We are given a N*M grid, print the number of rectangles in it.
Examples:
Input : N = 2, M = 2
Output : 9
There are 4 rectangles of size 1 x 1.
There are 2 rectangles of size 1 x 2
There are 2 rectangles of size 2 x 1
There is one rectangle of size 2 x 2.
Input : N = 5, M = 4
Output : 150
Input : N = 4, M = 3
Output: 60
Brute Force Approach :
- Iterate over all possible pairs of horizontal lines.
- Iterate over all possible pairs of vertical lines.
- count the number of rectangles that can be formed using these lines.
Below is the code for the above approach :
#include <bits/stdc++.h>
using namespace std;
int rectCount(int n, int m)
{
int count = 0;
for(int i=1; i<=n; i++) // iterating over all possible pairs of horizontal lines
{
for(int j=1; j<=m; j++) // iterating over all possible pairs of vertical lines
{
count += (n-i+1)*(m-j+1); // counting the number of rectangles that can be formed using these lines
}
}
return count;
}
/* driver code */
int main()
{
int n = 5, m = 4;
cout << rectCount(n, m);
return 0;
}
import java.util.Scanner;
public class RectangleCount {
public static int rectCount(int n, int m) {
int count = 0;
// Iterate over all possible pairs of horizontal lines
for (int i = 1; i <= n; i++) {
// Iterate over all possible pairs of vertical lines
for (int j = 1; j <= m; j++) {
// Count the number of rectangles that can be formed using these lines
count += (n - i + 1) * (m - j + 1);
}
}
return count;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = 5, m = 4;
// Calculate and print the count of rectangles
System.out.println("Count of Rectangles: " + rectCount(n, m));
}
}
def rect_count(n, m):
count = 0
for i in range(1, n + 1): # iterating over all possible pairs of horizontal lines
for j in range(1, m + 1): # iterating over all possible pairs of vertical lines
count += (n - i + 1) * (m - j + 1) # counting the number of rectangles that can be formed using these lines
return count
# driver code
def main():
n = 5
m = 4
print(rect_count(n, m))
if __name__ == "__main__":
main()
using System;
class Program
{
static int RectCount(int n, int m)
{
int count = 0;
for (int i = 1; i <= n; i++) // iterating over all possible pairs of horizontal lines
{
for (int j = 1; j <= m; j++) // iterating over all possible pairs of vertical lines
{
count += (n - i + 1) * (m - j + 1); // counting the number of rectangles that can be formed using these lines
}
}
return count;
}
// driver code
static void Main()
{
int n = 5, m = 4;
Console.WriteLine(RectCount(n, m));
}
}
// Function to count the number of rectangles that can be formed
function rectCount(n, m) {
let count = 0;
for (let i = 1; i <= n; i++) { // iterating over all possible pairs of horizontal lines
for (let j = 1; j <= m; j++) { // iterating over all possible pairs of vertical lines
count += (n - i + 1) * (m - j + 1); // counting the number of rectangles that can be formed using these lines
}
}
return count;
}
// Driver code
const n = 5;
const m = 4;
console.log(rectCount(n, m));
Output
150
Time Complexity : O(N^2)
Space Complexity : O(1)
We have discussed counting number of squares in a n x m grid,
Let us derive a formula for number of rectangles.
If the grid is 1x1, there is 1 rectangle.
If the grid is 2x1, there will be 2 + 1 = 3 rectangles
If it grid is 3x1, there will be 3 + 2 + 1 = 6 rectangles.
we can say that for N*1 there will be N + (N-1) + (n-2) ... + 1 = (N)(N+1)/2 rectangles
If we add one more column to N×1, firstly we will have as many rectangles in the 2nd column as the first,
and then we have that same number of 2×M rectangles.
So N×2 = 3 (N)(N+1)/2
After deducing this we can say
For N*M we'll have (M)(M+1)/2 (N)(N+1)/2 = M(M+1)(N)(N+1)/4
So the formula for total rectangles will be M(M+1)(N)(N+1)/4
.
Combinatorial Logic:
N*M grid can be represented as (N+1) vertical lines and (M+1) horizontal lines.
In a rectangle, we need two distinct horizontal and two distinct verticals.
So going by the logic of Combinatorial Mathematics we can choose 2 vertical lines and 2 horizontal lines to form a rectangle. And total number of these combinations is the number of rectangles possible in the grid.
Total Number of Rectangles in N*M grid: N+1C2 * M+1C2 = (N*(N+1)/2!)*(M*(M+1)/2!) = N*(N+1)*M*(M+1)/4
// C++ program to count number of rectangles
// in a n x m grid
#include <bits/stdc++.h>
using namespace std;
int rectCount(int n, int m)
{
return (m * n * (n + 1) * (m + 1)) / 4;
}
/* driver code */
int main()
{
int n = 5, m = 4;
cout << rectCount(n, m);
return 0;
}
// JAVA Code to count number of
// rectangles in N*M grid
import java.util.*;
class GFG {
public static long rectCount(int n, int m)
{
return (m * n * (n + 1) * (m + 1)) / 4;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 5, m = 4;
System.out.println(rectCount(n, m));
}
}
// This code is contributed by Arnav Kr. Mandal.
# Python3 program to count number
# of rectangles in a n x m grid
def rectCount(n, m):
return (m * n * (n + 1) * (m + 1)) // 4
# Driver code
n, m = 5, 4
print(rectCount(n, m))
# This code is contributed by Anant Agarwal.
// C# Code to count number of
// rectangles in N*M grid
using System;
class GFG {
public static long rectCount(int n, int m)
{
return (m * n * (n + 1) * (m + 1)) / 4;
}
// Driver program
public static void Main()
{
int n = 5, m = 4;
Console.WriteLine(rectCount(n, m));
}
}
// This code is contributed by Anant Agarwal.
<script>
// Javascript Code to count number
// of rectangles in N*M grid
function rectCount(n, m)
{
return parseInt((m * n * (n + 1) *
(m + 1)) / 4, 10);
}
let n = 5, m = 4;
document.write(rectCount(n, m));
</script>
<?php
// PHP program to count
// number of rectangles
// in a n x m grid
function rectCount($n, $m)
{
return ($m * $n *
($n + 1) *
($m + 1)) / 4;
}
// Driver Code
$n = 5;
$m = 4;
echo rectCount($n, $m);
// This code is contributed
// by ajit
?>
Output
150
Time complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
This article is contributed by Pranav.
