algoadvance

  1. What is the structure of the input?
    • This involves understanding the type of inputs the problem statement provides.
  2. What are the constraints and edge cases?
    • Are there any specific things I need to consider regarding the input sizes, values, or other particularities?
  3. Can there be multiple occurrences of the target?
    • If the target element occurs multiple times, or if we need to consider all occurrences, etc.
  4. What should be the output if the target is not present in the array?
    • This is to understand how to handle cases where the target does not exist in the array.

Provided Problem:

Given an integer array nums (0-indexed) and two integers target and start, find the minimum distance between start and any index i such that nums[i] == target.

Example:

nums = [1, 2, 3, 4, 5]
target = 5
start = 3
# Output: 1

Strategy:

  1. Initialize a variable to store the minimum distance found so far (min_distance). Another variable (distance) to calculate the absolute difference between start and i.

  2. Iterate through the nums array and check for elements equal to target.

  3. Update min_distance each time a closer index i to start is found.

  4. Return the minimum distance after the loop ends.

Let’s discuss the time complexity here. We need to iterate through the entire list once, making our solution run in O(n) time complexity, where n is the length of nums.

Code:

def getMinDistance(nums, target, start):
    """
    :type nums: List[int]
    :type target: int
    :type start: int
    :rtype: int
    """
    min_distance = float('inf')
    
    for i in range(len(nums)):
        if nums[i] == target:
            distance = abs(i - start)
            if distance < min_distance:
                min_distance = distance
    
    return min_distance

# Example usage:
nums = [1, 2, 3, 4, 5]
target = 5
start = 3
print(getMinDistance(nums, target, start))  # Output: 1

Explanation:

  1. Initialization: min_distance is set to a very high value initially (float('inf')), this will help in finding the minimum value during comparisons.

  2. Iterate through nums: Check each element of nums to see if it matches the target.

  3. Calculate the distance: Compute the absolute difference between the current index i and start.

  4. Update min_distance: Whenever a smaller distance is found, update min_distance.

  5. Return the smallest distance found.

This approach ensures we account for multiple occurrences of the target and handle possible edge cases while maintaining an O(n) time complexity.

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