# 754. Reach a Number

## Description

You are standing at position 0 on an infinite number line. There is a destination at position target.

You can make some number of moves numMoves so that:

• On each move, you can either go left or right.
• During the ith move (starting from i == 1 to i == numMoves), you take i steps in the chosen direction.

Given the integer target, return the minimum number of moves required (i.e., the minimum numMoves) to reach the destination.

Example 1:

Input: target = 2
Output: 3
Explanation:
On the 1st move, we step from 0 to 1 (1 step).
On the 2nd move, we step from 1 to -1 (2 steps).
On the 3rd move, we step from -1 to 2 (3 steps).


Example 2:

Input: target = 3
Output: 2
Explanation:
On the 1st move, we step from 0 to 1 (1 step).
On the 2nd move, we step from 1 to 3 (2 steps).


Constraints:

• -109 <= target <= 109
• target != 0

## Solutions

• class Solution {
public int reachNumber(int target) {
target = Math.abs(target);
int s = 0, k = 0;
while (true) {
if (s >= target && (s - target) % 2 == 0) {
return k;
}
++k;
s += k;
}
}
}

• class Solution {
public:
int reachNumber(int target) {
target = abs(target);
int s = 0, k = 0;
while (1) {
if (s >= target && (s - target) % 2 == 0) return k;
++k;
s += k;
}
}
};

• class Solution:
def reachNumber(self, target: int) -> int:
target = abs(target)
s = k = 0
while 1:
if s >= target and (s - target) % 2 == 0:
return k
k += 1
s += k


• func reachNumber(target int) int {
if target < 0 {
target = -target
}
var s, k int
for {
if s >= target && (s-target)%2 == 0 {
return k
}
k++
s += k
}
}

• /**
* @param {number} target
* @return {number}
*/
var reachNumber = function (target) {
target = Math.abs(target);
let [s, k] = [0, 0];
while (1) {
if (s >= target && (s - target) % 2 == 0) {
return k;
}
++k;
s += k;
}
};