Can You Detect the Missing Number? (LeetCode 268: Missing Number)

[@atharvarajadhyaksha4813]

Instead of running a loop for expectedSum, just use the formula n*(n+1)/2 where n is size of array. Significantly reduces the running time for larger arrays.

[@CodeWithZi]

You are correct that also works! Thanks for recommending it. However the time complexity is still O(n) in both cases

[@atharvarajadhyaksha4813]

@@CodeWithZi True. You still need a loop for actualSum. But for competitive programming it will always be better to avoid having an extra loop.

[@Madhava_P7]

set(range(n)-set(given))

[@YashGondane-p3f]

thank you so much for this cool approach

[@CodeWithZi]

You're very welcome!

[@danielhod53]

I got a bit misleaded because the example array in the beginning is sorted so my first instict was to loop through the array and calculate (array[i+1]-array[i]) and if it's not equal to 1 then the number missing is (array[i+1]+array[i]) / 2 but by seeing the question in LeetCode you can't assume that the array is sorted so it won't work

[@CodeWithZi]

great approach!

[@black_ice7822]

That’s pretty intuitive. Guess you could still do that if you sort the array first

[@heathens2867]

Can't we use i==arr[i] and if it's false, then i is the answer

[@gnanaharish]

Works only if array is sorted , here array is not sorted , so it will not work

[@atharvarajadhyaksha4813]

As the other comment said, you'll have the added burden of sorting the array and depending on what sort you use, you may end up with a worst case time complexity of O(n^2)

[@abhikannor9301]

Sum of size of total number of array - total sum of array

[@orderandchaos_at_work]

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