Minimum Cost to Make All Characters Equal

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ArraysDynamic ProgrammingGreedy Algorithms

You are given a 0-indexed binary string s of length n on which you can apply two types of operations:

Choose an index i and invert all characters from index 0 to index i (both inclusive), with a cost of i + 1
Choose an index i and invert all characters from index i to index n - 1 (both inclusive), with a cost of n - i

Return the minimum cost to make all characters of the string equal.

Invert a character means if its value is '0' it becomes '1' and vice-versa.

Example 1:

Input: s = "0011"
Output: 2
Explanation: Apply the second operation with i = 2 to obtain s = "0000" for a cost of 2. It can be shown that 2 is the minimum cost to make all characters equal.

Example 2:

Input: s = "010101"
Output: 9
Explanation: Apply the first operation with i = 2 to obtain s = "101101" for a cost of 3.
Apply the first operation with i = 1 to obtain s = "011101" for a cost of 2. 
Apply the first operation with i = 0 to obtain s = "111101" for a cost of 1. 
Apply the second operation with i = 4 to obtain s = "111110" for a cost of 2.
Apply the second operation with i = 5 to obtain s = "111111" for a cost of 1. 
The total cost to make all characters equal is 9. It can be shown that 9 is the minimum cost to make all characters equal.

Constraints:

1 <= s.length == n <= 10⁵
s[i] is either '0' or '1'

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 maximum length of the string `s`? What characters besides '0' and '1' might appear in the input (or is it strictly guaranteed to be only '0' and '1')?
Is the string `s` ever empty or null? If so, what should I return?
Should I return the minimum cost as an integer, a long, or some other numeric type? Is there a possibility of integer overflow?
If there are multiple ways to achieve the minimum cost (e.g., flipping to all '0' or flipping to all '1' resulting in the same cost), should I return any of those minimum costs, or is there a preferred outcome?
Could you provide a few examples of the input and corresponding expected output for edge cases to ensure my understanding?

Brute Force Solution

Approach

The brute force method aims to find the cheapest way to make all letters in a string the same. It essentially explores all possible combinations of changes to the letters.

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

Consider changing every letter one at a time to the target character and find the cost.
For each letter in the string, try flipping it, calculating the cost of that single change, and then reverting the flip.
Next, explore flipping two letters at a time, calculating the total cost of those two flips, and then reverting them.
Continue this pattern, considering all possible combinations of flips from single letters to pairs of letters and so on, up to flipping almost all letters.
For each of these combinations, calculate the total cost to make all letters the same, recording the cost of the flips.
After trying all the possible combinations of flips, compare the costs and identify the minimum cost to make all characters equal.

Code Implementation

def minimum_cost_to_make_all_characters_equal_brute_force(input_string):
    string_length = len(input_string)
    minimum_cost = float('inf')

    for i in range(1 << string_length):
        # Represents each possible combination of flips using bit manipulation.
        current_string = list(input_string)
        current_cost = 0

        indices_to_flip = []
        for j in range(string_length):
            if (i >> j) & 1:
                indices_to_flip.append(j)

        for index in indices_to_flip:
            # Keep track of the cost to flip characters.
            current_cost += 1
            current_string[index] = '0' if current_string[index] == '1' else '1'

        # Check if all characters are now equal.
        if all(character == current_string[0] for character in current_string):
            minimum_cost = min(minimum_cost, current_cost)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(2^n) – The provided solution considers all possible subsets of characters to flip in the string of length n. It evaluates single flips, pairs of flips, triplets, and so on, up to almost all characters being flipped. Essentially, for each of the n characters, we either flip it or we don't, resulting in 2^n possible combinations to check. Calculating the cost for each combination takes O(n) time, but the dominating factor is the generation of these 2^n combinations. Therefore, the overall time complexity is O(2^n).

