Max Points On a Line (Asked in Google, Apple, LinkedIn, Amazon) - Codexplained

On Episode 17 of Codexplained, we're going to be tackling the coding problem, "Max Points On a Line". Points scattered everywhere! Tis' confusion! It's time to connect the dots!
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[@acceleratorlevel645]

alright, you earned a sub, beautiful explanation of the problem before providing solution , that was enough for me to come up with the answer, this is what i wanted, amazing!

[@aryanyadav3926]

Thanks for the solution and explanation!

[@kewtomrao]

Those hints were enough to solve the problem. Thanks! Subscribed!

[@nani4027]

Where are you checking the common points in the code?

[@HackerEarthInc]

All lines passing through one point and having the same slope, lie on the same line (1:50).
Lines 12 and 14 in the code are used to calculate the number of points passing through point 'i', with slope 'slope'. In other words, the number of points that are colinear, with i as a common point.
Once the i iteration ends, the next i value will be different. Hence, we clear the dictionary in line 16.

[@sandeshsonawane4988]

why are we clearing hashmap each time

[@HackerEarthInc]

Good question - it's because of parallel lines. Say the points are:
(0,0), (1,1), (0,1), (1,2), (2.3)
If you plot them out on a graph, you will find that (0,0) and (1,1) lie on one line, and (0,1), (1,2) and (2,3) lie on another parallel line. Both have a slope of 1. Here, the answer must be 3.
When we complete i=0, the hashmap will say that slope 1 has 1 point.
If we do not clear the hashmap, when all iterations are complete, the hashmap will say that slope 1 has 4 points (1 from i=0 + 2 from i=2 +1 from i=3). That's because the slope is the same, and it will give us an incorrect answer of 5 (1 gets added at the end).
dict.clear ensures that the slope is reset after each point. Now, before we enter i=2, the value is reset, and when i=2 is complete, the hasmap will say that slope 1 has 2 points, which becomes 3 when we add 1 at the end.
To identify colinear points, the slopes have to be the same, and they have to pass through at least one common point (1:50). Since the dictionary/hashmap is only tracking the slope, it must be reset whenever we change that common point after each i iteration.

[@sandeshsonawane4988]

@@HackerEarthInc Great Explanation Thank you!!

[@me.deepaksharma]

@@HackerEarthInc Thanks.

[@two_wheel_tales_jsr]

Osm Explanation love it bro

[@satyamjha6268]

Nyc explanation

[@meetshah2288]

No need for max variable in the i loop. Changing max doesn't make a difference to the i value since range is precalculated taking max as 0 in the begining of the for loop.
So you are not doing any optimization by adding max in the range() of i loop

[@sunilpanchal1498]

You can add one condition inside of for loop if(n-max>=i) break then it will save your iteration

[@adyantgupta4661]

j should loop from 0 not from i+1

[@saiudaydasari]

parallel lines have same slope then ur adding all parallel line points to one point then ur code will fail

[@HackerEarthInc]

We are not counting the parallel line points, but rather the colinear points. The way to check if many points are colinear is if they have the same slope and pass through at least 1 common point.
In each j loop, we are considering the slopes of the lines passing through one common point. Parallel lines never intersect, so if they have the same slope and pass through a common point, it means that it must be same line!
After each j loop, we are clearing the dictionary using dict.clear. We are starting fresh, with another point. The counter gets reset to 0, so if there is another parallel line, it will be counted separately.

[@yogeshyts]

hindi bolne m sharm aati h ky

[@henrypaul5676]

angrezi seekhne me aalas aata hai kya? nhk bhai??

[@yogeshyts]

@@henrypaul5676 h

[@henrypaul5676]

To seekh le bhai. Dawai ki dukan pr to nahi bolta ki angreji naam wli dawa mt de wrna nahi khaoonga😂😂

[@yogeshyts]

@@henrypaul5676 English psnd ni