What is the time complexity of Minimum Distance to the Target Element?

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

What data structures are used to solve Minimum Distance to the Target Element?

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

What are the key insights for solving Minimum Distance to the Target Element?

Insight 1: A linear scan of the array is sufficient to find all occurrences of the target element. Insight 2: Keeping track of the minimum distance encountered so far is crucial. This can be achieved using a variable initialized to a large value (infinity or the maximum integer value).

Which companies ask LeetCode 1848. Minimum Distance to the Target Element in interviews?

This problem is frequently asked at Honeywell.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Distance to the Target Element - LeetCode 1848 Solution

Minimum Distance to the Target Element - Complete Solution Guide

Minimum Distance to the Target Element is LeetCode problem 1848, 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

Given an integer array nums (0-indexed) and two integers target and start , find an index i such that nums[i] == target and abs(i - start) is minimized . Note that abs(x) is the absolute value of x . Return abs(i - start) . It is guaranteed that target exists in nums . Example 1: Input: nums = [1,2,3,4,5], target = 5, start = 3 Output: 1 Explanation: nums[4] = 5 is the only value equal to target, so the answer is abs(4 - 3) = 1. Example 2: Input: nums = [1], target = 1, start = 0 Output: 0 Expla

Detailed Explanation

The problem asks to find the minimum distance between a given starting index and the index of a target element within an array. The input consists of an integer array `nums`, an integer `target` representing the value to search for, and an integer `start` representing the starting index. The output is the minimum absolute difference between `start` and the index of any occurrence of `target` in `nums`. It's guaranteed that `target` exists at least once in `nums`. For example, if `nums = [1,2,3,4,5]`, `target = 5`, and `start = 3`, the output is 1 because `nums[4] = 5`, and the absolute difference between 4 and 3 is 1. The problem emphasizes that the index is 0-indexed.

Solution Approach

The provided solutions use a brute-force approach. They iterate through the input array `nums`. For each element, they check if it's equal to the `target`. If it is, they calculate the absolute difference between the current index `i` and the `start` index. They then update the `min_distance` variable to store the minimum distance found so far. Finally, they return the `min_distance`. This approach works because it systematically checks every element and guarantees finding the minimum distance.

Step-by-Step Algorithm

Step 1: Initialize a variable `min_distance` to a large value (positive infinity or the maximum integer value depending on the language used). This ensures that the first calculated distance will always be smaller.
Step 2: Iterate through the `nums` array using a loop. In each iteration, check if the current element `nums[i]` is equal to the `target`.
Step 3: If `nums[i]` equals `target`, calculate the absolute difference `abs(i - start)` between the current index `i` and the starting index `start`.
Step 4: Update `min_distance` with the minimum value between the current `min_distance` and the newly calculated absolute difference. This ensures that `min_distance` always holds the smallest distance found so far.
Step 5: After iterating through all elements, return the final value of `min_distance`.

Key Insights

Insight 1: A linear scan of the array is sufficient to find all occurrences of the target element.
Insight 2: Keeping track of the minimum distance encountered so far is crucial. This can be achieved using a variable initialized to a large value (infinity or the maximum integer value).
Insight 3: The `abs()` function is necessary to handle cases where the target element might appear before or after the starting index.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Honeywell.

Minimum Distance to the Target Element | LeetCode 1848

Easy

Array

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

Solution Code

Available in:

class Solution {
    public int getMinDistance(int[] nums, int target, int start) {
        int minDistance = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                minDistance = Math.min(minDistance, Math.abs(i - start));
            }
        }
        return minDistance;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1848: Minimum Distance to the Target Element - Detailed explanation and solution approaches.

Topics

Array

Companies

Honeywell

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1848

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

5 steps

1
Step 1: Initialize a variable `min_distance` to a large value (positive infinity or the maximum integer value depending on the language used). This ensures that the first calculated distance will always be smaller.
2
Step 2: Iterate through the `nums` array using a loop. In each iteration, check if the current element `nums[i]` is equal to the `target`.
3
Step 3: If `nums[i]` equals `target`, calculate the absolute difference `abs(i - start)` between the current index `i` and the starting index `start`.
4
Step 4: Update `min_distance` with the minimum value between the current `min_distance` and the newly calculated absolute difference. This ensures that `min_distance` always holds the smallest distance found so far.
5
Step 5: After iterating through all elements, return the final value of `min_distance`.

💡

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 time complexity is O(n) because the algorithm iterates through the array once in the worst case (where the target is at the end of the array or appears only once). The space complexity is O(1) because it uses only a constant amount of extra space to store variables like `min_distance`, `i`, etc., 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

Optimized Search (if target distribution is known)

If the target element's distribution within the array is known (e.g., it's clustered around a certain area), binary search or other optimized search methods can be considered. This would reduce the search space if the target is likely to be near the start.

Time: O(log n) in best case (for binary search if applicable), O(n) in worst caseSpace: O(1)

Trade-offs: This approach is only advantageous if you have information about the target's distribution. If the target is randomly placed, this approach doesn't provide a significant advantage over the linear scan.

✨

Key Insights

1Insight 1: A linear scan of the array is sufficient to find all occurrences of the target element.
2Insight 2: Keeping track of the minimum distance encountered so far is crucial. This can be achieved using a variable initialized to a large value (infinity or the maximum integer value).
3Insight 3: The `abs()` function is necessary to handle cases where the target element might appear before or after the starting index.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The target element is at the starting index. The solution correctly handles this, resulting in a minimum distance of 0.
Edge case 2: The target element appears multiple times in the array. The solution correctly finds the minimum distance among all occurrences.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize `min_distance` to a sufficiently large value. If it's not initialized correctly, the minimum distance calculation will be flawed.

✗Common mistake 2: Incorrectly handling the absolute difference calculation. Using `i - start` directly instead of `abs(i - start)` could lead to incorrect results.

✅

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: Finding the closest sensor to an event. Given an array of sensor locations and an event location, find the sensor with the minimum distance to the event.

🏦Real-world scenario 2: Optimizing resource allocation. If you have a list of resources and a location, this algorithm can efficiently identify the closest resource to a given point.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: July 19, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Minimum Distance to the Target Element"?

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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?