# Count frequency of k in a matrix of size n where matrix(i, j) = i+j

Given a matrix of size n*n. Count the frequency of given element k in that matrix. Here base index is 1.

Examples:

```Input : n = 4, k = 7
Output : 2
Explanation
The matrix will be
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
in the given matrix where M(i, j) = i+j,
frequency of 7 is 2

Input : n = 5, k = 4
Output : 3
Explanation
The matrix will be
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
Explanation
In the given matrix where M(i, j) = i+j,
frequency of 4 is 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

First method
1) Construct a matrix of size n*n.
2) Fill the value with M(i, j)=i+j.(recall that here base index is 1)
3) Iteratively traverse the matrix and and count the frequency of given element.
This method is not that much efficient because if the matrix size is very large it’s will result in Time limit exceed.time complexity will be O(n^2).

Efficient method
In this method we avoid creating matrix of size n*n.
for example
if n = 10 the matrix will be

```2  3  4  5  6  7  8  9  10 11
3  4  5  6  7  8  9  10 11 12
4  5  6  7  8  9  10 11 12 13
5  6  7  8  9  10 11 12 13 14
6  7  8  9  10 11 12 13 14 15
7  8  9  10 11 12 13 14 15 16
8  9  10 11 12 13 14 15 16 17
9  10 11 12 13 14 15 16 17 18
10 11 12 13 14 15 16 17 18 19
11 12 13 14 15 16 17 18 19 20  ```

Now, notice how the values are same in the secondary Diagonal, and we can also find a pattern in the count it increases like 1, 2, 3, 4,
here we can see that
if (n+1)>=k then frequency of k is k-1
else frequency will be 2*n+1-k

 `// CPP program to find the frequency of k  ` `// in matrix where m(i, j)=i+j ` `#include ` `using` `namespace` `std; ` `int` `find(``int` `n, ``int` `k) ` `{ ` `    ``if` `(n + 1 >= k) ` `        ``return` `(k - 1); ` `    ``else` `        ``return` `(2 * n + 1 - k); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 4, k = 7; ` `    ``int` `freq = find(n, k); ` `    ``if` `(freq < 0) ` `        ``cout << ``" element not exist \n "``; ` `    ``else` `        ``cout << ``" Frequency of "` `<< k  ` `             ``<< ``" is "` `<< freq << ``"\n"``; ` `    ``return` `0; ` `} `

 `// Java program to find the  ` `// frequency of k in matrix ` `// in matrix where m(i, j)=i+j ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `public` `class` `GfG{ ` ` `  `    ``public` `static` `int` `find(``int` `n, ``int` `k) ` `    ``{ ` `        ``if` `(n + ``1` `>= k) ` `            ``return` `(k - ``1``); ` `        ``else` `            ``return` `(``2` `* n + ``1` `- k); ` `    ``} ` `         `  `    ``// Driver function  ` `    ``public` `static` `void` `main(String argc[]) ` `    ``{ ` `        ``int` `n = ``4``, k = ``7``; ` `        ``int` `freq = find(n, k); ` `        ``if` `(freq < ``0``) ` `            ``System.out.print(``" element"` `            ``+ ``" not exist \n "``); ` `        ``else` `            ``System.out.print(``" Frequency"` `            ``+ ``" of "` `+ k + ``" is "` `+ ` `            ``freq + ``"\n"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Sagar Shukla `

 `# Python program to find ` `# the frequency of k  ` `# in matrix where ` `# m(i, j)=i+j ` ` `  `import` `math ` ` `  `def` `find( n,  k): ` ` `  `    ``if` `(n ``+` `1` `>``=` `k): ` `        ``return` `(k ``-` `1``) ` `    ``else``: ` `        ``return` `(``2` `*` `n ``+` `1` `-` `k) ` ` `  `  `  `# Driver Code ` `n ``=` `4` `k ``=` `7` ` `  `freq ``=` `find(n, k) ` ` `  `if` `(freq < ``0``): ` `    ``print` `( ``" element not exist"``) ` `else``: ` `    ``print``(``" Frequency of "` `, k ,``" is "` `, freq ) ` ` `  `# This code is contributed ` `# by Gitanjali. `

 `// C# program to find the  ` `// frequency of k in matrix ` `// in matrix where m(i, j)=i+j ` `using` `System; ` ` `  `public` `class` `GfG{ ` ` `  `    ``public` `static` `int` `find(``int` `n, ``int` `k) ` `    ``{ ` `        ``if` `(n + 1 >= k) ` `            ``return` `(k - 1); ` `        ``else` `            ``return` `(2 * n + 1 - k); ` `    ``} ` `         `  `    ``// Driver function  ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4, k = 7; ` `        ``int` `freq = find(n, k); ` `        ``if` `(freq < 0) ` `            ``Console.WriteLine(``" element"` `            ``+ ``" not exist "``); ` `        ``else` `            ``Console.WriteLine(``" Frequency"` `            ``+ ``" of "` `+ k + ``" is "` `+ ` `            ``freq ); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m `

 `= ``\$k``) ` `        ``return` `(``\$k` `- 1); ` `    ``else` `        ``return` `(2 * ``\$n` `+ 1 - ``\$k``); ` `} ` ` `  `    ``// Driver Code ` `    ``\$n` `= 4;  ` `    ``\$k` `= 7; ` `    ``\$freq` `= find(``\$n``, ``\$k``); ` `    ``if` `(``\$freq` `< 0) ` `        ``echo` `" element not exist \n "``; ` `    ``else` `        ``echo` `" Frequency of "` `, ``\$k` `            ``, ``" is "` `, ``\$freq` `, ``"\n"``; ` `             `  `// This code is contributed by anuj_67. ` `?> `

Output:
```Frequency of 7 is 2
```

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