Leetcode #2712: Minimum Cost to Make All Characters Equal
In this guide, we solve Leetcode #2712 Minimum Cost to Make All Characters Equal in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
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
Complexity
The time complexity is , where is the length of the string . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.