Lecture_Notes/imperfect_notes/pragy/Binary Search 1.md

Binary Search

• needs ordering (not necessarily sorting, since orderings can be of more than one type)
  • basically, some proposition which makes

  N N N N N N Y Y Y Y Y Y Y
  

• ask about how many people are encountering Binary Search for the first time
• quickly explain Binary Search. Ask if everyone can implement it
• Recurrennce implementation

Implementation of Binary Search

• loop till low <= high
• Computing mid:

  m = (l+h) / 2
  

  vs

  m = l + (h-l)/2
  

Time complexity

• Recurrence for time complexity: T(n) = T(n/2) + O(1)
• Solve this recurrence to show complexity is O(\log_2 n)

Ternary Search

• Ternary Search: T(n) = T(n/3) + O(1)
• Ternary Search isn't really faster. Since O(1) of binary search is less than O(2) of ternary search. 1 \times \log_2 n < k \times \log_k n. Extreme Case: k=n, n-ary search, which is same as linear search

Eventual value of low and high if element is not present?

• if searching for x, after termination, A_l >= x while A_h <= x
• Note that the roles are reversed

Repeated elements, Find first occurance of element

• when array[mid] == x, either we've hit the first occurance or hit the not-first occurance.
• if array[mid] == x then check if array[mid - 1] == x. If yes, do high = mid - 1 and continue, since this is not the first occurance.
• edge case: if mid = 0, then make sure that you don't check array[-1]

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Rotated array:

• if shifted to the right
  • 6 7 8 9 0 1 2 3 4 5
  • A[0] >= A[-1]. If this is not true, there was no rotation
  • Now, we have mid, let's say 2
  • Now, if 2 is in the first half, then it must be greater than last element
  • Else it should be smaller than last element.
  • So search for rotation point based on this

Find any peak element in unsorted array

• Peak element is one that is not smaller than its neighbors
• 1, 5, 3, 2, 1, 6, 5
• \log n time using binary search
• if sorted ascending, peak is last
• if sorted descending, first element is peak
• algo:
  • check mid.
  • 1 6 5
    • if a[m-1] <= a[m] >= a[m+1], then peak
  • 7 6 5
    • if a[m-1] > a[m] >= a[m+1], then move left
    • theorem: one peak exists to the left
    • 90 80 70 60 50
    • ofcourse, there could be a peak to the right, but we can safely go left
  • 7 6 8
    • move in either of the parts, since peaks exist on both sides

All but one element comes twice, find this element. Sorted array

• 0 0 4 4 5 7 7 8 8
• get mid. find its first occurance in log n time.
• check if number of elements from mid to last is even or odd.
  • odd: singular exists in latter part
  • even: singular exists in former part

Family of strings

• s_0 = 0
• s_i = s_{i-1} \cdot 0 \cdot (s_{i-1})'

s0 =        0            1
s1 =       001           3
s2 =     0010110         7
s3 = 001011001101001    15

• Given N, k, find kth character of S_n.
• Length of string: |S_n| = 2^{n+1} - 1
• Draw tree. Show toggling of bits

Staircase

          __
       __|__|
    __|__|__|
 __|__|__|__|
|__|__|__|__|

• heights: 1, 2, ... h
• Given n bricks, find the max height of staircase that can be built
• Let ans be h. Then number of bricks needed = h(h+1) / 2
• Now, binary search on h. Min = 0, max = n.
• if f(h) > n, go left
• if f(h) < n and f(h+1) >= n. Stop
• if f(h) < n and f(h+1) < n, go right

MxN Matrix. Each row sorted. MxN is odd. Find overall median

M \times N matrix. Note: M \times N is odd.

        4 6 7
        1 7 9
        2 2 3

actual answer: 4 1 2 2 3 4 6 7 7 9

• Vanilla - concat and sort arrays. O(mn \log{mn})
• Median of Medians doesn't work, 7 is not the correct answer
• Note: lowest element is from 1st col, and largest is from last col (since rows are sorted)
• M rows. Find low and high in O(m) time
• Now, if x is median, then 1/2 elements should be smaller, 1/2 should be bigger
• more correctly, if x is repeated, then if first occurance of x is l, last occurance is r, then if l <= n/2 <= r, then x is median
• now, in each row, find how many >x and <x.
• if l <= n/2 <= r - stop
• if r < n/2, increase low
• if l > n/2, decrease high
• Time complexity: O(m \log n \log{(\max - \min)})
• why odd? 1 4 6 9, and searching for 5. both side are n/2 so we don't know what to increase/decrease

Any doubts, will you be able to code it?

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• 4 things
  • lecture 1 hour
  • assignment (4 questions) same which were discussed in class
  • standup session (homework problem / doubts)
  • homework problem
• 2 links generated - standup lecture and lecture
• operations guy will help setting up of lectures
• TAs will take a standup session, not me
• share questions and homeworks