# Count of Rectangles with area K made up of only 1s from given Binary Arrays

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 ` `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`

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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