Find the number of jumps to reach X in the number line from zero

Given an integer X. The task is to find the number of jumps to reach a point X in the number line starting from zero. 
Note: The first jump made can be of length one unit and each successive jump will be exactly one unit longer than the previous jump in length. It is allowed to go either left or right in each jump. 
Examples:

Input : X = 8
Output : 4
Explanation : 
0 -> -1 -> 1 -> 4-> 8 are possible stages.

Input : X = 9
Output : 5
Explanation : 
0 -> -1 -> -3 -> 0 -> 4-> 9 are 
possible stages

Approach: On observing carefully, it can be said easily that: 

• If you have always jumped in the right direction then after n jumps you will be at the point p = 1 + 2 + 3 + 4 + … + n.
• In any of these n jumps, if instead of jumping right, you jumped left in the kth jump (k<=n), you would be at point p – 2k.
• Moreover, by carefully choosing which jumps to go left and which to go right, after n jumps, you can be at any point between n * (n + 1) / 2 and – (n * (n + 1) / 2) with the same parity as n * (n + 1) / 2.

Keeping the above points in mind, what you must do is simulate the jumping process, always jumping to the right, and if at some point, you’ve reached a point that has the same parity as X and is at or beyond X, you’ll have your answer.
Below is the implementation of the above approach: 

C++

// C++ program to find the number of jumps
// to reach X in the number line from zero

#include <bits/stdc++.h>
using namespace std;

// Utility function to calculate sum
// of numbers from 1 to x
int getsum(int x)
{
    return (x * (x + 1)) / 2;
}

// Function to find the number of jumps
// to reach X in the number line from zero
int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = abs(n);

    // To store required answer
    int ans = 0;

    // Continue till number is lesser or not in same parity
    while (getsum(ans) < n or (getsum(ans) - n) & 1)
        ans++;

    // Return the required answer
    return ans;
}

// Driver code
int main()
{
    int n = 9;

    cout << countJumps(n);

    return 0;
}

Java

// Java program to find the number of jumps
// to reach X in the number line from zero

class GFG
{

// Utility function to calculate sum
// of numbers from 1 to x
static int getsum(int x)
{
    return (x * (x + 1)) / 2;
}

// Function to find the number of jumps
// to reach X in the number line from zero
static int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = Math.abs(n);

    // To store required answer
    int ans = 0;

    // Continue till number is lesser
    // or not in same parity
    while (getsum(ans) < n ||
         ((getsum(ans) - n) & 1) > 0)
        ans++;

    // Return the required answer
    return ans;
}

// Driver code
public static void main(String args[])
{
    int n = 9;

    System.out.println(countJumps(n));
}
}

// This code is contributed by Ryuga

Python3

# Python 3 program to find the number of jumps
# to reach X in the number line from zero

# Utility function to calculate sum
# of numbers from 1 to x
def getsum(x):
    return int((x * (x + 1)) / 2)

# Function to find the number of jumps
# to reach X in the number line from zero
def countJumps(n):

    # First make number positive
    # Answer will be same either it is
    # Positive or negative
    n = abs(n)

    # To store the required answer
    ans = 0

    # Continue till number is lesser
    # or not in same parity
    while (getsum(ans) < n or
          (getsum(ans) - n) & 1):
        ans += 1

    # Return the required answer
    return ans

# Driver code
if __name__ == '__main__':
    n = 9

    print(countJumps(n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find the number of jumps
// to reach X in the number line from zero
using System;

class GFG
{

// Utility function to calculate sum
// of numbers from 1 to x
static int getsum(int x)
{
    return (x * (x + 1)) / 2;
}

// Function to find the number of jumps
// to reach X in the number line from zero
static int countJumps(int n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = Math.Abs(n);

    // To store required answer
    int ans = 0;

    // Continue till number is lesser or not in same parity
    while (getsum(ans) < n || ((getsum(ans) - n) & 1)>0)
        ans++;

    // Return the required answer
    return ans;
}

// Driver code
static void Main()
{
    int n = 9;

    Console.WriteLine(countJumps(n));
}
}

// This code is contributed by mits

PHP

<?php
// PHP program to find the number of jumps
// to reach X in the number line from zero

// Utility function to calculate sum
// of numbers from 1 to x
function getsum($x)
{
    return ($x * ($x + 1)) / 2;
}

// Function to find the number of jumps
// to reach X in the number line from zero
function countJumps($n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    $n = abs($n);

    // To store required answer
    $ans = 0;

    // Continue till number is lesser
    // or not in same parity
    while (getsum($ans) < $n or
          (getsum($ans) - $n) & 1)
        $ans++;

    // Return the required answer
    return $ans;
}

// Driver code
$n = 9;

echo countJumps($n);

// This code is contributed by Akanksha Rai
?>

Javascript

<script>

// Javascript program to find the number of jumps
// to reach X in the number line from zero

// Utility function to calculate sum
// of numbers from 1 to x
function getsum(x)
{
    return (x * (x + 1)) / 2;
}

// Function to find the number of jumps
// to reach X in the number line from zero
function countJumps(n)
{
    // First make number positive
    // Answer will be same either it is
    // Positive or negative
    n = Math.abs(n);

    // To store required answer
    let ans = 0;

    // Continue till number is lesser
    // or not in same parity
    while (getsum(ans) < n ||
         ((getsum(ans) - n) & 1) > 0)
        ans++;

    // Return the required answer
    return ans;
}

// Driver Code

      let n = 9;

    document.write(countJumps(n));

</script>

Output:

5

Time Complexity: O(n)

Auxiliary Space: O(1)