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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

• Java
• C++
• Python
• Go
• Javascript

• 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; } }; 

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