Space Complexity

O(1) – The brute force solution described primarily involves iterative flipping and reverting of characters within the input string. While it explores numerous combinations, it doesn't explicitly create auxiliary data structures that scale with the input size N (the length of the string). The cost calculation and temporary flips happen in place or with a few constant-sized variables, leading to constant auxiliary space. Thus, the space complexity remains O(1).

Optimal Solution

Approach

To minimize the cost, we should focus on finding the most efficient way to change groups of characters. We can achieve this by focusing on changing the smaller segments of the string, choosing the side with fewer changes.

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

Look at the string character by character, focusing on places where the character changes to a different character.
At each change point, calculate the cost to make all characters to the left the same, and the cost to make all characters to the right the same.
Choose the cheaper option between changing the left side or the right side.
Add the cost of the cheaper option to your total cost.
Continue through the string, making these decisions at each change point.
The final total cost will be the minimum cost needed to make all the characters the same.

Code Implementation

def minimum_cost_to_make_all_characters_equal(input_string):
    total_cost = 0
    string_length = len(input_string)

    for i in range(1, string_length):
        if input_string[i] != input_string[i - 1]:
            # Calculate the cost to change left and right sides.
            cost_to_change_left = i
            cost_to_change_right = string_length - i

            # Choose the cheaper option.
            total_cost += min(cost_to_change_left, cost_to_change_right)

    return total_cost

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n character by character. At each position where the character changes, it calculates the cost to change the left and right sides. The cost calculation involves iterating through the left side and the right side segments. In the worst case, these left and right iterations at each change point sum up to at most a constant multiple of n, as each character is visited at most a fixed number of times when evaluating change costs. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The described algorithm iterates through the input string, but it doesn't create any auxiliary data structures like lists, arrays, or hash maps to store intermediate results. The cost calculation at each change point involves only a few constant-size variables for comparison and summing up the total cost. Therefore, the space required by the algorithm remains constant regardless of the input string's length, N.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0, as no flips are needed on an empty string, and null should be treated as empty in some languages.
String of length 1	Return 0, as a single character string already satisfies the condition.
String with all '0's or all '1's	Return 0, as no flips are needed if all characters are already equal.
Very long string (potential for integer overflow when calculating costs)	Use long integer type to store costs to avoid overflow, especially when string length approaches integer limits.
String alternating between '0' and '1' (e.g., '010101')	The solution should efficiently calculate the cost for flipping to all '0's versus all '1's and choose the minimum.
String with a large cluster of '0's followed by a small cluster of '1's (or vice versa)	The algorithm should correctly sum the costs associated with flipping the smaller cluster, and compare flipping strategies.
String that starts with a long sequence of one char, and then immediately flips to different char for all the other values	The solution should consider whether flipping a lot of higher indices or flipping a few low indices will have a smaller cost.
String of maximum allowed length, filled with alternating chars	Ensure the algorithm's time and space complexity allows execution within reasonable bounds, avoiding timeouts or memory errors.

Null or empty input string

How to Handle:

Return 0, as no flips are needed on an empty string, and null should be treated as empty in some languages.

String of length 1

How to Handle:

Return 0, as a single character string already satisfies the condition.

String with all '0's or all '1's

How to Handle:

Return 0, as no flips are needed if all characters are already equal.

Very long string (potential for integer overflow when calculating costs)

How to Handle:

Use long integer type to store costs to avoid overflow, especially when string length approaches integer limits.

String alternating between '0' and '1' (e.g., '010101')

How to Handle:

The solution should efficiently calculate the cost for flipping to all '0's versus all '1's and choose the minimum.

String with a large cluster of '0's followed by a small cluster of '1's (or vice versa)

How to Handle:

The algorithm should correctly sum the costs associated with flipping the smaller cluster, and compare flipping strategies.

String that starts with a long sequence of one char, and then immediately flips to different char for all the other values

How to Handle:

The solution should consider whether flipping a lot of higher indices or flipping a few low indices will have a smaller cost.

String of maximum allowed length, filled with alternating chars

How to Handle:

Ensure the algorithm's time and space complexity allows execution within reasonable bounds, avoiding timeouts or memory errors.