Find minimum moves to reach target on an infinite line

Given a target position on infinite number line, i.e -infinity to +infinity. Starting form 0 you have to reach the target by moving as described : In ith move you can take i steps forward or backward. Find the minimum number of moves require to reach the target.

Examples:

Input : target = 3
Output : 2
Explanation:
On the first move we step from 0 to 1.
On the second step we step from 1 to 3.

Input: target = 2
Output: 3
Explanation:
On the first move we step from 0 to 1.
On the second move we step  from 1 to -1.
On the third move we step from -1 to 2.

We have discussed a naive recursive solution in below post. 
Minimum steps to reach a destination
If target is negative, we can take it as positive because we start from 0 in symmetrical way. 
Idea is to move in one direction as long as possible, this will give minimum moves. Starting at 0 first move takes us to 1, second move takes us to 3 (1+2) position, third move takes us to 6 (1+2+3) position, ans so on; So for finding target we keep on adding moves until we find the nth move such that 1+2+3+…+n>=target. Now if sum (1+2+3+…+n) is equal to target the our job is done, i.e we’ll need n moves to reach target. Now next case where sum is greater than target. Find the difference by how much we are ahead, i.e sum – target. Let the difference be d = sum – target.  

Example: if the target is 2 then, the number of steps should be 3. Lets see how we get that,

1-2+3=2 ( if we start from 0 and move 1 step forward and then 2 steps backward (we land on -1) and further if take 3 steps forward we land on 2.)

1-2+3 can also be written as (1+2+3) -2*(2).

that means if we take the i_th move backward then the new sum will become (sum – 2i).

Now if sum-2i = target then our job is done. Since, sum – target = 2i, i.e difference should be even as we will get an integer i flipping which will give the answer. So following cases arise. 
Case 1 : Difference is even then answer is n, (because we will always get a move flipping which will lead to target). 
Case 2 : Difference is odd, then we take one more step, i.e add n+1 to sum and now again take the difference. If difference is even the n+1 is the answer else we would have to take one more move and this will certainly make the difference even then answer will be n+2.
Explanation : Since difference is odd. Target is either odd or even. 
case 1: n is even (1+2+3+…+n) then adding n+1 makes the difference even. 
case 2: n is odd then adding n+1 doesn’t makes difference even so we would have to take one more move, so n+2.
Example: 
target = 5. 
we keep on taking moves until we reach target or we just cross it. 
sum = 1 + 2 + 3 = 6 > 5, step = 3. 
Difference = 6 – 5 = 1. Since the difference is an odd value, we will not reach the target by flipping any move from +i to -i. So we increase our step. We need to increase step by 2 to get an even difference (since n is odd and target is also odd). Now that we have an even difference, we can simply switch any move to the left (i.e. change + to -) as long as the summation of the changed value equals to half of the difference. We can switch 1 and 4 or 2 and 3 or 5. 

C++

// CPP program to find minimum moves to
// reach target if we can move i steps in
// i-th move.
#include <iostream>
using namespace std;

int reachTarget(int target)
{
    // Handling negatives by symmetry
    target = abs(target);

    // Keep moving while sum is smaller or difference
    // is odd.
    int sum = 0, step = 0;
    while (sum < target || (sum - target) % 2 != 0) {
        step++;
        sum += step;
    }
    return step;
}

// Driver code
int main()
{
    int target = 5;
    cout << reachTarget(target);
    return 0;
}

Java

// Java program to find minimum
//moves to reach target if we can
// move i steps in i-th move.
import java.io.*;
import java.math.*;

class GFG {

    static int reachTarget(int target)
    {
        // Handling negatives by symmetry
        target = Math.abs(target);

        // Keep moving while sum is smaller
        // or difference is odd.
        int sum = 0, step = 0;

        while (sum < target || (sum - target) % 2
                                        != 0) {
            step++;
            sum += step;
        }
        return step;
    }

    // Driver code
    public static void main(String args[])
    {
       int target = 5;
       System.out.println(reachTarget(target));
    }
}

// This code is contributed by Nikita tiwari.

Python 3

# Python 3 program to find minimum
# moves to reach target if we can
# move i steps in i-th move.


def reachTarget(target) :

    # Handling negatives by symmetry
    target = abs(target)

    # Keep moving while sum is
    # smaller or difference is odd.
    sum = 0
    step = 0
    while (sum < target or (sum - target) %
                                  2 != 0) :
        step = step + 1
        sum = sum + step

    return step


# Driver code
target = 5
print(reachTarget(target))


# This code is contributed by Nikita Tiwari

C#

// C# program to find minimum
//moves to reach target if we can
// move i steps in i-th move.
using System;

class GFG {

    static int reachTarget(int target)
    {
        // Handling negatives by symmetry
        target = Math.Abs(target);

        // Keep moving while sum is smaller
        // or difference is odd.
        int sum = 0, step = 0;

        while (sum < target ||
              (sum - target) % 2!= 0)
        {
            step++;
            sum += step;
        }
        return step;
    }

    // Driver code
    public static void Main()
    {
    int target = 5;
    Console.WriteLine(reachTarget(target));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find
// minimum moves to reach
// target if we can move i
// steps in i-th move.

function reachTarget($target)
{
    // Handling negatives
    // by symmetry
    $target = abs($target);

    // Keep moving while sum is
    // smaller or difference is odd.
    $sum = 0; $step = 0;
    while ($sum < $target or
          ($sum - $target) % 2 != 0)
          {
            $step++;
            $sum += $step;
          }
    return $step;
}

// Driver code
$target = 5;
echo reachTarget($target);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// JavaScript program to find minimum
//moves to reach target if we can
// move i steps in i-th move.

function reachTarget(target)
    {
        // Handling negatives by symmetry
        target = Math.abs(target);

        // Keep moving while sum is smaller
        // or difference is odd.
        let sum = 0, step = 0;

        while (sum < target || (sum - target) % 2
                                        != 0) {
            step++;
            sum += step;
        }
        return step;
    }

// Driver code

    let target = 5;
      document.write(reachTarget(target));

</script>

Output:

5

Time Complexity: O(n), where n is the number of steps 
Auxiliary Space: O(1), since no extra space has been taken.

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above