754

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/
// Author: github.com/lzl124631x
// 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;
    }
};