Minimum Distance to the Target Element

Let’s break down the problem and the solution into simple terms:

Problem Description

You have an array of integers nums, an integer target, and another integer start. Your task is to find an index i in the array where nums[i] is equal to target, and the absolute difference between i and start is minimized.

Solution

Here’s a step-by-step guide to solving this problem:

  1. Initialize a Variable: Start with a variable to keep track of the minimum distance found so far.

  2. Iterate Through the Array: Go through each element in the array and check if it is equal to the target.

  3. Calculate Absolute Difference: If you find an element equal to the target, calculate the absolute difference between its index and the start value.

  4. Update Minimum Distance: Compare the absolute difference with the minimum distance found so far and update it if a smaller value is found.

  5. Return the Result: Return the minimum distance after iterating through the entire array.

Code

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class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        min_distance = float('inf') # Step 1: Initialize the minimum distance with a large value

        # Step 2: Iterate through the array
        for i in range(len(nums)):
            if nums[i] == target: # Check if the element is equal to the target
                # Step 3: Calculate absolute difference between the current index and start
                distance = abs(i - start)

                # Step 4: Update minimum distance if a smaller value is found
                min_distance = min(min_distance, distance)

        # Step 5: Return the result
        return min_distance

Key Takeaways

  • We need to find an index in the array that matches the target, and the absolute difference between that index and the start index is minimized.
  • By iterating through the array and calculating the absolute difference, we can find the minimum distance.
  • Using a variable to keep track of the minimum distance helps to efficiently find the final result.