The One-Line Trick That Solves LeetCode 1290 in Minutes

Master this essential coding interview problem with our step-by-step approach and optimized C++ solution

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Problem Overview

LeetCode Problem 1290 is a fundamental coding interview question that tests your understanding of:

• Linked List traversal

• Binary number system

• Mathematical operations

• Bit manipulation concepts

This problem appears frequently in technical interviews at companies like Google, Amazon, and Microsoft, making it essential for any serious programmer to master.

Problem Statement

Given a head node of a singly-linked list where each node contains either 0 or 1, convert this binary representation to its decimal equivalent.

Example:

Input: head = [1,0,1]
Output: 5
Explanation: (101) in base 2 = (5) in base 10

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Understanding the Challenge

What Makes This Problem Interesting?

Unlike working with arrays or strings, linked lists don't provide random access to elements. This means we can't simply:

• Jump to any position

• Know the length beforehand

• Traverse backwards easily

The key insight is recognizing that we're dealing with a Most Significant Bit (MSB) first representation, which actually makes our solution more elegant.

Binary to Decimal Conversion Basics

Before diving into the solution, let's refresh binary-to-decimal conversion:

Binary: 1 0 1
Powers: 2² 2¹ 2⁰
Values: 4  0  1
Result: 4 + 0 + 1 = 5

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The Intuitive Approach

My Thought Process

When I first encountered this problem, I considered several approaches:

1. Traverse twice: First to find length, then to calculate

2. Use extra space: Store values in an array

3. Mathematical trick: Build result incrementally

The third approach turned out to be the most elegant and efficient.

The "Shift and Add" Strategy

The breakthrough came when I realized: every time we encounter a new bit, we can shift our current result left by 1 position and add the new bit.

This is equivalent to:

result = result * 2 + current_bit

Let's trace through an example:

Binary: 1 → 0 → 1
Step 1: result = 0 * 2 + 1 = 1
Step 2: result = 1 * 2 + 0 = 2  
Step 3: result = 2 * 2 + 1 = 5

Perfect! We get our answer in a single pass.

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Step-by-Step Solution

Algorithm Breakdown

Here's my systematic approach:

1. Initialize result to 0

2. Traverse the linked list from head to tail

3. For each node:

   • Shift current result left (multiply by 2)

   • Add the current bit value

4. Return the final result

Visual Representation

Linked List: [1] → [0] → [1] → null

Iteration 1: result = 0 * 2 + 1 = 1
Iteration 2: result = 1 * 2 + 0 = 2
Iteration 3: result = 2 * 2 + 1 = 5

Final Answer: 5

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Code Implementation

Optimized C++ Solution

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    int getDecimalValue(ListNode* head) {
        int result = 0;  // Initialize result to store decimal value

        // Traverse the linked list from head to tail
        while (head != nullptr) {
            // Shift result left by 1 bit (multiply by 2) and add current bit
            result = result * 2 + head->val;

            // Move to the next node
            head = head->next;
        }

        return result;  // Return the final decimal value
    }
};

Code Explanation

Line-by-line breakdown:

• int result = 0;: Start with zero as our base

• while (head != nullptr): Continue until we reach the end

• result = result * 2 + head->val;: The magic formula that builds our decimal number

• head = head->next;: Move to the next node

• return result;: Return our calculated decimal value

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Complexity Analysis

Time Complexity: O(n)

• n = number of nodes in the linked list

• We visit each node exactly once

• Each operation inside the loop is O(1)

Space Complexity: O(1)

• Only using a single integer variable result

• No additional data structures needed

• Memory usage doesn't grow with input size

Why This is Optimal

This solution is optimal because:

1. Minimal time: Single pass through the data

2. Minimal space: Constant extra memory

3. Clean code: Easy to understand and maintain

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Alternative Approaches

Approach 1: Using Built-in Functions

// Less efficient but more readable
int getDecimalValue(ListNode* head) {
    string binary = "";
    while (head) {
        binary += to_string(head->val);
        head = head->next;
    }
    return stoi(binary, 0, 2);
}

Pros: Very readable Cons: O(n) extra space, slower execution

Approach 2: Recursive Solution

// Recursive approach
int helper(ListNode* head, int& result) {
    if (!head) return 0;
    int length = helper(head->next, result) + 1;
    result += head->val * pow(2, length - 1);
    return length;
}

Pros: Demonstrates recursion skills Cons: O(n) stack space, more complex

Why the Iterative Approach Wins

The iterative "shift and add" method is superior because:

• Efficiency: O(1) space, O(n) time

• Simplicity: Easy to understand and implement

• Interview-friendly: Shows mathematical insight

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Practice Problems

Related LeetCode Problems

To master this concept, try these similar problems:

1. [LeetCode 2]: Add Two Numbers - Linked list arithmetic

2. [LeetCode 19]: Remove Nth Node - Linked list manipulation

3. [LeetCode 206]: Reverse Linked List - Fundamental linked list operations

4. [LeetCode 1009]: Complement of Base 10 Integer - Binary operations

Variations to Consider

• What if the linked list represented a number in base 3?

• How would you handle negative numbers?

• Can you solve it recursively?

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Key Takeaways

Essential Concepts Mastered

✅ Linked List Traversal: Single-pass iteration technique ✅ Binary Number System: Understanding MSB-first representation ✅ Mathematical Optimization: Using shift-and-add strategy ✅ Space Efficiency: Achieving O(1) space complexity

Interview Tips

1. Start with examples: Always trace through the algorithm with sample input

2. Discuss trade-offs: Mention alternative approaches and their complexities

3. Handle edge cases: Consider empty lists, single nodes, all zeros

4. Clean code: Write readable, well-commented solutions

Common Mistakes to Avoid

• Overthinking: Don't try to calculate powers of 2 manually

• Extra space: Avoid storing the entire binary string

• Off-by-one errors: Be careful with indexing and powers

• Null pointer: Always check for nullptr before accessing nodes

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Final Thoughts

LeetCode 1290 is more than just a coding problem—it's a gateway to understanding how mathematical insights can lead to elegant algorithmic solutions. The "shift and add" technique demonstrated here applies to many other problems involving number systems and bit manipulation.

Remember: The best solutions often come from stepping back and thinking about the problem from a mathematical perspective rather than just trying to implement the most obvious approach.

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About the Author

This solution guide was crafted by experienced software engineers who have successfully interviewed at top tech companies. We specialize in breaking down complex algorithms into digestible, interview-ready explanations.

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Tags: #LeetCode #CodingInterview #LinkedList #BinaryNumbers #Algorithms #DataStructures #TechInterview #Programming #CPlusPlus