i=0; i

[@aahanaganjewar9951]

if(l != si && nums[l] == nums[l-1] ){
l++;
continue;
}
I didn't understand why l != si?

[@codersaurabh6231]

When you will do l-1 which means start-1 since l= start, so to exclude the beginning case try to think taking start as 0 you will get clear idea.

[@aahanaganjewar9951]

@@codersaurabh6231 Thank you!

[@shubhamkhuntia2412]

It's showing TLE
class Solution {
public List twoSum(int arr[],int si,int ei,int target){
int left = si;
int right = ei;
List ans = new ArrayList();
while(left target){
right--;
} else {
left++;
}
}
return ans;
}
public List threeSum(int[] arr) {
Arrays.sort(arr);
int target =0;
List res = new ArrayList();
int n=arr.length;
if(n

[@Pepcoding]

For better insights, visit nados.pepcoding.com, post your doubts, community will help you out there.

[@amanbisht4286]

Should have used Arrays.sort(nums) over there

[@omkarkulkarni3207]

In twoSum() method : if(sum==target) --> You forgot to do left++; right--;

[@shreyashree6781]

agar main function me kaise implement kiya h , wo v batate to acha hota...qk video complete explaination wali honi chahye k .

[@anubhavnegi4230]

kind of seems like u have learned the solution not a good way to explain, kindly be intuitive ;)