754. Reach a Number

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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

Solution 1: Mathematical Analysis

Due to symmetry, each time we can choose to move left or right, so we can take the absolute value of $\textit{target}$.

Define $s$ as the current position, and use the variable $k$ to record the number of moves. Initially, both $s$ and $k$ are $0$.

We keep adding to $s$ in a loop until $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0$. At this point, the number of moves $k$ is the answer, and we return it directly.

Why? Because if $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0$, we only need to change the sign of the positive integer $\frac{s - \textit{target}}{2}$ to negative, so that $s$ equals $\textit{target}$. Changing the sign of a positive integer essentially means changing the direction of the move, but the actual number of moves remains the same.

The time complexity is $O(\sqrt{\left | \textit{target} \right | })$, and the space complexity is $O(1)$.

Python3

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

Java

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;
        }
    }
}

C++

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;
        }
    }
};

Go

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
	}
}

JavaScript

/**
 * @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;
    }
};