# Count the number of times graph crosses X-axis

Given an integer array arr[] of size N, the task is to find the number of times graph crosses X-axis, where positive number means going above its current position by that value and a negative number means going down by that value. Initially, the current position is at the origin.

Examples:

Input: arr[] = {4, -6, 2, 8, -2, 3, -12}
Output: 3
Explanation: So the graph crosses X-axis 3 times

Input: arr[] = {1, 1, -3, 2}
Output: 2

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

Approach: Iterate over the array and keep the value of the previous level and current level into two variables. Initially, both the levels are zero. Increase / Decrease the level by the value given in the array and increase the count in below two cases.

• If the previous level is less than zero and the current level is greater than or equal to zero.
• If the previous level is greater than zero and the current level is less than or equal to zero.

Below is the implementation of the above approach.

## C++

 `// C++ implementation to count the ` `// number of times the graph ` `// crosses the x-axis. ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to to count the ` `// number of times the graph ` `// crosses the x-axis. ` `int` `times(``int` `steps[], ``int` `n) ` `{ ` ` `  `    ``int` `current_level = 0; ` `    ``int` `previous_level = 0; ` `    ``int` `count = 0; ` ` `  `    ``// Iterate over the steps array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Update the previous level and ` `        ``// current level by value given ` `        ``// in the steps array ` `        ``previous_level = current_level; ` `        ``current_level = current_level ` `                        ``+ steps[i]; ` ` `  `        ``// Condition to check that the ` `        ``// graph crosses the origin. ` `        ``if` `((previous_level < 0 ` `             ``&& current_level >= 0) ` `            ``|| (previous_level > 0 ` `                ``&& current_level <= 0)) { ` `            ``count++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `steps = { 1, -1, 0, 0, 1, 1, -3, 2 }; ` `    ``int` `n = ``sizeof``(steps) / ``sizeof``(``int``); ` ` `  `    ``cout << times(steps, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to count the  ` `// number of times the graph  ` `// crosses the x-axis.  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to to count the  ` `    ``// number of times the graph  ` `    ``// crosses the x-axis.  ` `    ``static` `int` `times(``int` `[]steps, ``int` `n)  ` `    ``{  ` `        ``int` `current_level = ``0``;  ` `        ``int` `previous_level = ``0``;  ` `        ``int` `count = ``0``;  ` `     `  `        ``// Iterate over the steps array  ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `     `  `            ``// Update the previous level and  ` `            ``// current level by value given  ` `            ``// in the steps array  ` `            ``previous_level = current_level;  ` `            ``current_level = current_level + steps[i];  ` `     `  `            ``// Condition to check that the  ` `            ``// graph crosses the origin.  ` `            ``if` `((previous_level < ``0` `&&  ` `                ``current_level >= ``0``)  ` `                ``|| (previous_level > ``0` `                ``&& current_level <= ``0``))  ` `            ``{  ` `                ``count++;  ` `            ``}  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `steps[] = { ``1``, -``1``, ``0``, ``0``, ``1``, ``1``, -``3``, ``2` `};  ` `        ``int` `n = steps.length;  ` `     `  `        ``System.out.println(times(steps, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation to count the ` `# number of times the graph ` `# crosses the x-axis. ` ` `  `# Function to to count the ` `# number of times the graph ` `# crosses the x-axis. ` `def` `times(steps, n): ` ` `  `    ``current_level ``=` `0` `    ``previous_level ``=` `0` `    ``count ``=` `0` ` `  `    ``# Iterate over the steps array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# Update the previous level and ` `        ``# current level by value given ` `        ``#in the steps array ` `        ``previous_level ``=` `current_level ` `        ``current_level ``=` `current_level``+` `steps[i] ` ` `  `        ``# Condition to check that the ` `        ``# graph crosses the origin. ` `        ``if` `((previous_level < ``0` `            ``and` `current_level >``=` `0``) ` `            ``or` `(previous_level > ``0` `                ``and` `current_level <``=` `0``)): ` `            ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver Code ` `steps ``=` `[``1``, ``-``1``, ``0``, ``0``, ``1``, ``1``, ``-``3``, ``2``] ` `n ``=` `len``(steps) ` ` `  `print``(times(steps, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation to count the  ` `// number of times the graph  ` `// crosses the x-axis. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to to count the  ` `    ``// number of times the graph  ` `    ``// crosses the x-axis.  ` `    ``static` `int` `times(``int` `[]steps, ``int` `n)  ` `    ``{  ` `        ``int` `current_level = 0;  ` `        ``int` `previous_level = 0;  ` `        ``int` `count = 0;  ` `     `  `        ``// Iterate over the steps array  ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{  ` `     `  `            ``// Update the previous level and  ` `            ``// current level by value given  ` `            ``// in the steps array  ` `            ``previous_level = current_level;  ` `            ``current_level = current_level + steps[i];  ` `     `  `            ``// Condition to check that the  ` `            ``// graph crosses the origin.  ` `            ``if` `((previous_level < 0 &&  ` `                ``current_level >= 0)  ` `                ``|| (previous_level > 0 ` `                ``&& current_level <= 0))  ` `            ``{  ` `                ``count++;  ` `            ``}  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main () ` `    ``{  ` `        ``int` `[]steps = { 1, -1, 0, 0, 1, 1, -3, 2 };  ` `        ``int` `n = steps.Length;  ` `     `  `        ``Console.WriteLine(times(steps, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

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My Personal Notes arrow_drop_up I am a Computer Science Student at JIIT-Noida I have worked for Sanfoundry previously and contributed more than 50 articles

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Improved By : mohit kumar 29, AnkitRai01

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