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Count of Rectangles with area K made up of only 1s from given Binary Arrays
  • Difficulty Level : Hard
  • Last Updated : 07 May, 2021

Given two binary arrays A[] and B[], of length N and M respectively, the task is to find the number of rectangles of area K consisting of 1‘s in the matrix C[][] generated by multiplying the two arrays such that, C[i][j] = A[i] * B[j] (1< i < n, 1< j < m).

Examples:

Input: N= 3, M = 3, A[] = {1, 1, 0}, B[] = {0, 1, 1}, K = 2 
Output:
Explanation: C[][] = {{0, 1, 1}, {0, 1, 1}, {0, 0, 0}}

0 1 1        0 1 1      0 1 1       0 1 1
0 1 1        0 1 1      0 1 1       0 1 1
0 0 0        0 0 0      0 0 0       0 0 0

Therefore, there are 4 possible rectangles of area 2 from the matrix.
Input: N = 4, M = 2, A[] = {0, 0, 1, 1}, B[] = {1, 0, 1}, K = 2 
Output:
Explanation: C[][] = {{0, 0, 0}, {0, 0, 0}, {1, 0, 1}, {1, 0, 1}} 

0 0 0        0 0 0
0 0 0        0 0 0
1 0 1        1 0 1
1 0 1        1 0 1

Therefore, there are 2 possible rectangles of area 2 in the matrix. 



Naive Approach: The simplest approach to solve the problem is to generate the required matrix by multiplying the two arrays and for every possible rectangle of area K, check if it consists of only 1’s or not. 
Time Complexity: O(N * M * K) 
Auxiliary Space: O(N * M)

Efficient Approach: To optimize the above approach, the following observations need to be made instead of generating the matrix: 

  • The area of a rectangle is equal to the product of its length and breadth.
  • Using this property, visualize the rectangle as a submatrix which contains only 1s. Therefore, this submatrix is the result of the product of two subarrays of length a, b where a * b = K.
  • Since the submatrix contains only 1‘s, it is obvious that these two subarrays also contain only 1‘s in them.

Therefore, the problem reduces to finding the subarrays consisting of only 1‘s of all possible lengths which are proper divisors of K, from the arrays A[] and B[]. Follow the steps below to solve the problem:

  • Precalculate the count of possible subarrays.
  • Iterate through all the divisors of K and for each possible pair (p, q) where p * q = K, check if there exist subarrays of length p, q in A[], and B[].
  • Increase the count of possible such subarrays accordingly and finally, print the obtained count.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the subarrays of
// all possible lengths made up of only 1s
vector<int> findSubarrays(vector<int>& a)
{
    int n = a.size();
 
    // Stores the frequency
    // of the subarrays
    vector<int> freq(n + 1);
 
    int count = 0;
 
    for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
 
            // Check if the previous
            // value was also 0
            if (count == 0)
                continue;
 
            // If the previous value was 1
            else {
 
                int value = count;
                for (int j = 1; j <= count; j++) {
 
                    // Find the subarrays of
                    // each size from 1 to count
                    freq[j] += value;
                    value--;
                }
 
                count = 0;
            }
        }
 
        else
            count++;
    }
 
    // If A[] is of the form ....111
    if (count > 0) {
        int value = count;
        for (int j = 1; j <= count; j++) {
            freq[j] += value;
            value--;
        }
    }
 
    return freq;
}
 
// Function to find the count
// of all possible rectangles
void countRectangles(vector<int>& a,
                     vector<int>& b, int K)
{
    // Size of each of the arrays
    int n = a.size();
 
    int m = b.size();
 
    // Stores the count of subarrays
    // of each size consisting of
    // only 1s from array A[]
 
    vector<int> subA
        = findSubarrays(a);
 
    // Stores the count of subarrays
    // of each size consisting of
    // only 1s from array B[]
    vector<int> subB
        = findSubarrays(b);
 
    int total = 0;
 
    // Iterating over all subarrays
    // consisting of only 1s in A[]
    for (int i = 1; i < subA.size(); i++) {
 
