How do you solve Minimum Cost to Make All Characters Equal on LeetCode?

Minimum Cost to Make All Characters Equal (LeetCode #2712) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the cost of operations is crucial for minimizing the total cost.

What is the brute force solution for Minimum Cost to Make All Characters Equal?

The brute force approach for Minimum Cost to Make All Characters Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible operation on every index to see how much it costs to make all characters equal. This is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Make All Characters Equal?

Minimum Cost to Make All Characters Equal uses the following concepts: String, Dynamic Programming, Greedy. The recommended patterns to study are: Dynamic Programming, Prefix Sum, Suffix Sum.

What are the interview tips for Minimum Cost to Make All Characters Equal?

Always discuss your thought process and approach before jumping into coding. Consider edge cases, such as strings that are already uniform. Practice breaking down problems into smaller components to simplify complex problems.

What are common mistakes when solving Minimum Cost to Make All Characters Equal?

Not considering both operations for each index, leading to suboptimal solutions. Overlooking the need to calculate costs for both '0' and '1' conversions.

#2712

Minimum Cost to Make All Characters Equal

Medium

StringDynamic ProgrammingGreedyDynamic ProgrammingPrefix SumSuffix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach calculates the cost to make all characters equal by leveraging prefix and suffix operations. This reduces the number of operations needed to find the minimum cost significantly.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays, prefix and suffix, to store the cost of making the prefix and suffix equal to '0' and '1'.
2Step 2: Fill the prefix array by iterating through the string and calculating the cost of converting the prefix to '0' and '1'.
3Step 3: Fill the suffix array similarly for the suffix.
4Step 4: Calculate the minimum cost by combining the prefix and suffix costs.

solution.py12 lines

1def minCost(s):
2    n = len(s)
3    prefix_cost = [0] * n
4    suffix_cost = [0] * n
5    for i in range(n):
6        prefix_cost[i] = (prefix_cost[i - 1] if i > 0 else 0) + (i + 1 if s[i] == '1' else 0)
7    for i in range(n - 1, -1, -1):
8        suffix_cost[i] = (suffix_cost[i + 1] if i < n - 1 else 0) + (n - i if s[i] == '0' else 0)
9    min_cost = float('inf')
10    for i in range(n):
11        min_cost = min(min_cost, prefix_cost[i] + suffix_cost[i])
12    return min_cost

ℹ

Complexity note: The complexity is O(n) because we make a single pass to compute the prefix costs and another to compute the suffix costs, resulting in linear time complexity.

1Understanding the cost of operations is crucial for minimizing the total cost.
2Breaking the problem into prefix and suffix calculations helps in optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.