LeetCode #754 — MEDIUM

Reach a Number

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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 i^th 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 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to -1 (2 steps).
On the 3^rd move, we step from -1 to 2 (3 steps).

Example 2:

Input: target = 3
Output: 2
Explanation:
On the 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to 3 (2 steps).

Constraints:

-10⁹ <= target <= 10⁹
target != 0

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Binary Search

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #754: Reach a Number
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;
        }
    }
}

// Accepted solution for LeetCode #754: Reach a Number
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
	}
}

# Accepted solution for LeetCode #754: Reach a Number
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

// Accepted solution for LeetCode #754: Reach a Number
struct Solution;

impl Solution {
    fn reach_number(target: i32) -> i32 {
        let target = target.abs();
        let n = (((2 * target as i64) as f64 + 0.25).sqrt() - 0.5).ceil() as i32;
        let sum = n * (n + 1) / 2;
        if sum == target {
            n
        } else {
            let diff = sum - target;
            if diff % 2 == 0 {
                n
            } else {
                if n % 2 == 0 {
                    n + 1
                } else {
                    n + 2
                }
            }
        }
    }
}

#[test]
fn test() {
    assert_eq!(Solution::reach_number(3), 2);
    assert_eq!(Solution::reach_number(2), 3);
    assert_eq!(Solution::reach_number(5), 5);
}

// Accepted solution for LeetCode #754: Reach a Number
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #754: Reach a Number
// 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;
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(sqrt\left | \textittarget \right |)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.