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Minimum Distance to the Target Element

Easy

Asked by:

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Topics:

Arrays

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
Explanation: nums[0] = 1 is the only value equal to target, so the answer is abs(0 - 0) = 0.

Example 3:

Input: nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0
Output: 0
Explanation: Every value of nums is 1, but nums[0] minimizes abs(i - start), which is abs(0 - 0) = 0.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
0 <= start < nums.length
target is in nums.

Try coding it in LeetCode

Minimum Distance to the Target Element

Easy

Asked by:

18 views

Topics:

Arrays

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
Explanation: nums[0] = 1 is the only value equal to target, so the answer is abs(0 - 0) = 0.

Example 3:

Input: nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0
Output: 0
Explanation: Every value of nums is 1, but nums[0] minimizes abs(i - start), which is abs(0 - 0) = 0.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
0 <= start < nums.length
target is in nums.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected return value if the target element is not present in the array?
Can the index `start` be outside the bounds of the array?
What are the possible data types and ranges for the elements in the array, the target value, and the start index?
If the target element appears multiple times in the array, should I return the minimum distance to any of those occurrences, or only the one closest to the starting index?
Can the input array be empty?

Brute Force Solution

Approach

The brute force approach to finding the minimum distance involves checking the distance between every element and the target element. We will examine each element, calculating its distance, and keeping track of the smallest distance found so far. We will repeat this for every element until we find the overall minimum distance.

Here's how the algorithm would work step-by-step:

Start at the beginning of the list of elements.
Calculate the distance between the current element and the target element.
If this distance is the first one you've calculated, or if it's smaller than the smallest distance you've found so far, remember this distance as the new smallest distance.
Move to the next element in the list.
Repeat steps two through four until you have checked every element in the list.
The smallest distance you remembered is the minimum distance to the target element.

Code Implementation

def minimum_distance_to_target(
        elements, target_element):

    minimum_distance = float('inf')

    for current_index in range(len(elements)):
        current_element = elements[current_index]
        # Calculate distance between element and target

        distance = abs(current_element - target_element)

        # Need to initialize minimum_distance at first
        if minimum_distance == float('inf'):
            minimum_distance = distance

        # Check if current distance is smaller
        elif distance < minimum_distance:
            minimum_distance = distance

    return minimum_distance

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n elements in the input array nums once. For each element, it calculates the absolute difference between the current element and the target value and updates the minimum distance if the current distance is smaller. Since the operation performed inside the loop takes constant time O(1), the overall time complexity is directly proportional to the number of elements, n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm only uses a single variable to store the minimum distance found so far. This variable consumes a constant amount of memory, irrespective of the number of elements in the list (N). No other data structures or significant memory allocations are involved in the process. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the location of a specific value that's closest to a given starting point. Instead of checking the entire set one by one, we can move outwards from the starting point in both directions simultaneously.

Here's how the algorithm would work step-by-step:

Begin by looking at the starting position. If the value we are looking for is at that position, we're done!
If not, check one position to the left and one position to the right of the starting position.
Continue expanding outwards, checking one position further to the left and one position further to the right in each step.
Keep going until you find the value you are looking for. The number of steps you took to reach it is the distance.
Since you're always checking the closest positions first, the first time you find the target value, you know that's the shortest distance to it.

Code Implementation

def find_closest_element(array, target, start_index):
    array_length = len(array)
    distance = 0

    while True:
        # Check the current distance from the start index
        left_index = start_index - distance
        right_index = start_index + distance

        # Check if the start index contains the target.
        if left_index == start_index and array[start_index] == target:
            return distance

        # Check the left index, making sure it's within bounds.
        if left_index >= 0:
            if array[left_index] == target:
                return distance

        # Check the right index, making sure it's within bounds.
        if right_index < array_length:
            if array[right_index] == target:
                return distance

        # If target is not found, increment distance.
        distance += 1

        # If distance is larger than the array length, target is not present
        if distance > array_length:
            return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm expands outwards from the start index, checking positions to the left and right. In the worst case, the target element is located at one of the extreme ends of the array, or not present at all. This forces the algorithm to iterate through, at most, all 'n' elements of the array. Therefore, the time complexity is O(n), as the number of operations grows linearly with the size of the input array.

Space Complexity

O(1) – The algorithm operates by incrementally checking positions to the left and right of the starting index. It does not use any auxiliary data structures like arrays, lists, or hash maps to store intermediate results or visited locations. The space used is limited to storing a few index variables (e.g., left_index, right_index) and the distance, which is independent of the input array size (N). Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 or throw an exception, as no distance can be computed.
Array with only one element	Return 0 if the single element is the target, otherwise -1.
Index out of bounds (negative or larger than array size)	Throw an IllegalArgumentException if the given start index is out of bounds.
Target value not found in the array	Iterate through the entire array and return -1 if the target is not found.
target index is at either end of the array	The algorithm should correctly calculate the distance from the edge to the target index.
Large array size leading to potential performance issues	Ensure the solution has linear time complexity (O(n)) for larger inputs.
Duplicate values of the target element in the array	Return the minimum distance among all occurrences of the target element.
Integer overflow if abs(index - start) is large	Use long type for intermediate calculations to avoid integer overflow.

