Sum of Absolute Differences in a Sorted Array - Leetcode 1685 - Python

[@aravindashokkumar5425]

Nice work! This problem cleverly uses what we know about prefixes and postfixes. Great job!

[@arihantbedagkar7678]

Time: O(n) and Space: O(1) diff = sum(nums) for i, num in enumerate(nums): yield ((i

[@shreehari2589]

Bro I was just solving this question, you uploaded this video, what a perfect timing

[@sanooosai]

thank your sir, learning a lot from you

[@sauravsingh4497]

The saviour has arrived 🙏

[@NeetCodeIO]

Will try to be consistent the rest of the year!

[@sauravsingh4497]

@@NeetCodeIO i know there must be reason for not uploading so no pressure upload whenever you feel like it. Love your explaination ❤️

[@soumyajitchatterjee5822]

Great Content as always.

[@thor1626]

My java solution : class Solution { public int[] getSumAbsoluteDifferences(int[] nums) { int len = nums.length; int[] prefixSum = new int[len + 1]; prefixSum[0] = 0; int sum = 0; int index = 1; for (int i = 0; i < len; i++) { sum += nums[i]; prefixSum[index] = sum; index++; } for (int i = 0; i < len; i++) { int x = (prefixSum[len] - prefixSum[i]) - (nums[i] * (len - i)); int y = (nums[i] * i) - prefixSum[i]; nums[i] = x + y; } return nums; } }

[@AbhishekMakwana-p1v]

Thankyou my niggahahahah loved your Solution it really helped ..may god make you a slave again in next life ...

[@NeetCodeIO]

Uh thanks 😅

[@leeroymlg4692]

wtf

[@pastori2672]

-Why does your code work but mine gets TLE they're almost identical: res = [0] * len(nums) s = sum(nums) prefix_sum = 0 postfix_sum = sum(nums) for i in range(len(nums)): postfix_sum -= nums[i] res[i] += postfix_sum - (len(nums[i + 1:]) * nums[i]) res[i] += (nums[i] * len(nums[:i])) - prefix_sum prefix_sum += nums[i] return res

[@AbhishekMakwana-p1v]

you are using len operator on update list every time i mean you are using it on sliced array so it is calculated every single time loop runs and value of i gets updated every time ..when you use len(nums) without slicing it does not count length every time it is inbuily & more efficient thana o(N) ,,when you use slicing the len has to be calculated every time ...

[@AbhishekMakwana-p1v]

by the i am a big fan of berserk ...griffith did nothing wrong i mean ...casca enjoyed it

[@pastori2672]

@@AbhishekMakwana-p1v ah i see

[@pastori2672]

@@AbhishekMakwana-p1v i'll touch you

[@1vader]

​@@AbhishekMakwana-p1vI'm quite certain len is never "calculated"/computed. A list always stores its length. But the slicing is indeed the issue because it creates a whole new list with the relevant elements which ofc is an O(n) operation.

[@shahwaliweb6505]

Look to my code. nums = [2,3,5] new_list = [] count = 0 length_nums = len(nums) for i in range(length_nums ): for j in range(length_nums ): new_list.append(nums[count] - nums[j]) count += 1 positive_list = [abs(x) for x in new_list] result = [sum(positive_list[i:i+length_nums]) for i in range(0, len(positive_list), length_nums)] print(result)

[@Vancha112]

Yes this is code

[@brahm_and_coding]

that's something man