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Minimum Cost to Make All Characters Equal
Find the minimum cost to make all characters of a binary string equal by performing two types of operations.
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Practice Focus
Medium · State transition dynamic programming
Answer-first summary
Find the minimum cost to make all characters of a binary string equal by performing two types of operations.
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The problem asks you to find the minimum cost to make all characters of a binary string equal. You can perform two types of operations: inverting a character or making a prefix equal to the character at an index. Using dynamic programming, we calculate the cost for each possible transition to determine the minimum cost.
Problem Statement
You are given a binary string s consisting of '0's and '1's, and two types of operations can be performed on it: invert a character or make a prefix equal to a certain character. The goal is to determine the minimum cost to make all characters of the string equal by using these operations efficiently.
For each index i, we need to calculate the number of operations required to make the prefix [0, i - 1] equal to the character at index i. The minimum cost is determined by considering all such possible transitions and selecting the one that leads to the least cost to make the entire string uniform.
Examples
Example 1
Input: s = "0011"
Output: 2
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
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'
Solution Approach
State Transition Dynamic Programming
Use dynamic programming to calculate the minimum cost required to make the entire string equal. At each index, calculate the cost of performing the operation to make the prefix equal to the character at that index. The state transition will depend on whether we choose to invert the character or make the prefix uniform.
Greedy Approach for Prefix Calculation
At each index, calculate how many operations are required to change all previous characters in the string to match the character at index i. This can be achieved by incrementing or decrementing the operation costs based on the pattern of characters encountered so far.
Optimizing with Cumulative Cost Calculation
To avoid redundant calculations, use a cumulative approach where costs of previous transitions are stored and reused for later calculations. This will improve efficiency by reducing the number of calculations, making the solution scalable even for large inputs.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time and space complexity depend on the approach used to calculate the prefix operations and state transitions. In general, a solution based on dynamic programming with cumulative cost calculation can have a time complexity of O(n) and space complexity of O(n).
What Interviewers Usually Probe
- The candidate efficiently applies dynamic programming principles to minimize the cost.
- They effectively manage the complexity of the problem by optimizing calculations for large inputs.
- The candidate demonstrates a clear understanding of state transitions and prefix operations for solving the problem.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to handle the cost calculation at each index and neglecting to track the prefix operations.
- Not optimizing the solution for larger inputs, leading to time limit exceed errors.
- Confusing the required operations for inverting and calculating prefix costs, leading to incorrect results.
Follow-up variants
- The problem can be extended to strings of different characters beyond '0' and '1', requiring a more generalized solution.
- A variation of the problem could involve allowing multiple operations on each character rather than just a prefix change or inversion.
- The cost structure could be modified, such as introducing different costs for different types of operations, changing the problem dynamics.
FAQ
What is the main algorithmic approach to solving the 'Minimum Cost to Make All Characters Equal' problem?
The main approach involves using state transition dynamic programming to calculate the cost at each index and minimize the total operations needed to make the string uniform.
How do we minimize the cost of making the string equal?
By calculating the number of operations required for each possible state transition and selecting the one that results in the minimum cost for the entire string.
What are the key challenges when solving the 'Minimum Cost to Make All Characters Equal' problem?
The primary challenges include managing the transition costs efficiently, optimizing the solution for large inputs, and ensuring accurate calculation of prefix operations.
How does the greedy approach fit into this problem?
The greedy approach is used to calculate the minimum cost for transforming each prefix of the string to match the character at the current index, minimizing operations.
Can this problem be solved with other algorithms besides dynamic programming?
While dynamic programming is the most efficient approach, other algorithms may be used, but they may not scale well for large inputs due to higher time complexity.
Solution
Solution 1: Greedy Algorithm
According to the problem description, if $s[i] \neq s[i - 1]$, an operation must be performed; otherwise, it's impossible to make all characters equal.
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 ansContinue Topic
string
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