2712. Minimum Cost to Make All Characters Equal

Problem Description

In this problem, we are dealing with a binary string s, which is a sequence of characters that can only be 0 or 1. The string uses a 0-based index, meaning that the first character of the string is at position 0, the next at position 1, and so on, up to the last character, which is at position n - 1. Our goal is to transform this binary string into a uniform string where all characters are the same, either all 0s or all 1s.

To achieve a uniform string, we can perform two types of operations:

1. Select an index i, and flip all the characters from the beginning of the string (index 0) up to and including i. For instance, if we flip characters from index 0 to index i in "01011", and i is 2, the string becomes "10100". This operation comes with a cost equal to i + 1.

2. Select an index i, and flip all the characters from that index i to the end of the string (index n - 1). For example, for the same string "01011", if we flip characters from index 2 to index 4 (n - 1), the string becomes "01100". This operation has a cost of n - i.

Each flip inverts the characters, which means 0 becomes 1 and 1 becomes 0. The challenge is to find the minimum total cost of such flipping operations that will result in a string composed entirely of the same character.

Intuition

The intuition behind the provided solution is to identify the positions in the string where flipping can actually contribute to making the string uniform and doing it at the lowest possible cost. If all characters are already equal, no operation is needed and the cost is zero. We only need to consider flipping at positions where a character differs from its preceding character, indicating a point of change in the string's structure.

For each character s[i] that differs from s[i - 1], there is a possibility of flipping either the sub-string before i or the sub-string after i to make the characters equal. We want to choose the index that minimizes the cost for each situation, which is given by min(i, n - i) where i is the cost for flipping from the start and n - i is the cost for flipping until the end. By iterating through the string and summing up the minimum cost at each change point, we accumulate the total minimum cost needed to make the string uniform.

Here is the step-by-step intuition for the provided solution:

• Initialize ans as 0, which will hold the overall minimum cost.
• Loop through each character in the string, starting from index 1 up to index n - 1.
• Check if the current character s[i] is different from the previous character s[i - 1].
• When a difference is detected, it means we have a potential flip point.
• Calculate the cost to flip either from the start up to i or from i to the end, by evaluating min(i, n - i).
• Add this cost to the ans total.
• After looping through the entire string, ans will contain the minimum cost to make the string uniform.
• Return the ans value.

The solution effectively avoids redundant operations and achieves minimal cost by flipping at the optimal points in the string.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation of the solution provided uses a simple linear scan algorithm which is both time and space efficient because it iterates over the string once and does not require any additional data structures. The key pattern used here is the iteration over changing points in the binary string.

Here's the breakdown of the solution approach:

1. Initialization: A variable ans is created to store the accumulated minimum cost and its initial value is set to 0. The length of the string n is computed to avoid recalculating it during each iteration.

   1ans, n = 0, len(s)

2. Looping through the changes: The program then iterates from 1 to n - 1, intentionally starting from 1 because we always compare the current character with the previous one to identify the change points.

   1for i in range(1, n):

3. Detecting change points: The condition inside the loop checks if the current character s[i] is different from the preceding character s[i - 1]. If it is, then a flip operation must be performed at this point, because we want to eliminate the discrepancy to move towards a uniform string.

   1if s[i] != s[i - 1]:

4. Calculating operation costs: For any change point, the cost of making the preceding or succeeding substring uniform is analyzed. Since the goal is to do this at the minimum cost, the function min(i, n - i) calculates the cheaper option between flipping the first i characters or the last n - i characters. This corresponds to choosing the shorter sub-string to flip which naturally costs less.

   1ans += min(i, n - i)

   The result is added to our running total ans.

5. Final result: After all potential change points have been processed, the variable ans now contains the minimum cost to make all characters of the string equal. The function finally returns ans.

   1return ans

By utilizing a single pass over the string and performing an efficient comparison and calculation at each step, the implementation ensures a minimal time complexity of O(n), where n is the length of the string. There's no use of additional data structures, so the space complexity is O(1). This approach ensures optimal performance for the problem at hand.

Example Walkthrough

Let's consider a simple example to illustrate the solution approach. We have the following binary string s = "0110101".

1. Initialization: Set the total minimum cost ans to 0 and compute the length of s to be n = 7.

   1ans, n = 0, len(s)  # ans = 0, n = 7

2. Looping through the changes: Start iterating from index 1 to n - 1, which goes from index 1 to index 6.

3. Detecting change points: We're looking for positions where s[i] != s[i - 1]. These are our points of change:

   • At i = 1, s[i] = "1" and s[i - 1] = "0". This is a change point; s[1] differs from s[0].
   • At i = 2, s[i] = "1" and s[i - 1] = "1". This is not a change point; s[2] is the same as s[1].
   • At i = 3, s[i] = "0" and s[i - 1] = "1". This is another change point; s[3] differs from s[2].
   • At i = 4, s[i] = "1" and s[i - 1] = "0". Another change point; s[4] differs from s[3].
   • At i = 5, s[i] = "0" and s[i - 1] = "1". Yet another change point; s[5] differs from s[4].
   • At i = 6, s[i] = "1" and s[i - 1] = "0". The last change point; s[6] differs from s[5].

4. Calculating operation costs: For every change point identified, calculate the cost:

   • At i = 1, the cost is min(1, 7 - 1) = 1. Add this to ans to get ans = 1.
   • At i = 3, the cost is min(3, 7 - 3) = 3. Now ans = 1 + 3 = 4.
   • At i = 4, the cost is min(4, 7 - 4) = 3. Update ans = 4 + 3 = 7.
   • At i = 5, the cost is min(5, 7 - 5) = 2. Update ans = 7 + 2 = 9.
   • At i = 6, the cost is min(6, 7 - 6) = 1. Finally, ans = 9 + 1 = 10.

5. Final result: After the loop, we have checked all change points. The total minimum cost ans is 10. Thus, this is the minimum cost required to make the string s = "0110101" uniform, using the flip operations described.

By sequentially applying the logic to each change point, the example demonstrates how we sum up the minimum of flipping costs at every change point to get the total minimal cost efficiently.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists of a single loop that iterates through the length of the input string s. During each iteration, it performs a constant number of operations (comparing characters and calculating the minimum of two numbers). Since the loop runs for n-1 iterations (where n is the length of s), the time complexity is O(n).

Space Complexity

The code uses a fixed number of integer variables (ans, n, and i). These variables do not depend on the size of the input string s, hence the space complexity is O(1) as no additional space is proportional to the input size.

Learn more about how to find time and space complexity quickly using problem constraints.