Number of ways to select equal sized subarrays from two arrays having atleast K equal pairs of elements

Given two arrays A[] and B[], and an integer K, the task is to find the number of ways to select two subarrays of same size, one from A and the other one from B such that the subarrays have at least K equal pairs of elements. (i.e. the number of pairs (A[i], B[j]) in the two selected subarrays such that A[i] = B[j] >= K).

Examples:

Input: A[] = {1, 2}, B[] = {1, 2, 3}, K = 1
Output: 4
The ways to select two subarrays are:

  1. [1], [1]
  2. [2], [2]
  3. [1, 2], [1, 2]
  4. [1, 2], [2, 3]

Input: A[] = {3, 2, 5, 21, 15, 2, 6}, B[] = {2, 1, 4, 3, 6, 7, 9}, K = 2
Output: 7

Approach:



  • Instead of dealing with the two arrays separately, lets combine them in form of a binary matrix such that:
    mat[i][j] = 0, if A[i] != B[j]
              = 1, if A[i] = B[j]
    
  • Now if we consider any submatrix of this matrix, say of size P x Q, it’s basically a combination of a subarray from A of size P and a subarray from B of size Q. Since we only want to check for equal sized subarrays, we will consider only square submatrices.
  • Let’s consider a square submatrix with top left corner as (i, j) and bottom right corner as (i + size,, j + size). This is equivalent to considering subarrays A[i: i + size] and B[j: j + size]. It can be observed that if these two subarrays will have x pairs of equal elements then the submatrix will have x 1’s in it.
  • So traverse over all the elements of matrix (i, j) and consider them to be the bottom right corner of the square. Now, one way is to traverse all the possible sizes of submatrix and find the sizes which have sum >= k, but this will be less efficient. It can be observed that say if a S x S submatrix with (i, j) as bottom right corner has sum >= k, then all square submatrices with size >= S and (i, j) as bottom right corner will follow the property.
  • So instead of iterating for all sizes at each (i, j), we will just apply binary search over the size of square submatrix and find the smallest size S such that it has sum >= K, and then simply add the matrices with greater side lengths.
    This article can be referred to see how submatrix sums are evaluated in constant time using 2D prefix sums.

Below is the implementation of the above approach:

C++

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// C++ implementation to count the 
// number of ways to select equal
// sized subarrays such that they
// have atleast K common elements
  
#include <bits/stdc++.h>
  
using namespace std;
  
// 2D prefix sum for submatrix 
// sum query for matrix
int prefix_2D[2005][2005];
  
// Function to find the prefix sum
// of the matrix from i and j
int subMatrixSum(int i, int j, int len)
{
    return prefix_2D[i][j] - 
           prefix_2D[i][j - len] - 
           prefix_2D[i - len][j] + 
           prefix_2D[i - len][j - len];
}
  
// Function to count the number of ways
// to select equal sized subarrays such
// that they have atleast K common elements
int numberOfWays(int a[], int b[], int n, 
                            int m, int k)
{
  
    // Combining the two arrays
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (a[i - 1] == b[j - 1])
                prefix_2D[i][j] = 1;
  
            else
                prefix_2D[i][j] = 0;
        }
    }
  
    // Calculating the 2D prefix sum
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            prefix_2D[i][j] += prefix_2D[i][j - 1];
        }
    }
  
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            prefix_2D[i][j] += prefix_2D[i - 1][j];
        }
    }
  
    int answer = 0;
  
    // iterating through all 
    // the elements of matrix
    // and considering them to 
    // be the bottom right
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
              
            // applying binary search 
            // over side length
            int low = 1;
            int high = min(i, j);
  
            while (low < high) {
                int mid = (low + high) >> 1;
  
                // if sum of this submatrix >=k then
                // new search space will be [low, mid]
                if (subMatrixSum(i, j, mid) >= k) {
                    high = mid;
                }
                // else new search space 
                // will be [mid+1, high]
                else {
                    low = mid + 1;
                }
            }
  
            // Adding the total submatrices
            if (subMatrixSum(i, j, low) >= k) {
                answer += (min(i, j) - low + 1);
            }
        }
    }
    return answer;
}
  
// Driver Code
int main()
{
    int N = 2, M = 3;
    int A[N] = { 1, 2 };
    int B[M] = { 1, 2, 3 };
  
    int K = 1;
  
    cout << numberOfWays(A, B, N, M, K);
    return 0;
}

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Java

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// Java implementation to count the 
// number of ways to select equal
// sized subarrays such that they
// have atleast K common elements
class GFG{
  
// 2D prefix sum for submatrix 
// sum query for matrix
static int [][]prefix_2D = new int[2005][2005];
  
// Function to find the prefix sum
// of the matrix from i and j
static int subMatrixSum(int i, int j, int len)
{
    return prefix_2D[i][j] - 
           prefix_2D[i][j - len] - 
           prefix_2D[i - len][j] + 
           prefix_2D[i - len][j - len];
}
  
