2712. Minimum Cost to Make All Characters Equal
Problem Description
In this problem, we are dealing with a binary string s
, which is a sequence of characters that can only be 0
or 1
. The string uses a 0-based index, meaning that the first character of the string is at position 0, the next at position 1, and so on, up to the last character, which is at position n - 1
. Our goal is to transform this binary string into a uniform string where all characters are the same, either all 0
s or all 1
s.
To achieve a uniform string, we can perform two types of operations:
-
Select an index
i
, and flip all the characters from the beginning of the string (index 0) up to and includingi
. For instance, if we flip characters from index 0 to indexi
in"01011"
, andi
is 2, the string becomes"10100"
. This operation comes with a cost equal toi + 1
. -
Select an index
i
, and flip all the characters from that indexi
to the end of the string (indexn - 1
). For example, for the same string"01011"
, if we flip characters from index 2 to index 4 (n - 1
), the string becomes"01100"
. This operation has a cost ofn - i
.
Each flip inverts the characters, which means 0
becomes 1
and 1
becomes 0
. The challenge is to find the minimum total cost of such flipping operations that will result in a string composed entirely of the same character.
Intuition
The intuition behind the provided solution is to identify the positions in the string where flipping can actually contribute to making the string uniform and doing it at the lowest possible cost. If all characters are already equal, no operation is needed and the cost is zero. We only need to consider flipping at positions where a character differs from its preceding character, indicating a point of change in the string's structure.
For each character s[i]
that differs from s[i - 1]
, there is a possibility of flipping either the sub-string before i
or the sub-string after i
to make the characters equal. We want to choose the index that minimizes the cost for each situation, which is given by min(i, n - i)
where i
is the cost for flipping from the start and n - i
is the cost for flipping until the end. By iterating through the string and summing up the minimum cost at each change point, we accumulate the total minimum cost needed to make the string uniform.
Here is the step-by-step intuition for the provided solution:
- Initialize
ans
as0
, which will hold the overall minimum cost. - Loop through each character in the string, starting from index 1 up to index
n - 1
. - Check if the current character
s[i]
is different from the previous characters[i - 1]
. - When a difference is detected, it means we have a potential flip point.
- Calculate the cost to flip either from the start up to
i
or fromi
to the end, by evaluatingmin(i, n - i)
. - Add this cost to the
ans
total. - After looping through the entire string,
ans
will contain the minimum cost to make the string uniform. - Return the
ans
value.
The solution effectively avoids redundant operations and achieves minimal cost by flipping at the optimal points in the string.
Learn more about Greedy and Dynamic Programming patterns.
Solution Approach
The implementation of the solution provided uses a simple linear scan algorithm which is both time and space efficient because it iterates over the string once and does not require any additional data structures. The key pattern used here is the iteration over changing points in the binary string.
Here's the breakdown of the solution approach:
-
Initialization: A variable
ans
is created to store the accumulated minimum cost and its initial value is set to0
. The length of the stringn
is computed to avoid recalculating it during each iteration.ans, n = 0, len(s)
-
Looping through the changes: The program then iterates from
1
ton - 1
, intentionally starting from1
because we always compare the current character with the previous one to identify the change points.for i in range(1, n):
-
Detecting change points: The condition inside the loop checks if the current character
s[i]
is different from the preceding characters[i - 1]
. If it is, then a flip operation must be performed at this point, because we want to eliminate the discrepancy to move towards a uniform string.if s[i] != s[i - 1]:
-
Calculating operation costs: For any change point, the cost of making the preceding or succeeding substring uniform is analyzed. Since the goal is to do this at the minimum cost, the function
min(i, n - i)
calculates the cheaper option between flipping the firsti
characters or the lastn - i
characters. This corresponds to choosing the shorter sub-string to flip which naturally costs less.ans += min(i, n - i)
The result is added to our running total
ans
. -
Final result: After all potential change points have been processed, the variable
ans
now contains the minimum cost to make all characters of the string equal. The function finally returnsans
.return ans
By utilizing a single pass over the string and performing an efficient comparison and calculation at each step, the implementation ensures a minimal time complexity of O(n), where n is the length of the string. There's no use of additional data structures, so the space complexity is O(1). This approach ensures optimal performance for the problem at hand.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Let's consider a simple example to illustrate the solution approach. We have the following binary string s = "0110101"
.
-
Initialization: Set the total minimum cost
ans
to0
and compute the length ofs
to ben = 7
.ans, n = 0, len(s) # ans = 0, n = 7
-
Looping through the changes: Start iterating from index
1
ton - 1
, which goes from index1
to index6
. -
Detecting change points: We're looking for positions where
s[i] != s[i - 1]
. These are our points of change:- At
i = 1
,s[i] = "1"
ands[i - 1] = "0"
. This is a change point;s[1]
differs froms[0]
. - At
i = 2
,s[i] = "1"
ands[i - 1] = "1"
. This is not a change point;s[2]
is the same ass[1]
. - At
i = 3
,s[i] = "0"
ands[i - 1] = "1"
. This is another change point;s[3]
differs froms[2]
. - At
i = 4
,s[i] = "1"
ands[i - 1] = "0"
. Another change point;s[4]
differs froms[3]
. - At
i = 5
,s[i] = "0"
ands[i - 1] = "1"
. Yet another change point;s[5]
differs froms[4]
. - At
i = 6
,s[i] = "1"
ands[i - 1] = "0"
. The last change point;s[6]
differs froms[5]
.
