Count Inversion

1. TECHNICAL INTERVIEW QUESTION on DATA STRUCTURES & ALGORITHMSCount the number of inversions in an array CodeGround Online Testing Platform is an online assessment and evaluation system that helps Recruiters conduct online screening tests to filter candidates before the interview process. CodeGround Recruitment Tests can be used during Campus Recruitment or screening walk-in candidates. CodeGround supports Aptitude Tests, English Communication Skills Assessments and Online Coding Contests in C, C++, Java, PHP, Ruby, Python and JavaScript. CodeGround also supports asynchronous automated interviews.

2. What is an inversion? Let A be an array of n distinct numbers. If i < j and A[i] > A[j], then the pair (i, j) is called an inversion of A. For example, the array {2,3,8,6,1} has 5 inversions: (2,1) (3,1) (8,6) (8,1) and (6,1) Trivial solution countInversions = 0; for i = 1 to N for j = i+1 to N if(A[i] > A[j]) countInversions++; Statement S1 will be executed (N-1) + (N-2) + … + 1 times = (N-1) * N / 2 Overall time complexity is O(n2) S1

3. Can you write an algorithm to count the number of inversions with time complexity O(n log n)? Hint: Merge Sort

4. Let’s look at Merge-Sort mid start end MERGE-SORT(A, start, end) { if(start < end) { mid = (start + end) / 2; MERGE-SORT(A, start, mid); MERGE-SORT(A, mid + 1, end); MERGE(A, start, mid, end); } } end mid+1 mid start

5. Merge Operation MERGE(A, start, mid, end) { firstArray = new Array of size mid – start + 2; secondArray = new Array of size end – mid + 1; Copy values from A[start: mid] to firstArray[0: mid-start]; Copy values from A[mid+1: end] to secondArray[0: end-mid-1]; firstArray[mid-start+1] = ∞ secondArray[end-mid] = ∞ i = 0, j = 0; for k = start to end { if(firstArray[i] <= secondArray[j]) { A[k] = firstArray[i]; i++; } else { A[k] = secondArray[j]; j++; } } } mid start end A end mid+1 mid start j ∞ i ∞ secondArray firstArray A k

6. Can you modify MergeSort to count the number of inversions?

7. Modified Merge-Sort mid start end int MERGE-SORT(A, start, end) { intcountInversions = 0; if(start < end) { mid = (start + end) / 2; countInversions += MERGE-SORT(A, start, mid); countInversions += MERGE-SORT(A, mid + 1, end); countInversions += MERGE(A, start, mid, end); } return countInversions; } end mid+1 mid start

8. Modified Merge Operation intMERGE(A, start, mid, end) { firstArray = new Array of size mid – start + 2; secondArray = new Array of size end – mid + 1; Copy values from A[start: mid] to firstArray[0: mid-start]; Copy values from A[mid+1: end] to secondArray[0: end-mid-1]; firstArray[mid-start+1] = ∞ secondArray[end-mid] = ∞ i = 0, j = 0, countInversions = 0; for k = start to end { if(firstArray[i] <= secondArray[j]) { A[k] = firstArray[i]; i++; } else { A[k] = secondArray[j]; j++; countInversions += firstArray.length – i - 1; } } return countInversions; } mid start end A end mid+1 mid start j ∞ i ∞ secondArray firstArray A k

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