What is the time complexity of Minimum Cost to Move Chips to The Same Position?

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

What data structures are used to solve Minimum Cost to Move Chips to The Same Position?

This problem uses Array, Math, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Cost to Move Chips to The Same Position?

**Parity is King, Distance is Irrelevant:** The cost structure dictates that moving `+/- 2` is free while `+/- 1` costs 1. This means two chips at positions of the same parity (e.g., 1 and 5, or 2 and 8) can always meet for free, regardless of their absolute distance, through a series of `+/- 2` moves. Conversely, moving between positions of different parities (e.g., 1 and 2, or 2 and 3) will always incur a cost of at least 1. **Two Isolated, Internally Free Groups:** Due to the parity rule, all chips starting at even positions form one group that can move freely amongst themselves. Similarly, all chips starting at odd positions form another group that can move freely amongst themselves. The only cost arises when a chip needs to move from an even position to an odd position, or vice-versa.

Which companies ask LeetCode 1217. Minimum Cost to Move Chips to The Same Position in interviews?

This problem is frequently asked at Morgan Stanley.

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Minimum Cost to Move Chips to The Same Position - LeetCode 1217 Solution

Minimum Cost to Move Chips to The Same Position - Complete Solution Guide

Minimum Cost to Move Chips to The Same Position is LeetCode problem 1217, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

We have n chips, where the position of the i th chip is position[i] . We need to move all the chips to the same position . In one step, we can change the position of the i th chip from position[i] to: position[i] + 2 or position[i] - 2 with cost = 0 . position[i] + 1 or position[i] - 1 with cost = 1 . Return the minimum cost needed to move all the chips to the same position. Example 1: Input: position = [1,2,3] Output: 1 Explanation: First step: Move the chip at position 3 to position 1 with cos

Detailed Explanation

Imagine you have a collection of chips scattered across various integer positions on a number line. Your ultimate goal is to gather *all* these chips at a *single, chosen position*, aiming to incur the least possible movement cost. The distinct twist in this problem comes from its unusual movement rules: you can shift any chip by two units (e.g., from `position[i]` to `position[i] + 2` or `position[i] - 2`) completely for free. However, moving a chip by just one unit (e.g., `position[i] + 1` or `position[i] - 1`) will cost you exactly 1.

Solution Approach

The provided solution elegantly cuts straight to the core of the problem: the parity of the chip positions. Since moving a chip by +/- 2 units is always free, this immediately tells us that any chip can be moved to *any other position of the same parity* for absolutely zero cost. For instance, a chip at position 1 can reach 3, 5, 7, -1, etc., all without spending a dime. Similarly, a chip starting at position 2 can effortlessly reach 0, 4, 6, -2, etc., also for free. This crucial insight effectively partitions our chips into two distinct, internally-free groups: those originating from even positions and those from odd positions. To gather *all* chips at a single location, we essentially have two optimal strategies for our target position's parity: 1. **Choose an even target position:** If we decide to move all chips to an even target (say, position 0 or 2), all chips that started at an even position can converge there for free. However, every single chip that originated from an odd position *must* make at least one +/- 1 move to reach any even target, incurring a cost of 1 per chip. The total cost in this scenario would simply be the count of all chips initially at odd positions. 2. **Choose an odd target position:** Conversely, if we target an odd position (say, position 1 or 3), all chips initially at odd positions move for free. But every chip from an even starting position will cost 1 to move. The total cost here would be the count of all chips initially at even positions. Since our objective is the *minimum* cost, we simply count the total number of chips in the 'even group' and the total number of chips in the 'odd group', and then return the smaller of these two counts. This smaller count represents the minimum number of chips that are forced to 'cross the parity line' and thus incur a cost of 1 to join the majority group.

Step-by-Step Algorithm

Step 1: Initialize two counters, `even_count` and `odd_count`, to 0.
Step 2: Iterate through the `position` array. For each position, check if it's even or odd using the modulo operator (%)
Step 3: If the position is even, increment `even_count`; otherwise, increment `odd_count`.
Step 4: After iterating through all positions, return the minimum of `even_count` and `odd_count` using `min()` function.

