Problem of The Day: 08/07/2023 | Find Triplets with Zero Sum | Abhinav Awasthi

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In this question we have to solve the problem in O(N^2) complexity, so I used this logic. if a+b+c=0, the a+b =-c.
We first sort the array. Then, we will find a+b such that it equals -c where -c is negative of a value in array. So, I run a loop from 0 to n-1; then, val= - arr[i];
Then, we traverse on the array. we have two pointer one front and another end. Now, we find the sum= arr[front] and arr[end];
here, front is i+1, and end=n-1. Since array is sorted, sum will be modified likewise. if target < sum: front++;

else if(target>sum) end--;
else{
return sum;
}
So, basically I used sorting, a little mathematics and then pointer to solve this question in given time and space constraints.

TusharSharmaBMT

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