The correct and optimal method starts at 8:06, for those who are short of time

thx. isn't safe to start the l index with 1 since min_sum was already evaluated using [0] and [1] indices before the loop began.

But if quick sort takes O(n^2) in worst case, how can we assume this as an optimal solution? Merge Sort is my choice

sir can you please explain this with program ..thank you for your solution

if sum is minimum then also we need to check l++ and r--, to compare with other min

Please make video with Java and Kotlin

Thanks for the solution but please try to explain program with the example. I have not understood program, so now I will write program and will try to understand it

@GeeksforGeeksVideos

7 жыл бұрын

You're welcome, Sanjay. We'll try to incorporate your input in our future videos.

@sanjaypardeshi1964

7 жыл бұрын

👍

Thanks for sharing this as other videos! You are awesome guys!

@GeeksforGeeksVideos

7 жыл бұрын

Thanks reyou7 for the appreciation!

-10 + (-80) = -90. Is it less close to zero than +5. I presume I did not understand the meaning of closest to zero

@spicytuna08

6 жыл бұрын

take absolute value. absolute value is making every non negative number to positive, if you take abs value of -90, it becomes +90.

class Solution { public: int closestToZero(int arr[], int n) { sort (arr, arr + n); // sorting the array int i = 0, j = n - 1; int sum = arr[i] + arr[j]; // initializing sum int diff = abs (sum); // initializing the result while (i { // if we have zero sum, there's no result better. Hence, we return if (arr[i] + arr[j] == 0) return 0; // if we get a better result, we update the difference if (abs (arr[i] + arr[j]) { diff = (arr[i] + arr[j]); sum = arr[i] + arr[j]; } else if(abs (arr[i] + arr[j]) == abs (diff)) { sum=max(sum,arr[i]+arr[j]); } // if the current sum is greater than zero, we search for a smaller sum if (arr[i] + arr[j] > 0) j--; // else, we search for a larger sum else i++; } return sum; } };

What if i want those two elements whose sum is equal to 0

@abhijeetjain2098

5 жыл бұрын

yes it can be found...

how is it that O(nlogn) + O(n) = O(nlogn)?

@kaifahmad4131

3 жыл бұрын

Please Study space and time Complexity