Problem Description

You have a binary string s (containing only '0's and '1's) that is 0-indexed with length n. Your goal is to make all characters in the string equal (either all '0's or all '1's) using the minimum total cost.

You can perform two types of operations:

1. Left-side inversion: Choose an index i and invert all characters from index 0 to index i (inclusive). This operation costs i + 1.

2. Right-side inversion: Choose an index i and invert all characters from index i to index n - 1 (inclusive). This operation costs n - i.

When you invert a character, '0' becomes '1' and '1' becomes '0'.

The key insight is that whenever two adjacent characters are different (s[i] ≠ s[i-1]), you must perform an operation to make them equal. At each such position, you have two choices:

• Invert the left portion s[0..i-1] with cost i
• Invert the right portion s[i..n-1] with cost n - i

The solution greedily chooses the minimum cost option at each position where adjacent characters differ, accumulating these costs to find the minimum total cost needed to make all characters equal.

For example, if s = "0011", the characters at positions 1 and 2 are different, so you need an operation there. You can either invert s[0..1] with cost 2, or invert s[2..3] with cost 2. Either way results in all characters being equal.

Intuition

The key observation is understanding when operations are necessary. If we scan through the string and find two adjacent characters that are different, we know that at least one operation must be performed to make them equal.

Consider what happens when we find s[i] ≠ s[i-1]. We have a "boundary" between different characters. To eliminate this boundary, we have two choices:

• Make all characters to the left match the right side by inverting s[0..i-1]
• Make all characters to the right match the left side by inverting s[i..n-1]

The crucial insight is that we don't need to decide which final character ('0' or '1') the entire string should become. We just need to eliminate all boundaries between different characters. Once all boundaries are eliminated, all characters will naturally be the same.

Why can we process each boundary independently and greedily choose the minimum cost? Because operations at different boundaries don't interfere with each other in terms of creating or removing boundaries. When we invert a segment, we're essentially "propagating" one side's value across the boundary. The order of operations doesn't matter for the final cost.

For each boundary at position i, the cost of fixing it is min(i, n-i). We choose whichever side is shorter to invert, minimizing the cost for that particular boundary. The total minimum cost is simply the sum of minimum costs for fixing all boundaries.

This greedy approach works because:

1. Each boundary must be fixed exactly once
2. The cost of fixing each boundary is independent
3. We can always choose the cheaper option without affecting other boundaries' costs

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation follows a straightforward greedy algorithm that processes the string in a single pass:

1. Initialize variables: Set ans = 0 to track the total minimum cost and n = len(s) to store the string length.

2. Iterate through the string: Loop through indices from 1 to n-1, checking each pair of adjacent characters.

3. Detect boundaries: For each position i, check if s[i] != s[i-1]. This condition identifies a boundary where adjacent characters differ.

4. Calculate minimum cost: When a boundary is found at position i, we have two options:

   • Invert the left portion s[0..i-1] with cost i
   • Invert the right portion s[i..n-1] with cost n - i

   We choose the minimum: min(i, n - i) and add it to our running total.

5. Return the result: After processing all positions, return ans as the minimum total cost.

The algorithm's efficiency comes from its simplicity:

• Time Complexity: O(n) - we make a single pass through the string
• Space Complexity: O(1) - we only use a constant amount of extra space

The code implementation:

class Solution:
    def minimumCost(self, s: str) -> int:
        ans, n = 0, len(s)
        for i in range(1, n):
            if s[i] != s[i - 1]:
                ans += min(i, n - i)
        return ans

This greedy approach guarantees the minimum cost because at each boundary, we independently choose the optimal (cheaper) operation without affecting the cost of handling other boundaries.

