-Given two positive integers **N** and **X**, the task is to count the occurrences of the given integer **X** in an **N**-length __square matrix__ generated such that each element of the matrix is equal to the product of its row and column indices (*1-based indexing*).

**Examples:**

Input:N = 5, X = 6Output:2Explanation:The 2D array formed is equal to the :1 2 3 4 52 4 6 8 103 6 9 12 154 8 12 16 205 10 15 20 25There are 2 occurrences of the element X(= 6) in the generated array.

Input:N = 7, X = 12Output:4

**Naive Approach:** Refer to __this article__ for the simplest approach to solve the problem by constructing the given matrix by multiplying the row and column indices to obtain each matrix element and count the number of occurrences of **X.**

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(N^{2})

**Efficient Approach: **The idea is based on the observation that the **i ^{th}**

^{ }row contains all the multiples of

**i**in the range

**[1, N]**. Therefore,

**X**occurs in the

**i**

^{th}^{ }row if and only if

**X**is exactly divisible by

**i**and

**X / i**should be less than or equal to

**N**. If found to be true, increment the count by 1. Follow the steps below to solve the problem:

- Initialize a variable, say
**count**, to store the count of occurrences of**X**in the generated matrix. __Iterate over the range__**[1, N]**using the variable**i**and perform the following steps:- If
**X**is divisible by**i**, store the quotient of**X**/**i**in a variable, say**b**. - If the value of
**b**falls in the range**[1, N]**, then increase the**count**by**1**.

- If
- After completing the above steps, print the value of
**count**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count the occurrences` `// of X in the generated square matrix` `int` `countOccurrences(` `int` `n, ` `int` `x)` `{` ` ` `// Store the required result` ` ` `int` `count = 0;` ` ` `// Iterate over the range [1, N]` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `// Check if x is a` ` ` `// multiple of i or not` ` ` `if` `(x % i == 0) {` ` ` `// Check if the other multiple` ` ` `// exists in the range or not` ` ` `if` `(x / i <= n)` ` ` `count++;` ` ` `}` ` ` `}` ` ` `// Print the result` ` ` `cout << count;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 7, X = 12;` ` ` `countOccurrences(N, X);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `class` `GFG{` `// Function to count the occurrences` `// of X in the generated square matrix` `static` `void` `countOccurrences(` `int` `n, ` `int` `x)` `{` ` ` `// Store the required result` ` ` `int` `count = ` `0` `;` ` ` `// Iterate over the range [1, N]` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) {` ` ` `// Check if x is a` ` ` `// multiple of i or not` ` ` `if` `(x % i == ` `0` `) {` ` ` `// Check if the other multiple` ` ` `// exists in the range or not` ` ` `if` `(x / i <= n)` ` ` `count++;` ` ` `}` ` ` `}` ` ` `// Print the result` ` ` `System.out.print(count);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `7` `, X = ` `12` `;` ` ` `countOccurrences(N, X);` `}` `}` `// This code contributed by shikhasingrajput` |

## Python3

`# Python3 program for the above approach` `# Function to count the occurrences` `# of X in the generated square matrix` `def` `countOccurrences(n, x):` ` ` ` ` `# Store the required result` ` ` `count ` `=` `0` ` ` `# Iterate over the range [1, N]` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `# Check if x is a` ` ` `# multiple of i or not` ` ` `if` `(x ` `%` `i ` `=` `=` `0` `):` ` ` `# Check if the other multiple` ` ` `# exists in the range or not` ` ` `if` `(x ` `/` `/` `i <` `=` `n):` ` ` `count ` `+` `=` `1` ` ` `# Print the result` ` ` `print` `(count)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `N ` `=` `7` ` ` `X ` `=` `12` ` ` ` ` `countOccurrences(N, X)` `# This code is contributed by ukasp` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to count the occurrences` ` ` `// of X in the generated square matrix` ` ` `static` `void` `countOccurrences(` `int` `n, ` `int` `x)` ` ` `{` ` ` `// Store the required result` ` ` `int` `count = 0;` ` ` `// Iterate over the range [1, N]` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `// Check if x is a` ` ` `// multiple of i or not` ` ` `if` `(x % i == 0) {` ` ` `// Check if the other multiple` ` ` `// exists in the range or not` ` ` `if` `(x / i <= n)` ` ` `count++;` ` ` `}` ` ` `}` ` ` `// Print the result` ` ` `Console.WriteLine(count);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `int` `N = 7, X = 12;` ` ` `countOccurrences(N, X);` ` ` `}` `}` `// This code is contributed by splevel62.` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to count the occurrences` `// of X in the generated square matrix` `function` `countOccurrences(n, x)` `{` ` ` `// Store the required result` ` ` `var` `count = 0;` ` ` `// Iterate over the range [1, N]` ` ` `for` `(` `var` `i = 1; i <= n; i++) {` ` ` `// Check if x is a` ` ` `// multiple of i or not` ` ` `if` `(x % i == 0) {` ` ` `// Check if the other multiple` ` ` `// exists in the range or not` ` ` `if` `(x / i <= n)` ` ` `count++;` ` ` `}` ` ` `}` ` ` `// Print the result` ` ` `document.write( count);` `}` `// Driver Code` `var` `N = 7, X = 12;` `countOccurrences(N, X);` `</script>` |

**Output:**

4

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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