- At
-
Calculating operation costs: For every change point identified, calculate the cost:
- At
i = 1
, the cost ismin(1, 7 - 1) = 1
. Add this toans
to getans = 1
. - At
i = 3
, the cost ismin(3, 7 - 3) = 3
. Nowans = 1 + 3 = 4
. - At
i = 4
, the cost ismin(4, 7 - 4) = 3
. Updateans = 4 + 3 = 7
. - At
i = 5
, the cost ismin(5, 7 - 5) = 2
. Updateans = 7 + 2 = 9
. - At
i = 6
, the cost ismin(6, 7 - 6) = 1
. Finally,ans = 9 + 1 = 10
.
- At
-
Final result: After the loop, we have checked all change points. The total minimum cost
ans
is10
. Thus, this is the minimum cost required to make the strings = "0110101"
uniform, using the flip operations described.
By sequentially applying the logic to each change point, the example demonstrates how we sum up the minimum of flipping costs at every change point to get the total minimal cost efficiently.
Solution Implementation
1class Solution:
2 def minimumCost(self, s: str) -> int:
3 """
4 Calculate the minimum cost of processing the string where
5 cost is defined as the minimum distance from either end of the string
6 to the position where the character differs from its adjacent one.
7
8 :param s: The input string consisting of characters.
9 :type s: str
10 :return: The minimum cost of processing the string.
11 :rtype: int
12 """
13 # Initialize the answer variable to store the total cost
14 total_cost = 0
15
16 # Get the length of the string
17 n = len(s)
18
19 # Iterate through the string starting from the second character
20 for i in range(1, n):
21 # Check if the current character is different from the previous character
22 if s[i] != s[i - 1]:
23 # If so, add the minimum distance from either end of the string
24 total_cost += min(i, n - i)
25
26 # Return the calculated total cost
27 return total_cost
28
1class Solution {
2
3 /**
4 * Calculates the minimum cost to ensure no two adjacent characters are the same.
5 *
6 * @param s The input string.
7 * @return The total minimum cost.
8 */
9 public long minimumCost(String s) {
10 long totalCost = 0; // Holds the running total cost
11 int lengthOfString = s.length(); // Stores the length of the string once for efficiency
12
13 // Loop through each character in the string, starting from 1 as we're comparing it with the previous character
14 for (int i = 1; i < lengthOfString; ++i) {
15 // Check if the current character is different from the previous one
16 if (s.charAt(i) != s.charAt(i - 1)) {
17 // Calculate minimum cost for this char to be either from the start or end of the string
18 totalCost += Math.min(i, lengthOfString - i);
19 }
20 }
21
22 return totalCost; // Return the calculated total cost
23 }
24}
25
1class Solution {
2public:
3 // Function to calculate the minimum cost of operations to make the string stable
4 long long minimumCost(string s) {
5 long long cost = 0; // Initialize the total cost to 0
6 int length = s.size(); // Get the length of the string
7
8 // Iterate through the string starting from the second character
9 for (int i = 1; i < length; ++i) {
10 // Check if the current character is different from the previous one
11 if (s[i] != s[i - 1]) {
12 // If different, add the minimum of 'i' or 'length - i' to the cost
13 // 'i' represents the cost to change all characters to the left to match the current one
14 // 'length - i' represents the cost to change all characters to the right
15 cost += min(i, length - i);
16 }
17 }
18 // Return the calculated cost
19 return cost;
20 }
21};
22
1// Function to calculate the minimum cost to make a binary string beautiful.
2// A binary string is considered beautiful if it does not contain any substring "01" or "10".
3function minimumCost(s: string): number {
4 // Initialize the answer, which will store the minimum cost.
5 let cost = 0;
6 // Get the length of the string.
7 const lengthOfString = s.length;
8
9 // Iterate through the string characters starting from the second character.
10 for (let index = 1; index < lengthOfString; ++index) {
11 // If the current and previous characters are different,
12 // It implies a "01" or "10" substring, which is not beautiful.
13 if (s[index] !== s[index - 1]) {
14 // The cost to remove this not-beautiful part is the minimum
15 // of either taking elements from the left or the right of it.
16 cost += Math.min(index, lengthOfString - index);
17 }
18 }
19
20 // Return the calculated minimum cost to make the string beautiful.
21 return cost;
22}
23
Time and Space Complexity
Time Complexity
The given code consists of a single loop that iterates through the length of the input string s
. During each iteration, it performs a constant number of operations (comparing characters and calculating the minimum of two numbers). Since the loop runs for n-1
iterations (where n
is the length of s
), the time complexity is O(n)
.
Space Complexity
The code uses a fixed number of integer variables (ans
, n
, and i
). These variables do not depend on the size of the input string s
, hence the space complexity is O(1)
as no additional space is proportional to the input size.
Learn more about how to find time and space complexity quickly using problem constraints.
A person thinks of a number between 1 and 1000. You may ask any number questions to them, provided that the question can be answered with either "yes" or "no".
What is the minimum number of questions you needed to ask so that you are guaranteed to know the number that the person is thinking?
Recommended Readings
Greedy Introduction div class responsive iframe iframe src https www youtube com embed WTslqPbj7I title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture web share allowfullscreen iframe div When do we use greedy Greedy algorithms tend to solve optimization problems Typically they will ask you to calculate the max min of some value Commonly you may see this phrased in the problem as max min longest shortest largest smallest etc These keywords can be identified by just scanning
What is Dynamic Programming Prerequisite DFS problems dfs_intro Backtracking problems backtracking Memoization problems memoization_intro Pruning problems backtracking_pruning Dynamic programming is an algorithmic optimization technique that breaks down a complicated problem into smaller overlapping sub problems in a recursive manner and uses solutions to the sub problems to construct a solution
LeetCode Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way we
Want a Structured Path to Master System Design Too? Donāt Miss This!