754. Reach a Number

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].

Approach #1: Math. [Java]

class Solution {
    public int reachNumber(int target) {
        int sum = 0;
        int steps = 1;
        int count = 0;        target = Math.abs(target);        while (sum < target || (sum - target) % 2 != 0) {
            sum += steps;
            steps++;
            count++;
        }        return count++;
    }
}

Analysis:

Step 0: Get positive target value (step to get negative target is the same as to get positive value due to symmetry).

Step 1: Find the smallest step that the summation from 1 to step just exceeds or equalstarget.

Step 2: find the difference between sum and target. The goal is to get rid of the difference to reach target. For i-th move, if we switch the right move to the left, the change in summation will be 2*i less. Now the difference between sum and target has to be an even number in order for the math to check out.

Step 2.1: If the difference value is even, we can return the current step.

Step 2.2: If the difference value is odd, we need to increase the step untill the difference is even (at most 2 more steps needed).

Eg:

target = 5

Step 0: target = 5.

Step 1: sum = 1 + 2 + 3 = 6 > 5, step = 3.

Step 2: Difference = 6 – 5 = 1. Since the difference is an odd value, we will not reach the target by swirching any right move to the left. So we increase our step.

Step 2.2: We need to increase step by 2 to get an even difference (i.e. i + 2 + 3 + 4 + 5 = 15, now step = 5, difference = 15 – 5 = 10). 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.

Reference:

https://leetcode.com/problems/reach-a-number/discuss/112968/Short-JAVA-Solution-with-Explanation