// Function to count the number of ways
// to select equal sized subarrays such
// that they have atleast K common elements
static int numberOfWays(int a[], int b[], int n, 
                        int m, int k)
{
      
    // Combining the two arrays
    for(int i = 1; i <= n; i++) 
    {
       for(int j = 1; j <= m; j++)
       {
          if (a[i - 1] == b[j - 1])
              prefix_2D[i][j] = 1;
          else
              prefix_2D[i][j] = 0;
       }
    }
  
    // Calculating the 2D prefix sum
    for(int i = 1; i <= n; i++)
    {
       for(int j = 1; j <= m; j++) 
       {
          prefix_2D[i][j] += prefix_2D[i][j - 1];
       }
    }
  
    for(int i = 1; i <= n; i++) 
    {
       for(int j = 1; j <= m; j++) 
       {
          prefix_2D[i][j] += prefix_2D[i - 1][j];
       }
    }
  
    int answer = 0;
  
    // Iterating through all 
    // the elements of matrix
    // and considering them to 
    // be the bottom right
    for(int i = 1; i <= n; i++)
    {
       for(int j = 1; j <= m; j++)
       {
            
          // Applying binary search 
          // over side length
          int low = 1;
          int high = Math.min(i, j);
            
          while (low < high)
          {
              int mid = (low + high) >> 1;
                
              // If sum of this submatrix >=k then
              // new search space will be [low, mid]
              if (subMatrixSum(i, j, mid) >= k)
              {
                  high = mid;
              }
                
              // Else new search space 
              // will be [mid+1, high]
              else
              {
                  low = mid + 1;
              }
          }
            
          // Adding the total submatrices
          if (subMatrixSum(i, j, low) >= k)
          {
              answer += (Math.min(i, j) - low + 1);
          }
       }
    }
    return answer;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 2, M = 3;
    int A[] = { 1, 2 };
    int B[] = { 1, 2, 3 };
  
    int K = 1;
  
    System.out.print(numberOfWays(A, B, N, M, K));
}
}
  
// This code is contributed by Princi Singh

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C#

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// C# implementation to count the 
// number of ways to select equal
// sized subarrays such that they
// have atleast K common elements
using System;
  
class GFG{
  
// 2D prefix sum for submatrix 
// sum query for matrix
static int [,]prefix_2D = new int[2005, 2005];
  
// Function to find the prefix sum
// of the matrix from i and j
static int subMatrixSum(int i, int j, int len)
{
    return prefix_2D[i, j] - 
           prefix_2D[i, j - len] - 
           prefix_2D[i - len, j] + 
           prefix_2D[i - len, j - len];
}
  
// Function to count the number of ways
// to select equal sized subarrays such
// that they have atleast K common elements
static int numberOfWays(int []a, int []b, int n, 
                        int m, int k)
{
      
    // Combining the two arrays
    for(int i = 1; i <= n; i++) 
    {
       for(int j = 1; j <= m; j++)
       {
          if (a[i - 1] == b[j - 1])
              prefix_2D[i, j] = 1;
          else
              prefix_2D[i, j] = 0;
       }
    }
  
    // Calculating the 2D prefix sum
    for(int i = 1; i <= n; i++)
    {
       for(int j = 1; j <= m; j++)
       {
          prefix_2D[i, j] += prefix_2D[i, j - 1];
             
       }
    }
  
    for(int i = 1; i <= n; i++) 
    {
       for(int j = 1; j <= m; j++) 
       {
          prefix_2D[i, j] += prefix_2D[i - 1, j];
       }
    }
  
    int answer = 0;
  
    // Iterating through all 
    // the elements of matrix
    // and considering them to 
    // be the bottom right
    for(int i = 1; i <= n; i++)
    {
       for(int j = 1; j <= m; j++)
       {
             
          // Applying binary search 
          // over side length
          int low = 1;
          int high = Math.Min(i, j);
          while (low < high)
          {
              int mid = (low + high) >> 1;
                
              // If sum of this submatrix >=k then
              // new search space will be [low, mid]
              if (subMatrixSum(i, j, mid) >= k)
              {
                  high = mid;
              }
                
              // Else new search space 
              // will be [mid+1, high]
              else
              {
                  low = mid + 1;
              }
          }
            
          // Adding the total submatrices
          if (subMatrixSum(i, j, low) >= k)
          {
              answer += (Math.Min(i, j) - low + 1);
          }
       }
    }
    return answer;
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 2, M = 3;
    int []A = { 1, 2 };
    int []B = { 1, 2, 3 };
  
    int K = 1;
  
    Console.Write(numberOfWays(A, B, N, M, K));
}
}
  
// This code is contributed by Princi Singh

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Output:

4

Time Complexity: O(N * M * log(max(N, M)))

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