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

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Problem Link: leetcode.com/problems/sum-of-...
0:00 - Read the problem
0:30 - Drawing Explanation
8:20 - Coding Explanation
leetcode 1685
#neetcode #leetcode #python

[@aravindashokkumar5425]

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

[@shreehari2589]

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

[@soumyajitchatterjee5822]

Great Content as always.

[@arihantbedagkar7678]

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

[@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 ❤️

[@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;
}
}

[@user-mh4lq5ce6u]

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

[@user-mh4lq5ce6u]

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 ...

[@user-mh4lq5ce6u]

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

[@pastori2672]

@@user-mh4lq5ce6u ah i see

[@pastori2672]

@@user-mh4lq5ce6u i'll touch you

[@1vader]

​@@user-mh4lq5ce6uI'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