TECHNICAL INTERVIEW QUESTION on DATA STRUCTURES & ALGORITHMS
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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.


What is an inversion
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


Can you write an algorithm to count the number of inversions with time complexity o n log n
Can you write an algorithm to count the number of inversions with time complexity O(n log n)?

Hint: Merge Sort


Let s look at merge sort
Let’s look at Merge-Sort with time complexity O(n log n)?

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


Merge operation
Merge Operation with time complexity O(n log n)?

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


Can you modify mergesort to count the number of inversions
Can you modify with time complexity O(n log n)?MergeSort to count the number of inversions?


Modified merge sort
Modified Merge-Sort with time complexity O(n log n)?

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


Modified merge operation
Modified Merge Operation with time complexity O(n log n)?

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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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.


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