        // If i is a factor of K, then
        // there is a subarray of size K/i in B[]
        if (K % i == 0 and (K / i) <= m) {
            total = total + subA[i] * subB[K / i];
        }
    }
 
    cout << total;
}
 
// Driver Code
int main()
{
    vector<int> a = { 0, 0, 1, 1 };
 
    vector<int> b = { 1, 0, 1 };
    int K = 2;
 
    countRectangles(a, b, K);
 
    return 0;
}

Java




// Java Program to implement
// the above approach
class GFG{
 
    // Function to find the subarrays of
    // all possible lengths made up of only 1s
    static int[] findSubarrays(int[] a)
    {
        int n = a.length;
 
        // Stores the frequency
        // of the subarrays
        int[] freq = new int[n + 1];
        int count = 0;
          for (int i = 0; i < n; i++)
        {
            if (a[i] == 0)
            {
 
                // Check if the previous
                // value was also 0
                if (count == 0)
                    continue;
 
                // If the previous value was 1
                else
                {
                    int value = count;
                    for (int j = 1; j <= count; j++)
                    {
 
                        // Find the subarrays of
                        // each size from 1 to count
                        freq[j] += value;
                        value--;
                    }
                    count = 0;
                }
            }
            else
                count++;
        }
 
        // If A[] is of the form ....111
        if (count > 0)
        {
            int value = count;
            for (int j = 1; j <= count; j++)
            {
                freq[j] += value;
                value--;
            }
        }
        return freq;
    }
 
    // Function to find the count
    // of all possible rectangles
    static void countRectangles(int[] a, int[] b, int K)
    {
        // Size of each of the arrays
        int n = a.length;
        int m = b.length;
 
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array A[]
        int[] subA = findSubarrays(a);
 
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array B[]
        int[] subB = findSubarrays(b);
 
        int total = 0;
 
        // Iterating over all subarrays
        // consisting of only 1s in A[]
        for (int i = 1; i < subA.length; i++)
        {
 
            // If i is a factor of K, then
            // there is a subarray of size K/i in B[]
            if (K % i == 0 && (K / i) <= m)
            {
                total = total + subA[i] * subB[K / i];
            }
        }
        System.out.print(total);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] a = {0, 0, 1, 1};
        int[] b = {1, 0, 1};
        int K = 2;
        countRectangles(a, b, K);
    }
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program to implement
# the above approach
 
# Function to find the subarrays of
# all possible lengths made up of only 1s
def findSubarrays(a):
 
    n = len(a)
 
    # Stores the frequency
    # of the subarrays
    freq = [0] * (n + 1)
 
    count = 0
 
    for i in range(n):
        if (a[i] == 0):
 
            # Check if the previous
            # value was also 0
            if (count == 0):
                continue
 
            # If the previous value was 1
            else:
                value = count
                for j in range(1, count + 1):
 
                    # Find the subarrays of
                    # each size from 1 to count
                    freq[j] += value
                    value -= 1
                 
                count = 0
                 
        else:
            count += 1
     
    # If A[] is of the form ....111
    if (count > 0):
        value = count
        for j in range(1, count + 1):
            freq[j] += value
            value -= 1
         
    return freq
 
# Function to find the count
# of all possible rectangles
def countRectangles(a, b, K):
 
    # Size of each of the arrays
    n = len(a)
    m = len(b)
 
    # Stores the count of subarrays
    # of each size consisting of
    # only 1s from array A[]
    subA = []
    subA = findSubarrays(a)
 
    # Stores the count of subarrays
    # of each size consisting of
    # only 1s from array B[]
    subB = []
    subB = findSubarrays(b)
 
    total = 0
 
    # Iterating over all subarrays
    # consisting of only 1s in A[]
    for i in range(1, len(subA)):
         
        # If i is a factor of K, then
        # there is a subarray of size K/i in B[]
        if (K % i == 0 and (K // i) <= m):
            total = total + subA[i] * subB[K // i]
         
    print(total)
 
