# Count entries equal to x in a special matrix

• Difficulty Level : Easy
• Last Updated : 20 Jul, 2022

You are given a square matrix (matrix[][]) of order n, where matrix[i][j] = i*j. Find the number of cells which have entry equal to a given number x.
NOte : Indexing of matrix starts from 1, i.e. 1<= i,j <= n.

Examples :

```Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 4
Output : 3

Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 12
Output : 2```

A simple Approach is to traverse the whole of matrix and check whether cell value is equal to given x and then increase count value accordingly. Time complexity in this approach is quite high and is equal to O(n^2) and space complexity O(1).

```// traverse and check whole matrix
for (int i=1; i<=n ; i++)
{
for (int j=1; j<=n; j++)
if (i * j == x)
count++;
}
// return count
return count;```

An efficient approach is to only find the number of factors of given x in the range 0 to x and also those factors (including divisor and quotient ) must be less than or equal to n (order of matrix). In this case time complexity will be O(n) and space complexity O(1).

```// traverse and find the factors
for (int i=1; i<=n && i<=x ; i++)
{
// x%i == 0 means i is factor of x
// x/i <= n means i and j are <= n (for i*j=x)
if ( x/i <= n && x%i ==0)
count++;
}
// return count
return count;```

Implementation:

## C++

 `// CPP program for counting number of cell``// equals to given x``#include``using` `namespace` `std;` `// function to count factors as number of cell``int` `count (``int` `n, ``int` `x)``{``    ``int` `count=0;``    ``// traverse and find the factors``    ``for` `(``int` `i=1; i<=n && i<=x ; i++)``    ``{``        ``// x%i == 0 means i is factor of x``        ``// x/i <= n means i and j are <= n (for i*j=x)``        ``if` `( x/i <= n && x%i ==0)``            ``count++;``    ``}``    ``// return count``    ``return` `count;``}` `// driver program``int` `main()``{``    ``int` `n = 8;``    ``// we can manually assume matrix of order 8*8``    ``// where mat[i][j] = i*j , 0

## Java

 `// Java program for counting number of``// cell equals to given x``class` `GFG``{``    ``// function to count factors as``    ``// number of cell``    ``static` `int` `count (``int` `n, ``int` `x)``    ``{``        ``int` `count = ``0``;``    ` `        ``// traverse and find the factors``        ``for` `(``int` `i = ``1``; i <= n && i <= x ;``                                    ``i++)``        ``{``            ``// x%i == 0 means i is factor``            ``// of x. x/i <= n means i and``            ``// j are <= n (for i*j=x)``            ``if` `( x / i <= n && x % i == ``0``)``                ``count++;``        ``}``        ` `        ``// return count``        ``return` `count;``    ``}` `    ``// driver program``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``8``;``        ` `        ``// we can manually assume matrix``        ``// of order 8*8 where``        ``// mat[i][j] = i*j , 0

## Python3

 `# Python 3 program for counting``# number of cell equals to given x` `# function to count factors``# as number of cell``def` `count(n, x):``    ``cnt ``=` `0` `    ``# traverse and find the factors``    ``for` `i ``in` `range``(``1``, n ``+` `1``):` `        ``# // x%i == 0 means i is factor of x``        ``# x/i <= n means i and j are <= n (for i*j=x)``        ``if` `i <``=` `x:``            ``if` `x ``/``/` `i <``=` `n ``and` `x ``%` `i ``=``=` `0``:``                ``cnt ``+``=` `1``    ``return` `cnt` `# Driver code``n ``=` `8``x ``=` `24``print``(count(n, x))` `# This code is contributed by Shrikant13`

## C#

 `// C# program for counting number``// of cell equals to given x``using` `System;` `class` `GFG {``    ` `    ``// function to count factors as``    ``// number of cell``    ``static` `int` `count (``int` `n, ``int` `x) {``        ` `        ``int` `count = 0;``    ` `        ``// traverse and find the factors``        ``for` `(``int` `i = 1; i <= n &&``             ``i <= x ; i++)``        ``{``            ` `            ``// x%i == 0 means i is factor``            ``// of x. x/i <= n means i and``            ``// j are <= n (for i*j=x)``            ``if` `( x / i <= n && x % i == 0)``                ``count++;``        ``}``        ` `        ``// return count``        ``return` `count;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 8;``        ` `        ``// we can manually assume matrix``        ``// of order 8*8 where``        ``// mat[i][j] = i*j , 0

## PHP

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## Javascript

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Output

`4`

Time Complexity: O(n)
Auxiliary Space: O(1)

This article is contributed by Aarti_Rathi and Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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