# Count Inversion - PowerPoint PPT Presentation

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Count Inversion

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### TECHNICAL INTERVIEW QUESTION on DATA STRUCTURES & ALGORITHMSCount the number of inversions in an array

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### 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

Hint: Merge Sort

### 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

### 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

### 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

### 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