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Formatted question description: https://leetcode.ca/all/754.html

# 754. Reach a Number (Easy)

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

On each move, you can either go left or right. During the n-th move (starting from 1), you take n steps.

Return the minimum number of steps required to reach the destination.

Example 1:

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.


Example 2:

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.


Note:

• target will be a non-zero integer in the range [-10^9, 10^9].
• Companies:
InMobi

Related Topics:
Math

## Solution 1.

Try to find pattern:

Firstly, for numbers n(n+1)/2, they require n steps, where n >= 1. For example, 1 = 1, 3 = 1 + 2, 6 = 1 + 2 + 3

How to get the numbers inbetween?

Think of 2: the smallest n(n+1)/2 greater than 2 is 3. If we invert 3’s component 1 or 2, we actually subtract 1*2 or 2*2 from 3. Notice the subtraction is using even number. But 3 - 2 = 1 is an odd number. So we can’t get 2 from 3.

Consider the next n(n+1)/2, which is 6. 6 - 2 = 4. So we can invert 4 / 2 = 2 in equation 6 = 1 + 2 + 3 to get 2 = 1 - 2 + 3.

Let’s examine our algorithm with two more test cases.

For 4, the next n(n+1)/2 is 6. Since 6 - 4 = 2, so we can invert 1 in 6 = 1 + 2 + 3 to get 4 = -1 + 2 + 3.

For 5:

• check 6, 6 - 5 = 1, doesn’t work.
• check 10 = 1+..+4, 10 - 5 = 5, doesn’t work.
• check 15 = 1+..+5, 15 - 5 = 10. Works! 5 = 1 + 2 + 3 + 4 - 5.

In sum, keep computing sum = n(n+1)/2. If sum is greater than or equal to target and (sum - target) % 2 == 0, the corresponding n is the answer.

// OJ: https://leetcode.com/problems/reach-a-number/
// Time: O(sqrt(T))
// Space: O(1)
class Solution {
public:
int reachNumber(int target) {
int i = 0, sum = 0;
target = abs(target);
while (sum < target || (sum - target) % 2 != 0) {
sum += ++i;
}
return i;
}
};

• class Solution {
public int reachNumber(int target) {
target = Math.abs(target);
if (target == 0)
return 0;
int position = 0;
int step = 0;
while (position < target || position % 2 != target % 2) {
step++;
position += step;
}
return step;
}
}

############

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;
}
}
}

• // OJ: https://leetcode.com/problems/reach-a-number/
// Time: O(sqrt(T))
// Space: O(1)
class Solution {
public:
int reachNumber(int target) {
int i = 0, sum = 0;
target = abs(target);
while (sum < target || (sum - target) % 2 != 0) {
sum += ++i;
}
return i;
}
};

• 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

############

class Solution(object):
def reachNumber(self, target):
"""
:type target: int
:rtype: int
"""
target = abs(target)
k = 0
sum = 0
while sum < target:
k += 1
sum += k
d = sum - target
if d % 2 == 0:
return k
return k + 1 + (k % 2)

• 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;
}
};