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:Â
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Examples:Â
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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
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Approach: By sketching a small diagram we can see the two cases:Â
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- 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:Â
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C++
// C++ program to find the number of steps// required to reach (x, y) from (0, 0) following// a zig-zag pathÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to return the required positionint countSteps(int x, int y){Â Â Â Â if (x < y) {Â Â Â Â Â Â Â Â return x + y + 2 * ((y - x) / 2);Â Â Â Â }Â Â Â Â else {Â Â Â Â Â Â Â Â return x + y + 2 * (((x - y) + 1) / 2);Â Â Â Â }}Â
// Driver Codeint main(){Â Â Â Â int x = 4, y = 3;Â Â Â Â cout << countSteps(x, y);Â
    return 0;} |
Java
// Java program to find the number of steps // required to reach (x, y) from (0, 0) following // a zig-zag path Â
class GfG {Â
// 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);     } } Â
// Driver Code public static void main(String[] args) { Â Â Â Â int x = 4, y = 3; Â Â Â Â System.out.println(countSteps(x, y)); }} Â
// 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) Â
# Driver Code if __name__ == "__main__": Â Â Â Â Â Â x, y = 4, 3Â Â Â Â print(countSteps(x, y)) Â Â Â # This code is contributed by Rituraj Jain |
C#
// C# program to find the number of steps // required to reach (x, y) from (0, 0)Â // following a zig-zag path using System;Â
class GfG {Â
// 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);     } } Â
// Driver Code public static void Main() { Â Â Â Â int x = 4, y = 3; Â Â Â Â Console.WriteLine(countSteps(x, y)); }} Â
// This code is contributed by Code_Mech. |
PHP
<?php// PHP program to find the number of steps // required to reach (x, y) from (0, 0) // following a zig-zag path Â
// Function to return the required position function countSteps($x, $y) {     if ($x < $y)     {         return $x + $y + 2 *              (($y - $x) / 2);     }     else    {         return $x + $y + 2 *             ((($x - $y) + 1) / 2);     } } Â
// Driver Code $x = 4; $y = 3; echo(countSteps($x, $y)); Â
// This code is contributed // by Code_Mech.?> |
Javascript
<script>Â
    // Javascript program to find the number of steps    // required to reach (x, y) from (0, 0) following    // a zig-zag pathÂ
    // Function to return the required position    function countSteps(x, y)    {      if (x < y) {        return x + y + 2 * parseInt((y - x) / 2);      }      else {        return x + y + 2 * parseInt(((x - y) + 1) / 2);      }    }Â
    // Driver Code    var x = 4, y = 3;    document.write(countSteps(x, y));Â
// This code is contributed by rrrtnx.  </script> |
Output:Â
9
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Time Complexity: O(1)
Auxiliary Space: O(1)
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