Jobs

Interviews

Discussions

Minimum Cost to Move Chips to The Same Position

Easy

Morgan Stanley

1 view

Topics:

ArraysGreedy Algorithms

We have n chips, where the position of the i^th chip is position[i].

We need to move all the chips to the same position. In one step, we can change the position of the i^th chip from position[i] to:

position[i] + 2 or position[i] - 2 with cost = 0.
position[i] + 1 or position[i] - 1 with cost = 1.

Return the minimum cost needed to move all the chips to the same position.

Example 1:

Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.

Example 2:

Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at position  3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

Input: position = [1,1000000000]
Output: 1

Constraints:

1 <= position.length <= 100
1 <= position[i] <= 10^9

Try coding it in LeetCode

Minimum Cost to Move Chips to The Same Position

Easy

Morgan Stanley

1 view

Topics:

ArraysGreedy Algorithms

We have n chips, where the position of the i^th chip is position[i].

We need to move all the chips to the same position. In one step, we can change the position of the i^th chip from position[i] to:

position[i] + 2 or position[i] - 2 with cost = 0.
position[i] + 1 or position[i] - 1 with cost = 1.

Return the minimum cost needed to move all the chips to the same position.

Example 1:

Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.

Example 2:

Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at position  3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

Input: position = [1,1000000000]
Output: 1

Constraints:

1 <= position.length <= 100
1 <= position[i] <= 10^9

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 are the constraints on the size of the `positions` array, and the range of values within it?
Can the values in the `positions` array be negative, zero, or non-integer?
If the input array is empty or null, what should I return?
Are there any specific performance considerations or memory constraints I should be aware of, given a very large input array?
Is there a requirement to return the target position, or just the minimum cost itself?

Brute Force Solution

Approach

The brute force approach tries every possible location to move all the chips to. For each location, we calculate the total cost of moving all chips there and then pick the location with the lowest cost.

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

Consider each position as a possible final destination.
For each possible final destination, calculate the cost to move each chip individually to that destination. Moving a chip one position costs one unit.
Sum up the cost for moving all chips to that particular destination. This gives you the total cost for that location.
Repeat this process for every possible final destination.
Compare the total costs obtained for all the possible destinations.
Select the destination that resulted in the minimum total cost. That's your answer.

Code Implementation

def min_cost_to_move_chips_brute_force(positions):
    minimum_total_cost = float('inf')

    # Consider each position as a potential destination
    for destination in positions:

        total_cost_for_destination = 0
        for chip_position in positions:

            # Calculate cost to move this chip to destination
            cost_to_move = abs(chip_position - destination)
            total_cost_for_destination += cost_to_move

        # Keep track of the minimum cost seen so far
        minimum_total_cost = min(minimum_total_cost, total_cost_for_destination)

    return minimum_total_cost

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each chip position in the input array of size n, considering each position as a potential target location. For each of these n potential target locations, the algorithm calculates the cost of moving every chip (again n chips) to that target. Therefore, the total operations are approximately n * n. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach, as described, iterates through possible final destination positions. For each position, it calculates the cost to move all chips, sums these costs, and compares the totals to find the minimum. The algorithm only uses a few variables to store the current destination, the cost to move each chip, the total cost for a given destination, and the overall minimum cost found so far. The number of variables is independent of the input size N (the number of chips). Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The key insight is that moving a chip two positions away costs nothing. Therefore, we only need to consider moving chips either one position away. The optimal solution involves counting how many chips are in even positions and how many are in odd positions and then selecting the smaller count as the answer.

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

Separate the chips into two groups: those at even-numbered locations and those at odd-numbered locations.
Count how many chips are in the even group and how many are in the odd group.
Choose the smaller of the two counts. This is the minimum cost to move all the chips to the same position.
The idea is that the chips in the larger group are moved to the position of chips in the smaller group by at most one position away from each other (since even moves are free).
For example, if there are fewer chips at even locations, we'll move the chips in odd locations to an adjacent even spot. This costs 1 per chip.