Null or empty input array

How to Handle:

Return -1 or throw an exception, as no distance can be computed.

Array with only one element

How to Handle:

Return 0 if the single element is the target, otherwise -1.

Index out of bounds (negative or larger than array size)

How to Handle:

Throw an IllegalArgumentException if the given start index is out of bounds.

Target value not found in the array

How to Handle:

Iterate through the entire array and return -1 if the target is not found.

target index is at either end of the array

How to Handle:

The algorithm should correctly calculate the distance from the edge to the target index.

Large array size leading to potential performance issues

How to Handle:

Ensure the solution has linear time complexity (O(n)) for larger inputs.

Duplicate values of the target element in the array

How to Handle:

Return the minimum distance among all occurrences of the target element.

Integer overflow if abs(index - start) is large

How to Handle:

Use long type for intermediate calculations to avoid integer overflow.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected return value if the target element is not present in the array?
Can the index `start` be outside the bounds of the array?
What are the possible data types and ranges for the elements in the array, the target value, and the start index?
If the target element appears multiple times in the array, should I return the minimum distance to any of those occurrences, or only the one closest to the starting index?
Can the input array be empty?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the list of elements.
Calculate the distance between the current element and the target element.
If this distance is the first one you've calculated, or if it's smaller than the smallest distance you've found so far, remember this distance as the new smallest distance.
Move to the next element in the list.
Repeat steps two through four until you have checked every element in the list.
The smallest distance you remembered is the minimum distance to the target element.

Code Implementation

def minimum_distance_to_target(
        elements, target_element):

    minimum_distance = float('inf')

    for current_index in range(len(elements)):
        current_element = elements[current_index]
        # Calculate distance between element and target

        distance = abs(current_element - target_element)

        # Need to initialize minimum_distance at first
        if minimum_distance == float('inf'):
            minimum_distance = distance

        # Check if current distance is smaller
        elif distance < minimum_distance:
            minimum_distance = distance

    return minimum_distance

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Begin by looking at the starting position. If the value we are looking for is at that position, we're done!
If not, check one position to the left and one position to the right of the starting position.
Continue expanding outwards, checking one position further to the left and one position further to the right in each step.
Keep going until you find the value you are looking for. The number of steps you took to reach it is the distance.
Since you're always checking the closest positions first, the first time you find the target value, you know that's the shortest distance to it.

Code Implementation

def find_closest_element(array, target, start_index):
    array_length = len(array)
    distance = 0

    while True:
        # Check the current distance from the start index
        left_index = start_index - distance
        right_index = start_index + distance

        # Check if the start index contains the target.
        if left_index == start_index and array[start_index] == target:
            return distance

        # Check the left index, making sure it's within bounds.
        if left_index >= 0:
            if array[left_index] == target:
                return distance

        # Check the right index, making sure it's within bounds.
        if right_index < array_length:
            if array[right_index] == target:
                return distance

        # If target is not found, increment distance.
        distance += 1

        # If distance is larger than the array length, target is not present
        if distance > array_length:
            return -1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 or throw an exception, as no distance can be computed.
Array with only one element	Return 0 if the single element is the target, otherwise -1.
Index out of bounds (negative or larger than array size)	Throw an IllegalArgumentException if the given start index is out of bounds.
Target value not found in the array	Iterate through the entire array and return -1 if the target is not found.
target index is at either end of the array	The algorithm should correctly calculate the distance from the edge to the target index.
Large array size leading to potential performance issues	Ensure the solution has linear time complexity (O(n)) for larger inputs.
Duplicate values of the target element in the array	Return the minimum distance among all occurrences of the target element.
Integer overflow if abs(index - start) is large	Use long type for intermediate calculations to avoid integer overflow.

Null or empty input array

How to Handle:

Return -1 or throw an exception, as no distance can be computed.

Array with only one element

How to Handle:

Return 0 if the single element is the target, otherwise -1.

Index out of bounds (negative or larger than array size)

How to Handle:

Throw an IllegalArgumentException if the given start index is out of bounds.

Target value not found in the array

How to Handle:

Iterate through the entire array and return -1 if the target is not found.

target index is at either end of the array

How to Handle:

The algorithm should correctly calculate the distance from the edge to the target index.

Large array size leading to potential performance issues

How to Handle:

Ensure the solution has linear time complexity (O(n)) for larger inputs.

Duplicate values of the target element in the array

How to Handle:

Return the minimum distance among all occurrences of the target element.

Integer overflow if abs(index - start) is large

How to Handle:

Use long type for intermediate calculations to avoid integer overflow.