Example Walkthrough

Let's walk through the solution with s = "00110":

Initial state: s = "00110", n = 5, ans = 0

Step 1: Check position i = 1

• Compare s[1] = '0' with s[0] = '0'
• They're equal, no boundary here, continue

Step 2: Check position i = 2

• Compare s[2] = '1' with s[1] = '0'
• They're different! We found a boundary
• Cost options:
  • Invert left s[0..1]: cost = i = 2
  • Invert right s[2..4]: cost = n - i = 5 - 2 = 3
• Choose minimum: min(2, 3) = 2
• Update: ans = 0 + 2 = 2

Step 3: Check position i = 3

• Compare s[3] = '1' with s[2] = '1'
• They're equal, no boundary here, continue

Step 4: Check position i = 4

• Compare s[4] = '0' with s[3] = '1'
• They're different! Another boundary
• Cost options:
  • Invert left s[0..3]: cost = i = 4
  • Invert right s[4..4]: cost = n - i = 5 - 4 = 1
• Choose minimum: min(4, 1) = 1
• Update: ans = 2 + 1 = 3

Result: Return ans = 3

To verify: If we apply these operations:

1. At position 2, invert left s[0..1]: "00110" → "11110"
2. At position 4, invert right s[4..4]: "11110" → "11111"

All characters are now equal with total cost = 3, which matches our calculated minimum cost.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the string s. The algorithm iterates through the string once with a single for loop from index 1 to n-1, performing constant time operations (comparison and minimum calculation) at each iteration.

Space Complexity: O(1). The algorithm only uses a fixed amount of extra space for variables ans, n, and the loop variable i, regardless of the input size. No additional data structures that scale with input size are created.

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

Common Pitfalls

1. Misunderstanding the Operation Effects

A common mistake is thinking that we need to track the actual state of the string after each operation. Developers might try to simulate the flips and maintain the modified string state:

# INCORRECT approach - unnecessarily complex
class Solution:
    def minimumCost(self, s: str) -> int:
        s = list(s)  # Convert to list for mutation
        total_cost = 0
      
        for i in range(1, len(s)):
            if s[i] != s[i - 1]:
                if i <= len(s) - i:
                    # Flip left side
                    for j in range(i):
                        s[j] = '1' if s[j] == '0' else '0'
                    total_cost += i
                else:
                    # Flip right side
                    for j in range(i, len(s)):
                        s[j] = '1' if s[j] == '0' else '0'
                    total_cost += len(s) - i
      
        return total_cost

Why this is wrong: The operations don't need to be actually performed. The key insight is that whenever we encounter a boundary (different adjacent characters), we must perform an operation there. The actual final state doesn't matter - we only care about the cost.

2. Incorrect Cost Calculation

Another pitfall is miscalculating the operation costs, especially confusing 0-based indexing with the actual cost:

# INCORRECT - wrong cost calculation
class Solution:
    def minimumCost(self, s: str) -> int:
        ans, n = 0, len(s)
        for i in range(1, n):
            if s[i] != s[i - 1]:
                # Wrong! Should be min(i, n - i), not min(i + 1, n - i + 1)
                ans += min(i + 1, n - i + 1)
        return ans

The correct understanding:

• Left-side inversion from index 0 to i-1 costs i (not i+1)
• Right-side inversion from index i to n-1 costs n - i (not n-i+1)

3. Overthinking the Final State

Some might worry about whether the final string will be all '0's or all '1's and try to calculate both scenarios:

# INCORRECT - overthinking the problem
class Solution:
    def minimumCost(self, s: str) -> int:
        # Try to make all '0's
        cost_all_zeros = self.calculateCost(s, '0')
        # Try to make all '1's
        cost_all_ones = self.calculateCost(s, '1')
        return min(cost_all_zeros, cost_all_ones)

Why this is unnecessary: The greedy approach automatically finds the minimum cost regardless of whether the final string is all '0's or all '1's. The boundaries between different characters are what matter, not the target character.

Solution to Avoid These Pitfalls:

1. Focus on boundaries only: Remember that you only need to make decisions at positions where adjacent characters differ.

2. Use the simple formula: At each boundary position i, the cost is simply min(i, n - i).

3. Don't simulate operations: The actual string transformations don't need to be tracked - only the accumulated cost matters.

4. Trust the greedy approach: Each local optimal choice (choosing the cheaper flip at each boundary) leads to the global optimum.

The correct, simple implementation avoids all these pitfalls:

class Solution:
    def minimumCost(self, s: str) -> int:
        ans, n = 0, len(s)
        for i in range(1, n):
            if s[i] != s[i - 1]:
                ans += min(i, n - i)
        return ans