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The problem “Reach a Number” on LeetCode is stated as follows:

You are standing at position 0 on an infinite number line. There is a destination at position target.

You can make a move of +1, -1, +2, -2, +3, -3, …, etc.

In other words, on the i-th move, you can choose to move either i steps to the right or i steps to the left.

Return the minimum number of moves required to reach the target.

Clarifying Questions

1. Negative Target: Can the target be negative?
   • Yes, you can treat the problem symmetrically because moving to a negative target is the same as moving to the positive one due to symmetry.
2. Complexity Constraints: Is there any constraint on the value of the target?
   • The value of target is an integer and may be large, a brute-force approach may be infeasible.
3. Starting Point: Do we always start from position 0?
   • Yes.

Strategy

The task is to minimize the number of steps required to reach the target position.

Steps to Approach:

1. Symmetry: Since the problem is symmetric around 0, we can always consider target as positive. If the target is negative, we can take its absolute value.

2. Sum of First N Naturals: We consider making steps of 1, 2, 3, ..., k by summing these steps. This series sum is S = k * (k + 1) / 2.

3. Find Minimum k: Find the smallest k such that S is greater than or equal to the target. However, S should be adjusted to ensure that the difference between S and target is an even number, because only then can the positive and negative step adjustments ultimately balance out to reach the exact target.

Code

class Solution:
    def reachNumber(self, target: int) -> int:
        target = abs(target)  # Treat as positive for symmetry
        step = 0
        sum_steps = 0
        
        # Keep incrementing the step until the cumulative sum is at least the target,
        # and the difference between sum_steps and target is even.
        while sum_steps < target or (sum_steps - target) % 2 != 0:
            step += 1
            sum_steps += step
        
        return step

# Example usage:
sol = Solution()
print(sol.reachNumber(3))  # Output: 2
print(sol.reachNumber(2))  # Output: 3

Time Complexity

• Time Complexity: O(√target)
  • The while loop ultimately converges at a step that is roughly proportional to the square root of the target.
• Space Complexity: O(1)
  • The solution uses a constant amount of space for variables irrespective of the input size.

This approach efficiently calculates the minimum number of steps required to reach the target position on an infinite number line.

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