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++
#include <bits/stdc++.h>
using namespace std;
int getsum(int x)
{
return (x * (x + 1)) / 2;
}
int countJumps(int n)
{
n = abs(n);
int ans = 0;
while (getsum(ans) < n or (getsum(ans) - n) & 1)
ans++;
return ans;
}
int main()
{
int n = 9;
cout << countJumps(n);
return 0;
}
|
Java
class GFG
{
static int getsum(int x)
{
return (x * (x + 1)) / 2;
}
static int countJumps(int n)
{
n = Math.abs(n);
int ans = 0;
while (getsum(ans) < n ||
((getsum(ans) - n) & 1) > 0)
ans++;
return ans;
}
public static void main(String args[])
{
int n = 9;
System.out.println(countJumps(n));
}
}
|
Python3
def getsum(x):
return int((x * (x + 1)) / 2)
def countJumps(n):
n = abs(n)
ans = 0
while (getsum(ans) < n or
(getsum(ans) - n) & 1):
ans += 1
return ans
if __name__ == '__main__':
n = 9
print(countJumps(n))
|
C#
using System;
class GFG
{
static int getsum(int x)
{
return (x * (x + 1)) / 2;
}
static int countJumps(int n)
{
n = Math.Abs(n);
int ans = 0;
while (getsum(ans) < n || ((getsum(ans) - n) & 1)>0)
ans++;
return ans;
}
static void Main()
{
int n = 9;
Console.WriteLine(countJumps(n));
}
}
|
PHP
<?php
function getsum($x)
{
return ($x * ($x + 1)) / 2;
}
function countJumps($n)
{
$n = abs($n);
$ans = 0;
while (getsum($ans) < $n or
(getsum($ans) - $n) & 1)
$ans++;
return $ans;
}
$n = 9;
echo countJumps($n);
?>
|
Javascript
<script>
function getsum(x)
{
return (x * (x + 1)) / 2;
}
function countJumps(n)
{
n = Math.abs(n);
let ans = 0;
while (getsum(ans) < n ||
((getsum(ans) - n) & 1) > 0)
ans++;
return ans;
}
let n = 9;
document.write(countJumps(n));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)