Minimum Cost to Make All Characters Equal

1
2
3
4
5
6
7
8

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

10 Prerequisite LeetCode Problems

For this, the following are a good preparation:

  1. “136. Single Number” - Familiarizes you with bitwise operations and helps in understanding the binary representation.

  2. “190. Reverse Bits” - Teaches how to manipulate bits in a given number.

  3. “461. Hamming Distance” - Practices calculating the difference between two binary numbers.

  4. “338. Counting Bits” - Reinforces your understanding of binary numbers and bitwise operations.

  5. “476. Number Complement” - Provides practice on flipping bits, which is directly applicable to the original problem.

  6. “231. Power of Two” - Helps in understanding binary representation and bit manipulation.

  7. “371. Sum of Two Integers” - The skill to add numbers without an arithmetic operator is a nice tool to have in your arsenal.

  8. “389. Find the Difference” - It will familiarize you with the concepts of character counts and their manipulations.

  9. “268. Missing Number” - A binary manipulation problem that teaches you to handle bit operations for an array of numbers.

  10. “693. Binary Number with Alternating Bits” - A good problem to understand sequences in binary representation.

These cover handling binary numbers, bit manipulation, and operations related to binary sequences, which are relevant for the problem “2712. Minimum Cost to Make All Characters Equal”.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

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

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.

Example 1:

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. Example 2:

Input: s = “010101” Output: 9 Explanation: 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’

Problem Classification

The domain of this problem is Data Manipulation and Optimization. It involves manipulating a binary string through a set of operations and finding an optimized solution, in this case, the minimum cost.

‘What’ Components:

  1. Binary String (s) of length n: This is the main data object that we will be working with. The operations will be performed on this string.

  2. Operation 1: This operation inverts all characters from index 0 to i in the string. The cost of this operation is i + 1.

  3. Operation 2: This operation inverts all characters from index i to the end of the string. The cost of this operation is n - i.

  4. Cost: The aim is to find the minimum total cost required to make all characters of the string equal using the two types of operations.

This problem is an optimization problem within the domain of String Manipulation. It involves performing operations on a binary string and tracking the associated costs, with the goal of achieving a certain condition (all characters equal) for the minimum total cost. This type of problem often requires dynamic programming or greedy algorithms to solve efficiently.

Language Agnostic Coding Drills

  1. Dissect the Code:

This piece of code can be broken down into several coding concepts:

a. Variable and Data Type Handling: This involves defining variables and handling data types in Python. For instance, ‘ans’ is an integer variable, and ’s’ is a string.

b. Control Flow (Conditional Statements and Loops): The ‘if’ statement is used to compare two adjacent characters in the string, while the ‘for’ loop iterates over each character in the string.

c. String Manipulation: This concept involves accessing and comparing string characters using indexing.

d. Mathematical Operations: Basic addition and subtraction operations are performed to calculate the cost of operations on the string.

  1. List of Coding Concepts:

    a. Variable and Data Type Handling: This is a basic concept in almost all programming languages. It involves creating variables and assigning values to them.

    b. Control Flow (Conditional Statements and Loops): This concept involves controlling the order in which the code is executed based on certain conditions and repeating some part of the code multiple times.

    c. String Manipulation: This is a medium difficulty concept that involves working with strings, which includes accessing and comparing string characters.

    d. Mathematical Operations: This involves performing mathematical operations like addition and subtraction.

  2. Problem-solving Approach:

The problem-solving approach here is to iterate over the string and check each character with its next one. If they are different, then an operation is needed to make them equal. The cost of the operation is calculated based on which operation (from the two given ones) is less expensive. This cost is then added to the total cost. This process is repeated until the end of the string.

In terms of the coding drills:

a. We need to define variables to hold the total cost and the length of the string.

b. We use a ‘for’ loop to iterate through the string up to the second last character, checking each character with its next one using an ‘if’ statement.

c. Inside the ‘if’ statement, another ‘if’ condition checks which operation is less expensive and adds its cost to the total cost.

d. The result is the minimum cost, which is returned by the function.

Targeted Drills in Python

  1. Python Drills for Each Concept:

    a. Variable and Data Type Handling:

    1
2
3
4

    # Create an integer variable
my_int = 5
# Create a string variable
my_str = "Hello World"

    b. Control Flow (Conditional Statements and Loops):

    1
2
3
4
5
6
7
8
9

    # If-else condition
if my_int > 0:
    print("Positive number")
else:
    print("Negative number or Zero")

# For loop
for i in range(my_int):
    print(i)

    c. String Manipulation:

    1
2
3
4
5
6
7
8

    # Accessing string character
first_char = my_str[0]

# Comparing string characters
if my_str[0] == my_str[1]:
    print("Equal")
else:
    print("Not Equal")

    d. Mathematical Operations:

    1
2
3
4
5

    # Addition
sum = my_int + 3

# Subtraction
diff = my_int - 2

  2. Drills for Specific Needs of the Problem:

    a. Comparison and calculation using multiple conditions:

    1
2
3
4

    # Compare two numbers and assign the smaller one to a variable
num1 = 5
num2 = 3
smaller = num1 if num1 < num2 else num2

    b. Accessing consecutive characters in a string:

    1
2
3
4
5
6
7

    # Iterate over a string comparing each character to the next
my_str = "Hello"
for i in range(len(my_str) - 1):
    if my_str[i] == my_str[i + 1]:
        print("Equal")
    else:
        print("Not Equal")

  3. Merging These Drills for the Final Solution:

The solution begins by defining the variables that will be used to hold the total cost and the length of the string. The code then enters a loop where each character in the string is compared to the next one. If they are different, the cost of making them equal is calculated based on which of the two operations is cheaper. This cost is then added to the total cost. The process repeats until we’ve iterated through the entire string. Finally, the function returns the total cost as the result. Each of these steps utilizes the coding drills that we’ve discussed. The loop and conditional statements control the flow of the code, the mathematical operations calculate the cost, and the string manipulation allows us to compare characters.