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Data Structures and Algorithms
(CS210/ESO207/ESO211)
Lecture 14 Algorithm paradigm of Divide and Conquercontinued
Counting the number of inversions
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Divide and Conquer paradigm for
Algorithm Design
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Divide and Conquer paradigm
An Overview
A problem in this paradigm is solved in the following way.
1. FirstDividethe problem instance into two or more instancesof the same problem. Solve each smaller instances
recursively(base case suitably defined).
2. Combinethe solutions of the smaller instances to get thesolution of the original instance.
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This is usually the main nontrivialstep
in the design of an algorithm using
divide and conquer strategy
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Example 3
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Counting the number of inversions
in an array
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Counting Inversions in an arrayProblem description
Definition (Inversion): Given an array Aof size n, a pair (i,j), 0i
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Counting Inversions in an arrayProblem familiarization
Trivial-algo(A[0..n-1])
{count0;
For(j=1 to n-1) do
{ For( i=0toj-1)
{ If (A[i]>A[j]) countcount + 1;
}
}
}
Time complexity: O(
)Question: What can be the max. no. of inversions in an array A?
Answer:
, which is O().
Question: Is the algorithm given above optimal ?
Answer: No, our aim is notto report all inversions but to report the count.6
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Let us try to design a
Divide and Conquer based algorithm
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How do we approach using divide & conquer
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A 3 15 8 19 9 67 11 27
0 1 2 3 4 5 6 7
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Counting Inversions
Divide and Conquer based algorithm
CountInversion( A,i,k) // Counting no. of inversions in A[i..k]
If (i=k) return 0;
Else{ mid(i+k)/2;
CountInversion(A,i,mid);
CountInversion(A,mid+1,k);
. Code for .
return + + ;
}
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How to efficiently compute
(Inversions of type III)?
Aim:For each mid
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How to efficiently compute
(Inversions of type III)?
Key Observation: We have to perform n/2operations of the same kind:
How many elements in A[i..mid] aregreaterthanA[j] ?
Lesson from Data Structures :
We should build a suitable data structure storing elements of A[i..mid] so
that the above operation can be performed efficiently for anyj.
Question: What should be the data structure ?
Answer: Sorted subarray A[i..mid].
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Counting Inversions
First algorithm based on divide & conquer
CountInversion( A,i,k)
If (i=k) return 0;
Else{ mid(i+k)/2;
CountInversion(A,i,mid);
CountInversion(A,mid+1,k);
. Code for .
return + + ;
}
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Sort(A,i,mid);
Foreach mid
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Counting Inversions
First algorithm based on divide & conquer
Time complexity analysis:
If n= 1,
T(n) = cfor some constant c
If n> 1,T(n)= cn logn+ 2T(n/2)
= cnlogn + cn ((logn)-1) + T(n/)
= cnlogn + cn ((logn)-1) + cn ((logn)-2) + T(n/)
= O(nn)
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Can we improve it further ?
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Sequence of observations
To achieve better running time
The extra logn factor arises because for the combine step, we are
spending O(nlog n) time instead of O(n).
The reason for O(nlog n) time for the combine step is due to the
following tasks: SortingA[0..n/2] takesO(nlog n) time.
Doing Binary Search for n/2elements from A[n/2n-1]
Each of the above tasks have optimal running time.
So the only way to improve the running time of combine step is to look
for some new ideas to compute .
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Learn from the past knowledge
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Many of you noticed some similaritybetween
the code of the O(nn) time algorithm and
Merge Sort. Explore these similarities moreclosely.
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Revisiting MergeSortalgorithm
MSort(A,i,k)// Sorting A[i..k]
{ If(i< k)
{ mid(i+k)/2;
MSort(A,i,mid);
MSort(A,mid+1,k);
Create a temporary array C[0..k-i]
Merge(A,i,mid,k,C);
CopyC[0..k-i] to A[i..k]
}}
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We shall carefully look at the Merge()
procedure to find an efficient way to count
the number of elements from A[i..mid]
which are smaller than A[j] for any given
mid
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RelookMerging A[i..mid] and A[mid+1..k]
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A
Sorted Sorted
j
x
C
x
>x
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Pesudo-code for Merging two sorted arrays
Merge(A,i,mid,k,C)
pi; jmid+1; r0;
While(p mid andj k)
{ If(A[p]< A[j]) { C[r]A[p]; r++; p++ }
Else { C[r]A[j]; r++; j++ }
}
While(p mid) { C[k]A[i]; k++; i++ }
While(j k) { C[k]
A[j]; k++; j++ }returnC ;
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We shall makejust a slight change in the above
pseudo-code to achieve our main objective of
computing
. If you understood the discussion
of the previous slide, can you guess it now ?
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Pesudo-code for
Merging and counting inversions
Merge_and_CountInversion(A,i,mid,k, C)
pi; jmid+1; r0;
0;
While(p mid andj k)
{ If(A[p]< A[j]) { C[r]A[p]; r++; p++ }
Else { C[r]A[j]; r++; j++;
??
}
}While(p mid) { C[k]A[p]; k++; i++ }
While(j k) { C[k]A[j]; k++; j++ }
return;
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+ (mid-p+1);
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Counting Inversions
Final algorithm based on divide & conquer
Sort_and_CountInversion(A,i,k)
{ If(i=k) return 0;
else
{ mid(i+k)/2;
Sort_and_CountInversion (A,i,mid);
Sort_and_CountInversion (A,mid+1,k);
Create a temporary array C[0..k-i]
Merge_and_CountInversion(A,i,mid,k,C);
CopyC[0..k-i] to A[i..k];return + + ;
}
}
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2 T(n/2)
O(n)
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Counting Inversions
Final algorithm based on divide & conquer
Time complexity analysis:
If n= 1,
T(n) = cfor some constant c
If n> 1,T(n)= cn + 2T(n/2)
= O(nlogn)
Theorem:There is a divide and conquer based algorithm forcomputing the number of inversions in an array of size n. The
running time of the algorithm is O(nlogn).
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The second practice sheet of problems is available on
moodle. Make sincere attempts to solve its exercises.
