How do you solve Minimum Distance to the Target Element on LeetCode?

Minimum Distance to the Target Element (LeetCode #1848) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem guarantees that the target exists, so we don't need to handle cases where it doesn't.

What is the brute force solution for Minimum Distance to the Target Element?

The brute force approach for Minimum Distance to the Target Element is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks every index in the array to find the target and calculates the distance from the start index. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Distance to the Target Element?

Minimum Distance to the Target Element uses the following concepts: Array. The recommended patterns to study are: Array, Linear Search.

What are the interview tips for Minimum Distance to the Target Element?

Always clarify the constraints and guarantees of the problem before diving into coding. Think about edge cases, such as when the target is at the start or end of the array. Discuss your thought process and approach with the interviewer before writing code.

What are common mistakes when solving Minimum Distance to the Target Element?

Not considering that the target can appear multiple times in the array. Forgetting to update the minimum distance correctly when finding a target.

#1848

Minimum Distance to the Target Element

Easy

ArrayArrayLinear Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking every index, we can loop through the array once and keep track of the minimum distance as we find the target. This reduces unnecessary checks and is more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to store the minimum distance as infinity.
2Step 2: Loop through each index in the array.
3Step 3: If the current element equals the target, calculate the absolute distance from the start index and update the minimum distance if it's smaller.
4Step 4: Return the minimum distance.

solution.py6 lines

1def getMinDistance(nums, target, start):
2    min_distance = float('inf')
3    for i in range(len(nums)):
4        if nums[i] == target:
5            min_distance = min(min_distance, abs(i - start))
6    return min_distance

ℹ

Complexity note: The time complexity is O(n) because we check each element once. The space complexity is O(1) since we only use a few variables for tracking distances.

1The problem guarantees that the target exists, so we don't need to handle cases where it doesn't.
2The distance is always calculated as an absolute value, which simplifies our checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.