Number of rectangles in N*M grid
We are given a N*M grid, print the number of rectangles in it.
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
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 1×1, there is 1 rectangle.
If the grid is 2×1, there will be 2 + 1 = 3 rectangles
If it grid is 3×1, 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
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
Time complexity: O(1)
Auxiliary Space: O(1)
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