# Driver Code
a = [ 0, 0, 1, 1 ]
b = [ 1, 0, 1 ]
 
K = 2
 
countRectangles(a, b, K)
 
# This code is contributed by code_hunt

C#




// C# Program to implement
// the above approach
using System;
class GFG{
 
    // Function to find the subarrays of
    // all possible lengths made up of only 1s
    static int[] findSubarrays(int[] a)
    {
        int n = a.Length;
 
        // Stores the frequency
        // of the subarrays
        int[] freq = new int[n + 1];
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            if (a[i] == 0)
            {
                // Check if the previous
                // value was also 0
                if (count == 0)
                    continue;
 
                // If the previous value was 1
                else
                {
                    int value = count;
                    for (int j = 1; j <= count; j++)
                    {
                        // Find the subarrays of
                        // each size from 1 to count
                        freq[j] += value;
                        value--;
                    }
                    count = 0;
                }
            }
            else
                count++;
        }
 
        // If []A is of the form ....111
        if (count > 0)
        {
            int value = count;
            for (int j = 1; j <= count; j++)
            {
                freq[j] += value;
                value--;
            }
        }
        return freq;
    }
 
    // Function to find the count
    // of all possible rectangles
    static void countRectangles(int[] a, int[] b,
                                int K)
    {
        // Size of each of the arrays
        int n = a.Length;
        int m = b.Length;
 
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array []A
        int[] subA = findSubarrays(a);
 
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array []B
        int[] subB = findSubarrays(b);
 
        int total = 0;
 
        // Iterating over all subarrays
        // consisting of only 1s in []A
        for (int i = 1; i < subA.Length; i++)
        {
 
            // If i is a factor of K, then
            // there is a subarray of size K/i in []B
            if (K % i == 0 && (K / i) <= m)
            {
                total = total + subA[i] *
                        subB[K / i];
            }
        }
        Console.Write(total);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] a = {0, 0, 1, 1};
        int[] b = {1, 0, 1};
        int K = 2;
        countRectangles(a, b, K);
    }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// Javascript program to implement
// the above approach
     
    // Function to find the subarrays of
    // all possible lengths made up of only 1s
    function findSubarrays(a)
    {
        let n = a.length;
   
        // Stores the frequency
        // of the subarrays
        let freq = new Array(n+1).fill(0);
        let count = 0;
          for (let i = 0; i < n; i++)
        {
            if (a[i] == 0)
            {
   
                // Check if the previous
                // value was also 0
                if (count == 0)
                    continue;
   
                // If the previous value was 1
                else
                {
                    let value = count;
                    for (let j = 1; j <= count; j++)
                    {
   
                        // Find the subarrays of
                        // each size from 1 to count
                        freq[j] += value;
                        value--;
                    }
                    count = 0;
                }
            }
            else
                count++;
        }
   
        // If A[] is of the form ....111
        if (count > 0)
        {
            let value = count;
            for (let j = 1; j <= count; j++)
            {
                freq[j] += value;
                value--;
            }
        }
        return freq;
    }
   
    // Function to find the count
    // of all possible rectangles
    function countRectangles(a, b, K)
    {
        // Size of each of the arrays
        let n = a.length;
        let m = b.length;
   
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array A[]
        let subA = findSubarrays(a);
   
        // Stores the count of subarrays
        // of each size consisting of
        // only 1s from array B[]
        let subB = findSubarrays(b);
   
        let total = 0;
   
        // Iterating over all subarrays
        // consisting of only 1s in A[]
        for (let i = 1; i < subA.length; i++)
        {
   
            // If i is a factor of K, then
            // there is a subarray of size K/i in B[]
            if (K % i == 0 && (K / i) <= m)
            {
                total = total + subA[i] * subB[K / i];
            }
        }
        document.write(total);
    }
 
// Driver Code
 
        let a = [0, 0, 1, 1];
        let b = [1, 0, 1];
        let K = 2;
        countRectangles(a, b, K);
     
</script>
Output: 
2

Time Complexity: O(D) * O(N + M), where D is the number of divisors of K. 
Auxiliary Space: O(N + M)

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