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Minimum Distance to the Target Element

Difficulty: Easy

Problem Description

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. Return abs(i - start).

Key Insights

  • The problem involves finding the closest index of the target in the array to a specified starting index.
  • The absolute distance is calculated as abs(i - start), which is always non-negative.
  • It is guaranteed that the target exists in the array, so we do not need to handle cases where it is absent.

Space and Time Complexity

Time Complexity: O(n), where n is the number of elements in nums. We may need to traverse the entire array in the worst case. Space Complexity: O(1), as we are using a constant amount of space regardless of the input size.

Solution

To solve this problem, we will iterate through the array nums to find indices where the value equals target. For each of these indices, we will calculate the absolute distance from the start index. We will keep track of the minimum distance found during the iteration.

The data structure used here is a simple array, and the algorithm is a linear search through the array to find the target values.

Code Solutions

def minimumDistance(nums, target, start):
    min_distance = float('inf')  # Start with a large number
    for i in range(len(nums)):
        if nums[i] == target:  # Check if the current number is the target
            distance = abs(i - start)  # Calculate the distance
            min_distance = min(min_distance, distance)  # Update the minimum distance
    return min_distance  # Return the minimum distance found
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