Code Implementation

def minimum_cost_to_move_chips(position):
    even_position_count = 0
    odd_position_count = 0

    # Iterate through the positions array
    for current_position in position:
        # Count positions, even or odd
        if current_position % 2 == 0:
            even_position_count += 1
        else:
            odd_position_count += 1

    # We pick the minimum between even and odd counts, as we move
    # all chips to either all even or all odd positions. The moves between
    # the same parity are free
    if even_position_count < odd_position_count:
        return even_position_count
    else:
        return odd_position_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n (where n is the number of chips) to count the number of chips at even and odd positions. This involves a single pass through the array. Determining whether a number is even or odd is a constant time operation O(1). Thus, the overall time complexity is directly proportional to the number of chips, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm only uses two integer variables, one to count chips at even positions and another to count chips at odd positions. The number of chips, denoted as N, does not affect the space used because we only need to store these two counts. Therefore, the auxiliary space required remains constant, regardless of the input size. The space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array.	Return 0, as no moves are needed with no chips.
Array with a single chip.	Return 0, as the chip is already at a single position.
Array with two chips at adjacent positions.	The minimum cost is 0 to move them to the same position (either position).
Array with chips at positions differing by a multiple of 2.	Moving chips between these positions involves cost 1 for each move of two positions.
All chips are already at the same position.	The minimum cost is 0, as no moves are needed.
Array contains a mix of even and odd positions.	Count evens and odds, return the minimum of the two counts.
Input array with very large numbers.	The numbers themselves don't matter, only whether they are even or odd, so large numbers do not cause overflow.
Input with negative numbers.	Only the parity matters. The problem statement does not specify a range.

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 are the constraints on the size of the `positions` array, and the range of values within it?
Can the values in the `positions` array be negative, zero, or non-integer?
If the input array is empty or null, what should I return?
Are there any specific performance considerations or memory constraints I should be aware of, given a very large input array?
Is there a requirement to return the target position, or just the minimum cost itself?

Brute Force Solution

Approach

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

Consider each position as a possible final destination.
For each possible final destination, calculate the cost to move each chip individually to that destination. Moving a chip one position costs one unit.
Sum up the cost for moving all chips to that particular destination. This gives you the total cost for that location.
Repeat this process for every possible final destination.
Compare the total costs obtained for all the possible destinations.
Select the destination that resulted in the minimum total cost. That's your answer.

Code Implementation

def min_cost_to_move_chips_brute_force(positions):
    minimum_total_cost = float('inf')

    # Consider each position as a potential destination
    for destination in positions:

        total_cost_for_destination = 0
        for chip_position in positions:

            # Calculate cost to move this chip to destination
            cost_to_move = abs(chip_position - destination)
            total_cost_for_destination += cost_to_move

        # Keep track of the minimum cost seen so far
        minimum_total_cost = min(minimum_total_cost, total_cost_for_destination)

    return minimum_total_cost

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Separate the chips into two groups: those at even-numbered locations and those at odd-numbered locations.
Count how many chips are in the even group and how many are in the odd group.
Choose the smaller of the two counts. This is the minimum cost to move all the chips to the same position.
The idea is that the chips in the larger group are moved to the position of chips in the smaller group by at most one position away from each other (since even moves are free).
For example, if there are fewer chips at even locations, we'll move the chips in odd locations to an adjacent even spot. This costs 1 per chip.

Code Implementation

def minimum_cost_to_move_chips(position):
    even_position_count = 0
    odd_position_count = 0

    # Iterate through the positions array
    for current_position in position:
        # Count positions, even or odd
        if current_position % 2 == 0:
            even_position_count += 1
        else:
            odd_position_count += 1

    # We pick the minimum between even and odd counts, as we move
    # all chips to either all even or all odd positions. The moves between
    # the same parity are free
    if even_position_count < odd_position_count:
        return even_position_count
    else:
        return odd_position_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array.	Return 0, as no moves are needed with no chips.
Array with a single chip.	Return 0, as the chip is already at a single position.
Array with two chips at adjacent positions.	The minimum cost is 0 to move them to the same position (either position).
Array with chips at positions differing by a multiple of 2.	Moving chips between these positions involves cost 1 for each move of two positions.
All chips are already at the same position.	The minimum cost is 0, as no moves are needed.
Array contains a mix of even and odd positions.	Count evens and odds, return the minimum of the two counts.
Input array with very large numbers.	The numbers themselves don't matter, only whether they are even or odd, so large numbers do not cause overflow.
Input with negative numbers.	Only the parity matters. The problem statement does not specify a range.