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imperfect_notes/pragy/Binary Search 2.md

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+Binary Search 2
+---------------
+
+Median of two sorted arrays
+---------------------------
+- merge: O(n). Search O(log n) 
+- binary search: O(log n)
+- pick the smaller array
+- low = 0, high = length of smaller array
+- partition the other array so that number of elements in first of both = number of elements in last of both
+- now check if this partition is valid by cross comparison
+- change high and low accordingly
+- append -inf and +inf to start and ends of arrays
+
+Special Integer
+-------------------------------------
+> Unsorted array of size N.
+> +ve numbers.
+> Given X, find max value of K such that no subarray of length k has sum > x
+- binary search over k in 0..N
+- if any sum > X, k is not the answer
+- if sum <= x, k might be the answer if k+1 is not allowed
+- finding max sum for a given k takes O(n) time
+- todo: change to m and not m-1/m+1 loop till h-l>1
+-- --
+
+Sorted array of length N.
+-------------------------------------
+> numbers are from 1..N-1 (so contiguous).
+> One of the numbers is repeated.
+> Find it
+
+- 1 2 3 4 4 5 6 7 8  (number)
+- 0 1 2 3 4 5 6 7 8  (index)
+- N N N N Y Y Y Y Y  (proposition)
+
+Agressive Cows
+--------------
+
+> N stalls, n >= 2
+> C cows, c >= 2
+> locations of stalls, let's say 1, 2, 5, 7, 8
+> Maximize the minimum distance between any two cows
+
+- low = 1
+- high = N
+- calculate mid = distance, and check
+
+Smallest Good Base
+------------------
+> $N \in \{3, 10^{18}\}$
+> k >= 2
+> $(N)_k = 1111111\ldots$
+> Find smallest k
+
+- example: N = 7
+    - k = 2, $(7)_2 = 111$
+- finding the largest good base is trivial - N-1
+- low = 2, high = n-1
+- If $m$ is the answer, then $\exists i: 1 + m + m^2 + \cdots + m^i = N$
+- But we still can't binary search since we don't know $i$. $i$ will change for different numbers
+- Also, say we found a working or non working m. How do we change the low and high? Multiple possible answers, which one to chose?
+- Let's look at it from searching on i
+- We need to maximize i (thereby minimize k)
+- $(7)_6 = 11, (7)_2 = 111$, we need to output 2.
+- So, linear search over all possible number of bits from 63 to 2
+    - Then for a fixed (i), binary search for the value of m (or use formula of geometric series)
+- low, high for number of bits will be $1, \log_2 N$ (because smallest k is 2)
+- $2^{63} \approx 10^{18}$
+
+
+Smallest subarray
+-----------------
+
+> Array of length N
+> Not sorted
+> +ve numbers
+> Find length of smallest subarray such that sum >= k
+- calculate prefix sum (prefix sum is sorted)
+- can you calculate sum a[i] .. a[j]?
+    - a[i] .. a[j] = P[j] - P[i] + a[i]
+- now, let's say the answer starts from S and ends at E
+    - loop over all possible S
+        - binary search for E
+    - return S, E that has min distance
+
+Election
+--------
+
+> N People
+> influence[1..n]: 1 2 2 3 1    (+ve)
+> Voting: ith person votes for jth person if sum(influence[i+1:j-1]) <= influence[i]
+> Note: j can lie on the left side as well
+> Print the array of votes each person got
+
+- So, in 1 2 2 3 1, the number of people who can vote for a[i] are 2 3 2 3 3
+- now, if we try to find how many people vote for i, it is linear search, so total $O(n^2)$
+- instead, we can calculate prefix sum, and for each person, see what range of people it can vote for using binary search. So O(n \log n)
+- Now that we have ranges, we need to find the final array
+- O(n) algo to convert ranges to array by doing +1 on start, -1 on end, and then taking prefix sum
+