Number of steps required to reach point (x,y) from (0,0) using zig-zag way

Given a coordinate (x, y). The task is to calculate the number of steps required to reach point (x, y) from (0, 0) using zig-zag way and you cannot travel in straight line for more than 1 unit. Also, start moving along Y axis.

For example we can reach the Point denoted by red color in the respective ways as shown in the below diagram:

Examples:

```Input: x = 4, y = 4
Output: 8
In the diagram above the line is passing
using 8 steps.

Input: x = 4, y = 3
Output: 9

Input: x = 2, y = 1
Output: 5
```

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

Approach: By sketching a small diagram we can see the two cases:

• Case 1: If x is less than y then answer will always be x + y + 2*((y-x)/2).
• Case 2: If x is greater than equal to y then answer will always be x + y + 2*(((x-y)+1)/2).

Below is the implementation of the above approach:

C++

 `// C++ program to find the number of steps ` `// required to reach (x, y) from (0, 0) following ` `// a zig-zag path ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the required position ` `int` `countSteps(``int` `x, ``int` `y) ` `{ ` `    ``if` `(x < y) { ` `        ``return` `x + y + 2 * ((y - x) / 2); ` `    ``} ` `    ``else` `{ ` `        ``return` `x + y + 2 * (((x - y) + 1) / 2); ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `x = 4, y = 3; ` `    ``cout << countSteps(x, y); ` ` `  `    ``return` `0; ` `} `

Java

 `// Java program to find the number of steps  ` `// required to reach (x, y) from (0, 0) following  ` `// a zig-zag path  ` ` `  `class` `GfG  ` `{ ` ` `  `// Function to return the required position  ` `static` `int` `countSteps(``int` `x, ``int` `y)  ` `{  ` `    ``if` `(x < y)  ` `    ``{  ` `        ``return` `x + y + ``2` `* ((y - x) / ``2``);  ` `    ``}  ` `    ``else`  `    ``{  ` `        ``return` `x + y + ``2` `* (((x - y) + ``1``) / ``2``);  ` `    ``}  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `x = ``4``, y = ``3``;  ` `    ``System.out.println(countSteps(x, y));  ` `} ` `}  ` ` `  `// This code is contributed by Prerna Saini `

Python3

 `# Python3 program to find the number of  ` `# steps required to reach (x, y) from  ` `# (0, 0) following a zig-zag path  ` `   `  `# Function to return the required position  ` `def` `countSteps(x, y):  ` `  `  `    ``if` `x < y: ` `        ``return` `x ``+` `y ``+` `2` `*` `((y ``-` `x) ``/``/` `2``)  ` `      `  `    ``else``: ` `        ``return` `x ``+` `y ``+` `2` `*` `(((x ``-` `y) ``+` `1``) ``/``/` `2``)  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` `  `  `    ``x, y ``=` `4``, ``3`  `    ``print``(countSteps(x, y))  ` `   `  `# This code is contributed by Rituraj Jain `

C#

 `// C# program to find the number of steps  ` `// required to reach (x, y) from (0, 0)   ` `// following a zig-zag path  ` `using` `System; ` ` `  `class` `GfG  ` `{ ` ` `  `// Function to return the required position  ` `static` `int` `countSteps(``int` `x, ``int` `y)  ` `{  ` `    ``if` `(x < y)  ` `    ``{  ` `        ``return` `x + y + 2 * ((y - x) / 2);  ` `    ``}  ` `    ``else` `    ``{  ` `        ``return` `x + y + 2 * (((x - y) + 1) / 2);  ` `    ``}  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `x = 4, y = 3;  ` `    ``Console.WriteLine(countSteps(x, y));  ` `} ` `}  ` ` `  `// This code is contributed by Code_Mech. `

PHP

 ` `

Output:

```9
```

Time Complexity: O(1)

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