Question

ALGORITHM PROBLEM:

A) Significant Inversions: We are given a sequence of n arbitrary but distinct real numbers <a1 , a2 ,..., an>. We define a significant inversion to be a pair i < j such that ai > 2 aj . Design and analyze an O(n log n) time algorithm to count the number of significant inversions in the given sequence.

[Hint: Use divide-&-conquer. Do the “combine” step carefully]

B) The Maximum-Sum Monotone Sub-Array Problem: Input: An array A[1..n] of arbitrary positive integers. Output: The maximum-element-sum contiguous sub-array of A[1..n] whose entries form a monotone sequence (either ascending or descending)

[Hint: Use an incremental approach]

Answer 1

A.

We will divide the array into equal halves, we will count the number of significant inversions in each halves, and we will also count the number of significant inversion in which one of the element is in one half and another element is in another half. We will add these to get the result. So we will perform analogous operation to merge sort in which we will divide the array into halves, count the significant inversal in each half then combine the subarray to form sorted array but also note if at some index element in left half form significant inversal with element in right half, then all the elements in next index in left half will be significant inversal. We will use this property to count the number of significant inversal in O(n log n) time.

int Divide (A,begin,end) 1. count- keep count of significant inversal 2. If end-begin< 2: then if A[begin]>Alend] then if A[begin]>2 A[end] count-1 else count o swap (A[begin],A[end] return count 3. mid(begin+end)/2 count count+Divide (A,begin,mid) count significant inversal in left end count count+Divide (A,mid,mid+1) count significant inversal in right end count -count+Merge(A,begin,mid,) count significant inversal with elenents in different subarray 4. Return count

Now merge step is very important to observe.

Note in step 4 of merge where we are incrementing count very smartly.

Complexity = O(n log n) just like merge sort

B.

We will divide the array into equal halves, we will check the maximum sum of monotone array in each halves, and we will also check the maximum sum of monotone array which has some portion in left half and some portion in right half. We will return the result which is maximum of all three. Without loss of generality assume that array is of size which is some exponent of 2.

Below is the Merge procudure where we keep the name Merge just to say that we are searching for monotone sequence which passed through middle of the array.

Time complexity T(n) = 2T(n/2)+O(n) = O( n log n)

Please comment for any clarification