Lecture_Notes/imperfect_notes/pragy/Sorting 2.md

Sorting 2

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Merge Sort

• Divide and Conquer
  • didive into 2
  • sort individually
  • combine the solution
• Merging takes O(n+m) time.
  • needs extra space
• code for merging:

  // arr1[n1]
  // arr2[n2]
  int i = 0, j = 0, k = 0;
  
  // output[n1+n2]
  
  while (i<n1 && j <n2) {
      if (arr1[i] <= arr2[j])  // if <, then unstable
          output[k++] = arr1[i++];
      else
          output[k++] = arr2[j++];
  }
  // only one array can be non-empty
  while (i < n1)
      output[k++] = arr1[i++];
  
  while (j < n2)
      output[k++] = arr2[j++];
  

• stable? Yes
• in-place? No
• Time complexity recurrence: T(n) = 2T(n/2) + O(n)
• Solve by Master Theorem.
• Solve by algebra
• Solve by Tree height (\log n) * level complexity (O(n))

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Intersection of sorted arrays

2 sorted arrays

1 2 2 3 4 9
2 3 3 9 9

intersection: 2 3 9

• calculate intersection. Report an element only once
• Naive:
  • Search each element in the other array. O(n \log m)
• Optimied:
  • Use merge.
  • Ignore unequal.
  • Add equal.
  • Move pointer ahead till next element

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Merging without extra space

• can use extra time
• if a[i] < b[j], i++
• else: swap put b[i] in place of a[i]. Sorted insert a[i] in b array
• so, O(n^2) time

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Count inversions

inversion: i < j, but a[i] > a[j] (strict inequalities)

• naive: O(n^2)
• Split array into 2.
• Number of inversions = number of inversions in A + B + cross terms
• count the cross inversions by example
• does number of cross inversions change when sub-arrays are permuted?
• no
• can we permute so that it becomes easier to count cross inversions?
• sort both subarrays and count inversions in A, B recursively
• then, merge A and B and during the merge count the number of inversions
• A_i B_j
• if A[i] > B[j], then there are inversions
• num inversions for A[i], B[j] = |A| - i
• intra array inversions? Counted in recursive case.

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Find doubled-inversions

same as inversion just i < j, a[i] > 2 * a[j]

• same as previous. Split and recursilvely count
• while merging, for some b[j], I need to find how many elements in A are greater than 2 * b[j]
• linear search for that, but keep index
• linear search is better than binary search

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Sort n strings of length n each - T(n) = 2T(n/2) + O(n^2) = O(n^2) is wrong - T(n) = 2T(n/2) + O(n) * O(m) = O(nm\log n) is correct. Here m = the initial value of n

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I G N O R E

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Bounded Subarray Sum Count

given A[N] can have -ve given lower <= upper find numbe of subarrays such that lower <= sum <= upper

• naive: O(n^2) (keep prefix sum to calculate sum in O(1), n^2 loop)
• if only +ve, O(n\log n) using prefix sum
• but what if -ve?

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