Key Insights

**Parity is King, Distance is Irrelevant:** The cost structure dictates that moving `+/- 2` is free while `+/- 1` costs 1. This means two chips at positions of the same parity (e.g., 1 and 5, or 2 and 8) can always meet for free, regardless of their absolute distance, through a series of `+/- 2` moves. Conversely, moving between positions of different parities (e.g., 1 and 2, or 2 and 3) will always incur a cost of at least 1.
**Two Isolated, Internally Free Groups:** Due to the parity rule, all chips starting at even positions form one group that can move freely amongst themselves. Similarly, all chips starting at odd positions form another group that can move freely amongst themselves. The only cost arises when a chip needs to move from an even position to an odd position, or vice-versa.
**Minimize the 'Crossing Over' Chips:** To unite all chips at a single position, we must choose *one* final parity (either even or odd) for our target. If we pick an even target, all odd-positioned chips must pay 1 to become even. If we pick an odd target, all even-positioned chips must pay 1 to become odd. The minimum cost is therefore simply the count of the smaller group of chips, as these are the ones that *must* incur a cost to join the larger, cost-free group.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Greedy.

Companies

Asked at: Morgan Stanley.

Minimum Cost to Move Chips to The Same Position | LeetCode 1217

Easy

Array Math Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minCostToMoveChips(int[] position) {
        int even = 0;
        int odd = 0;
        for (int pos : position) {
            if (pos % 2 == 0) {
                even++;
            } else {
                odd++;
            }
        }
        return Math.min(even, odd);
    }
}

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Problem Info

Description

Problem 1217: Minimum Cost to Move Chips to The Same Position - Detailed explanation and solution approaches.

Topics

Array Math Greedy

Companies

Morgan Stanley

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1217

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

4 steps

1
Step 1: Initialize two counters, `even_count` and `odd_count`, to 0.
2
Step 2: Iterate through the `position` array. For each position, check if it's even or odd using the modulo operator (%)
3
Step 3: If the position is even, increment `even_count`; otherwise, increment `odd_count`.
4
Step 4: After iterating through all positions, return the minimum of `even_count` and `odd_count` using `min()` function.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the input array once, performing constant-time operations for each element. Therefore, the time complexity is linear with respect to the input size. The space used is constant because only a few integer variables are used, regardless of the input size.

📖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

Dynamic Programming (Inefficient)

A dynamic programming approach could be conceptually designed, but it would be highly inefficient. It would involve exploring all possible target positions, calculating the cost for each, and finding the minimum. This would result in a much higher time complexity.

Time: O(n*m), where m is the range of positionsSpace: O(m), where m is the range of positions

Trade-offs: This approach is highly impractical due to its significantly worse complexity compared to the greedy approach, especially for large ranges of positions.

✨

Key Insights

1**Parity is King, Distance is Irrelevant:** The cost structure dictates that moving `+/- 2` is free while `+/- 1` costs 1. This means two chips at positions of the same parity (e.g., 1 and 5, or 2 and 8) can always meet for free, regardless of their absolute distance, through a series of `+/- 2` moves. Conversely, moving between positions of different parities (e.g., 1 and 2, or 2 and 3) will always incur a cost of at least 1.
2**Two Isolated, Internally Free Groups:** Due to the parity rule, all chips starting at even positions form one group that can move freely amongst themselves. Similarly, all chips starting at odd positions form another group that can move freely amongst themselves. The only cost arises when a chip needs to move from an even position to an odd position, or vice-versa.
3**Minimize the 'Crossing Over' Chips:** To unite all chips at a single position, we must choose *one* final parity (either even or odd) for our target. If we pick an even target, all odd-positioned chips must pay 1 to become even. If we pick an odd target, all even-positioned chips must pay 1 to become odd. The minimum cost is therefore simply the count of the smaller group of chips, as these are the ones that *must* incur a cost to join the larger, cost-free group.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: An empty input array. The solution correctly handles this by returning 0 (min(0,0)=0) because there are no chips to move.
Edge case 2: An array with all even or all odd numbers. The solution correctly returns 0 because no cost is required for moving all chips to the same parity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to understand that moving an even number of spaces is free. Beginners might try to calculate the cost of moving every chip to every other position, resulting in a much more complex and inefficient solution.

✗Common mistake 2: Incorrectly calculating the minimum cost. Some might return the sum of `even_count` and `odd_count`, forgetting that we are allowed free movement of even distances.

✅

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: Optimizing the movement of robotic arms or machines in a manufacturing setting. The positions might represent locations where a robot needs to pick up items, and the goal is to minimize the total distance or movement time.

🏦Real-world scenario 2: Data center optimization. Minimizing data transfer between servers or minimizing energy consumption by grouping server operations based on their energy profiles (analogous to parity).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: October 16, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Cost to Move Chips to The Same Position"?

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

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