Sorting 3

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))

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

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

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.

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

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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