What is the time complexity of Reach a Number?

The optimal solution has a time complexity of O(sqrt(target)).

What data structures are used to solve Reach a Number?

This problem uses Math, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Reach a Number?

Insight 1: The absolute value of the target is all that matters because we can always change the sign of our movements to reach a negative target by flipping the direction of our moves. Insight 2: The sum of the first `k` natural numbers (1 + 2 + ... + k) represents the furthest distance we can reach in `k` moves if we only move to the right. We need to find the smallest `k` such that this sum is greater than or equal to the absolute value of the target.

Which companies ask LeetCode 754. Reach a Number in interviews?

This problem is frequently asked at Commvault, InMobi.

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Reach a Number - LeetCode 754 Solution

Reach a Number - Complete Solution Guide

Reach a Number is LeetCode problem 754, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 m

Detailed Explanation

The problem asks us to find the minimum number of moves needed to reach a target position on an infinite number line, starting from position 0. Each move `i` allows us to move `i` steps either left or right. The goal is to determine the fewest number of moves to arrive exactly at the target.

Solution Approach

The solution iteratively calculates the sum of the first `k` natural numbers while incrementing `k`. It continues this process until the sum is greater than or equal to the absolute value of the target AND the difference between the sum and the target is even. The even difference guarantees that we can flip some moves from positive to negative to reach the target exactly. The value of `k` at that point is the minimum number of moves required.

Step-by-Step Algorithm

Step 1: Take the absolute value of the target. This simplifies the problem as we can always convert positive moves to negative to reach negative targets.
Step 2: Initialize `k` to 0 and `current_sum` to 0. `k` represents the number of moves and `current_sum` represents the total distance traveled after `k` moves.
Step 3: Enter a `while` loop that continues as long as `current_sum` is less than the absolute value of `target` OR the difference `(current_sum - target)` is not even (i.e., the remainder when divided by 2 is not 0).
Step 4: Inside the loop, increment `k` by 1.
Step 5: Update `current_sum` by adding the current value of `k` to it.
Step 6: After the loop terminates, return the value of `k`. This is the minimum number of moves required to reach the target.

Key Insights

Insight 1: The absolute value of the target is all that matters because we can always change the sign of our movements to reach a negative target by flipping the direction of our moves.
Insight 2: The sum of the first `k` natural numbers (1 + 2 + ... + k) represents the furthest distance we can reach in `k` moves if we only move to the right. We need to find the smallest `k` such that this sum is greater than or equal to the absolute value of the target.
Insight 3: If the difference between the sum of the first `k` natural numbers and the target is even, we can reach the target. This is because each move can be flipped from positive to negative, effectively subtracting twice the value of that move from the total sum. If the difference is odd, we need to increment `k` until the difference becomes even. Incrementing `k` by one step will cause the difference to increase or decrease by an odd number.
Insight 4: We don't need to explicitly calculate all possible move combinations. We can cleverly check the parity of the difference between the current sum and the target. If the difference is even, it's possible to make the target.

Complexity Analysis

Time Complexity: O(sqrt(target))

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search.

Companies

Asked at: Commvault, InMobi.

Reach a Number | LeetCode 754

Medium

Math Binary Search

Time: O(sqrt(target))Space: O(1)

Solution Code

Available in:

class Solution {
    public int reachNumber(int target) {
        target = Math.abs(target);
        int k = 0;
        int current_sum = 0;
        while (current_sum < target || (current_sum - target) % 2 != 0) {
            k += 1;
            current_sum += k;
        }
        return k;
    }
}

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Problem Info

Description

Problem 754: Reach a Number - Detailed explanation and solution approaches.

Topics

Math Binary Search

Companies

CommvaultInMobi

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:754

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Take the absolute value of the target. This simplifies the problem as we can always convert positive moves to negative to reach negative targets.
2
Step 2: Initialize `k` to 0 and `current_sum` to 0. `k` represents the number of moves and `current_sum` represents the total distance traveled after `k` moves.
3
Step 3: Enter a `while` loop that continues as long as `current_sum` is less than the absolute value of `target` OR the difference `(current_sum - target)` is not even (i.e., the remainder when divided by 2 is not 0).
4
Step 4: Inside the loop, increment `k` by 1.
5
Step 5: Update `current_sum` by adding the current value of `k` to it.
6
Step 6: After the loop terminates, return the value of `k`. This is the minimum number of moves required to reach the target.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(sqrt(target))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of iterations of the while loop. The sum of integers from 1 to k is k(k+1)/2. Since we are iterating until this sum is >= abs(target), we can deduce that the number of iterations is roughly proportional to the square root of the target. The space complexity is constant because we are using a fixed number of variables (k, current_sum, target) irrespective of the input value.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

We can use binary search to find the smallest k such that the sum of the first k integers is greater than or equal to the absolute value of the target. After finding k, we check the parity of the difference between the sum and the target, and increment k if necessary.

Time: O(log(target))Space: O(1)

Trade-offs: While the binary search approach can have a logarithmic time complexity in terms of the target range, the constant factor overhead of binary search may not make it significantly faster than the iterative approach in practice for the typical input sizes, and the code might be more complex to write. The iterative approach is usually preferred for its simplicity and comparable performance in this scenario.

✨

Key Insights

1Insight 1: The absolute value of the target is all that matters because we can always change the sign of our movements to reach a negative target by flipping the direction of our moves.
2Insight 2: The sum of the first `k` natural numbers (1 + 2 + ... + k) represents the furthest distance we can reach in `k` moves if we only move to the right. We need to find the smallest `k` such that this sum is greater than or equal to the absolute value of the target.
3Insight 3: If the difference between the sum of the first `k` natural numbers and the target is even, we can reach the target. This is because each move can be flipped from positive to negative, effectively subtracting twice the value of that move from the total sum. If the difference is odd, we need to increment `k` until the difference becomes even. Incrementing `k` by one step will cause the difference to increase or decrease by an odd number.
4Insight 4: We don't need to explicitly calculate all possible move combinations. We can cleverly check the parity of the difference between the current sum and the target. If the difference is even, it's possible to make the target.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: target = 1. The solution will return 1, which is correct because one move is enough to reach the destination (0 + 1 = 1).
Edge case 2: target = 2. The solution will return 3, since moves of +1, -2, +3 bring you to 2.
Edge case 3: target = 3. The solution will return 2, since moves of +1, +2 bring you to 3.
Edge case 4: Large positive or negative targets. The `abs()` function ensures that the algorithm works correctly for both positive and negative targets. The time complexity O(sqrt(target)) means it scales acceptably even for the constraints -10^9 <= target <= 10^9.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the absolute value of the target. This will lead to incorrect results for negative targets if the logic is not handled properly.

✗Common mistake 2: Forgetting to check the parity (even/odd) of the difference between the sum and the target. If the difference is odd, simply reaching a sum greater than or equal to the target is not sufficient because we need to be able to reach it exactly by flipping the sign of certain moves.

✗Common mistake 3: Assuming that the minimum moves needed is always the smallest `k` for which the sum (1+2+...+k) is just greater than target. It requires considering the parity and further incrementing k if required

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning in robotics. Determining the optimal sequence of movements for a robot to reach a specific location, where each movement has an increasing cost (similar to the increasing step sizes in the problem).

🏦Real-world scenario 2: Resource allocation. Allocating resources in increasing amounts over time to reach a target goal, where the objective is to minimize the number of allocation periods.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reach a Number"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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