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Leetcode 754. Reach a Number

Problem Statement

The problem is to determine the minimum number of steps required to reach a target number starting from zero. In each step (i), you can either go (i) steps to the right (positive direction) or (i) steps to the left (negative direction).

Given an integer target, return the minimum number of steps to reach the target.

Clarifying Questions

1. Clarify the inputs and outputs:
   • Input: A single integer, target, which represents the target position on a number line.
   • Output: A single integer, which is the minimum number of steps required to reach the target.
2. Target Position:
   • Is the target always non-negative? For example, if target is -3, should we consider it the same as reaching 3 due to symmetry?

   Answer: Yes, the target can be both positive or negative, but due to symmetry, reaching -target is the same as reaching target. Therefore, we can work with the absolute value of target.

Strategy

1. Convert Target to Absolute Value: Since moving to -target is the same as moving to target, convert the target to its absolute value to simplify the problem.
2. Cumulative Sum: Begin summing steps starting from 1. At each step (i), add (i) to a cumulative sum. Continue this until the cumulative sum is greater than or equal to the target.
3. Overshoot Adjustment:
   • If the cumulative sum exactly matches the target, return the number of steps taken.
   • If not, calculate the difference between the cumulative sum and the target.
   • Check the parity (odd/even) of the difference.
     • If the difference is even, we can adjust the steps to reach the target by flipping the direction of one of the earlier steps (since flipping an even offset retains integrality).
     • If the difference is odd, we continue to the next step until the sum minus target difference becomes even, ensuring we can adjust direction to reach target.

Code Implementation

#include <cmath>

class Solution {
public:
    int reachNumber(int target) {
        target = std::abs(target);
        int sum = 0;
        int steps = 0;
        
        // Sum steps until we surpass or match the target
        while (sum < target || (sum - target) % 2 != 0) {
            steps++;
            sum += steps;
        }

        return steps;
    }
};

Time Complexity

• The time complexity is O(sqrt(n)) where (n) is the target. This is due to the fact that the sum of the first (n) integers grows quadratically (as per the formula ( \frac{n(n + 1)}{2} )) and thus the loop runs roughly the square root of the target number.

This approach efficiently calculates the minimum steps to reach the target, correctly handling both positive and negative targets.

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