Count of common subarrays in two different permutations of 1 to N
Last Updated :
17 Aug, 2021
Given two arrays A and B of the same length N, filled with a permutation of natural numbers from 1 to N, the task is to count the number of common subarrays in A and B.
Examples:
Input: A = [1, 2, 3], B = [2, 3, 1]
Output: 4
Explanation:
The common subarrays are [1], [2], [3], [2, 3]
Hence, total count = 4
Input: A = [1, 2, 3, 4, 5], B = [2, 3, 1, 4, 5]
Output: 7
Explanation:
The common subarrays are [1], [2], [3], [4], [5], [2, 3], [4, 5]
Hence, total count = 7
Naive Approach:
The idea is to generate all subarrays of A and B separately, which would take O(N2) for each array. Now, compare all subarrays of A with all subarrays of B and count common subarrays. It would take O(N4).
Efficient Approach:
The idea is to use Hashing to solve this problem efficiently.
- Create a Hash array H of size N+1.
- Represent all elements of A by their respective indices:
Element Representation
A[0] 0
A[1] 1
A[2] 2
.
.
and so on.
3. Use array H to store this representation, H[ A[ i ] ] = i
4. Update the elements of B according to this new representation using H, B[ i ] = H[ B[ i ] ]
5. Now, array A can be represented as [0, 1, 2, ..N], so simply count number of subarrays in B which have consecutive elements. Once we get length K of subarray of consecutive elements, count total possible subarray using following relation:
Total number of subarrays = (K * (K + 1)) / 2
Look at this example to understand this approach in detail:
Example:
A = [4, 3, 1, 2, 5]
B = [3, 1, 2, 4, 5]
Common subarrays are [1], [2], [3], [4], [5], [3, 1], [1, 2], [3, 1, 2] = 8
1. Represent A[i] as i, and store in H as H[A[i]] = i, Now array H from index 1 to N is,
H = [2, 3, 1, 0, 4]
2. Update B according to H, B[i] = H[B[i]]
B = [1, 2, 3, 0, 4]
3. Look for subarray in B with consecutive elements,
Subarray from index 0 to 2 is [1, 2, 3], consisting of consecutive elements with length K = 3
Element at index 3 forms a subarray [0] of length K = 1
Element at index 4 forms a subarray [4] of length K = 1
4. Total number of common subarrays =
(3*(3+1))/2 + (1*(1+1))/2 + (1*(1+1))/2 = 6 + 1 + 1 = 8
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int commonSubarrays( int *A, int *B, int N)
{
int Map[N + 1];
for ( int i = 0 ; i< N; i++)
Map[*(A + i)] = i;
for ( int i = 0; i < N; i++)
{
*(B + i) = Map[*(B + i)];
}
int count = 0;
int i = 0, K;
while (i < N)
{
K = 1;
i+= 1;
while (i < N && B[i] == B[i - 1] + 1)
{
i += 1;
K += 1;
}
count = count + ((K) * (K + 1)) / 2;
}
return count;
}
int main()
{
int N = 3;
int A[] = {1, 2, 3};
int B[] = {2, 3, 1};
cout << (commonSubarrays(A, B, N))
<< endl;
N = 5;
int C[] = {1, 2, 3, 4, 5};
int D[] = {2, 3, 1, 4, 5};
cout << (commonSubarrays(C, D, N));
}
|
Java
class GFG{
static int commonSubarrays( int []A,
int []B, int N)
{
int []Map = new int [N + 1 ];
for ( int i = 0 ; i< N; i++)
Map[A[i]] = i;
for ( int i = 0 ; i < N; i++)
{
B[i] = Map[B[i]];
}
int count = 0 ;
int i = 0 , K;
while (i < N)
{
K = 1 ;
i+= 1 ;
while (i < N && B[i] == B[i - 1 ] + 1 )
{
i += 1 ;
K += 1 ;
}
count = count + ((K) * (K + 1 )) / 2 ;
}
return count;
}
public static void main(String[] args)
{
int N = 3 ;
int A[] = { 1 , 2 , 3 };
int B[] = { 2 , 3 , 1 };
System.out.print(commonSubarrays(A, B, N));
System.out.print( "\n" );
N = 5 ;
int C[] = { 1 , 2 , 3 , 4 , 5 };
int D[] = { 2 , 3 , 1 , 4 , 5 };
System.out.print(commonSubarrays(C, D, N));
}
}
|
Python3
def commonSubarrays(A, B, N):
Map = [ 0 for i in range (N + 1 )]
for i in range (N):
Map [A[i]] = i
for i in range (N) :
B[i] = Map [B[i]]
count = 0
i = 0
while i<N:
K = 1
i + = 1
while i<N and B[i] = = B[i - 1 ] + 1 :
i + = 1
K + = 1
count = count + (
(K) * (K + 1 )) / / 2
return count
N = 3
A = [ 1 , 2 , 3 ]
B = [ 2 , 3 , 1 ]
print (commonSubarrays(A, B, N))
N = 5
A = [ 1 , 2 , 3 , 4 , 5 ]
B = [ 2 , 3 , 1 , 4 , 5 ]
print (commonSubarrays(A, B, N))
|
C#
using System;
class GFG{
static int commonSubarrays( int []A,
int []B,
int N)
{
int []Map = new int [N + 1];
for ( int i = 0; i < N; i++)
Map[A[i]] = i;
for ( int i = 0; i < N; i++)
{
B[i] = Map[B[i]];
}
int count = 0;
int a = 0, K;
while (a < N)
{
K = 1;
a += 1;
while (a < N && B[a] == B[a - 1] + 1)
{
a += 1;
K += 1;
}
count = count + ((K) * (K + 1)) / 2;
}
return count;
}
public static void Main()
{
int N = 3;
int []A = {1, 2, 3};
int []B = {2, 3, 1};
Console.Write(commonSubarrays(A, B, N));
Console.Write( "\n" );
N = 5;
int []C = {1, 2, 3, 4, 5};
int []D = {2, 3, 1, 4, 5};
Console.Write(commonSubarrays(C, D, N));
}
}
|
Javascript
<script>
function commonSubarrays(A, B, N)
{
let Map = Array.from({length: N+1}, (_, i) => 0);
for (let i = 0; i< N; i++)
Map[A[i]] = i;
for (let i = 0; i < N; i++)
{
B[i] = Map[B[i]];
}
let count = 0;
let i = 0, K;
while (i < N)
{
K = 1;
i+= 1;
while (i < N && B[i] == B[i - 1] + 1)
{
i += 1;
K += 1;
}
count = count + ((K) * (K + 1)) / 2;
}
return count;
}
let N = 3;
let A = [1, 2, 3];
let B = [2, 3, 1];
document.write(commonSubarrays(A, B, N));
document.write( "<br/>" );
N = 5;
let C = [1, 2, 3, 4, 5];
let D = [2, 3, 1, 4, 5];
document.write(commonSubarrays(C, D, N));
</script>
|
Time complexity: O(N)
Auxiliary Space: O(N)
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