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imperfect_notes/pragy/Greedy Algos.md

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+Greedy
+------
+
+$k$ negations
+-------------
+
+> Given A[N]
+> can have -ve can have 0s
+> k negations
+> can choose same element multiple times
+> find the maximum sum that you can achieve
+
+always find the smallest and negate it
+
+Sort: $O(n \log n + k)$
+Min Heap: $O(n+k \log n)$
+
+Note: 1 2 will have -1 2 or 1 2 as the final answer
+can keep toggling the smallest element
+-- --
+
+What is Greedy
+--------------
+
+Greedy - when optimal solution to a local problem is also the optimal solution to the global problem
+
+That is, if you can act greedily at every step, and still reach the optimal.
+If you don't have to think long term to achieve the optimal
+
+-- --
+
+Where will greedy not work?
+---------------------------
+
+![65112ffa.png](:storage/8d511c68-5ee9-423f-b945-bf7407cdb509/65112ffa.png)
+
+Local solution is not contributing to the global solution
+
+Whenever greedy works, it will be based on observations and intuition
+
+But have to be careful because can be subtle-non-greedy case
+
+-- --
+
+Max Sum Subset
+--------------
+
+> Given A[N] with -ves
+> Find the subset of size k with maximum sum
+> 
+
+Sort and take top k values
+
+$O(n \log n)$
+
+-- --
+
+Amazon - Min Subset Product
+----------------------------
+
+> Given A[N]
+> min possible subset product
+> 
+
+Examples
+
+> -1 -1 -2 4 5  = -24
+> -1 0 2  = -2
+> 0 1  = 0
+> 
+
+1. If no -ve, then choose min $\geq 0$
+2. If k -ve, then choose all $> 0$. don't choose 0
+    - if odd -ve, choose all
+    - if even -ve, choose the smallest odd -ves
+
+$O(n)$
+
+-- --
+
+
+Activity Selection
+------------------
+
+Multiple companies. Very important
+
+> ![96706e2d.png](:storage/8d511c68-5ee9-423f-b945-bf7407cdb509/96706e2d.png)
+> max no of activity that can be performed without overlapps
+
+**Incorrect:** min sized?
+```
+=============    ===============
+           ========
+```               
+
+**Incorrect:** start time
+
+**Correct:** least overlaps?
+
+**but easier:** first one to end
+
+sort by end time and execute
+
+**Proof:**
+
+Assume there exists a solution that doesn't include the min end time job.
+
+Add this job. Either it overlaps with 1, or with 0.
+Optimal in both cases
+
+So, for every optimal solution without the min end job, we also have an optimal solution with the min end job.
+
+So, optimal to always choose the min end job
+
+![8d6e9fbb.png](:storage/8d511c68-5ee9-423f-b945-bf7407cdb509/8d6e9fbb.png)
+-- --
+
+Flights
+-------
+
+> start, end time of flights
+> num of platforms
+> 
+
++1 -1, prefix sum max
+
+![03c1d903.png](:storage/8d511c68-5ee9-423f-b945-bf7407cdb509/03c1d903.png)
+
+-- --
+
+Job Scheduling + Profit
+-----------------------
+
+> day deadlines for jobs
+> can perform only 1 job per day
+
+delay the job
+
+$O(n \log n + n^2)$
+
+$n \log n$ to sort by profit
+$n^2$ because start at end time and move left to find an empty slot
+
+Can reduce to n log n using BBST
+
+we want the highest number in the set which is less than x
+we also want to remove any given number from set
+
+set - upperbound - google
+
+-- --
+
+
+Rats and Holes
+--------------
+
+
+> Rats
+> Holes
+> equal count
+> locations given for each
+> ech rat runs together
+> each must reach a hole
+> max time so that all reach hole
+> 
+
+sort, all go left or all go right
+
+cant cross because that will increase time
+
+```
+
+-------          ----------
+
+vs
+
+       ----------
+---------------------------
+
+```