Leetcode 2712: Minimum Cost to Make All Characters Equal
You are given a binary string
s of length n. You can perform two types of operations: (1) Flip all characters from index 0 to i (inclusive), with a cost of i + 1. (2) Flip all characters from index i to n - 1 (inclusive), with a cost of n - i. The goal is to return the minimum cost required to make all the characters in the string equal (either all 0s or all 1s).Problem
Approach
Steps
Complexity
Input: A binary string `s` of length `n`.
Example: s = "1101"
Constraints:
• 1 <= s.length <= 10^5
• s[i] is either '0' or '1'
Output: Return the minimum cost to make all characters equal in the string.
Example: 3
Constraints:
• The output should be a non-negative integer.
Goal: The goal is to compute the minimum cost by applying the two types of operations on the string `s`.
Steps:
• Iterate over the string to determine the best points to apply the flip operations.
• Apply the operations considering the costs and track the minimum cost.
Goal: The string length is at most 10^5, which means an O(n) or O(n log n) solution is preferable.
Steps:
• 1 <= s.length <= 10^5
• s[i] is either '0' or '1'
Assumptions:
• The input string is non-empty and contains only '0' and '1' characters.
• Input: Example 1
• Explanation: To make the string '1101' all '0's or all '1's, apply the first operation with `i = 1` (cost 2) and the second operation with `i = 2` (cost 1), giving a minimum total cost of 3.
• Input: Example 2
• Explanation: For the string '111000', applying the first operation with `i = 3` (cost 4) results in '111111', which is the minimum cost to make all characters equal.
Approach: The solution involves checking where to apply flip operations and computing the minimum cost for each scenario.
Observations:
• The operations can only flip from the beginning or from the end of the string.
• We need to identify points where flipping operations can be applied with minimum cost to make all characters equal.
Steps:
• Identify positions in the string where flip operations can change the string to all '0's or all '1's.
• Compute the cost of each operation and return the minimum total cost.
Empty Inputs:
• An empty string is not a valid input.
Large Inputs:
• For large strings (near the upper constraint), the solution should be efficient (O(n) or O(n log n) time complexity).
Special Values:
• The string can already be uniform (all '0's or all '1's), in which case no operations are needed.
Constraints:
• The solution should handle up to 100,000 characters efficiently.
long long minimumCost(string s) {
/*
0100111
0000000 3 + 2 + 1
1011111 4 + 2 + 1 if odd len choose middle
0101000
1011111 3 + 3 + 2 + 1
0100000 4 + 3 + 2 + 1 if odd len choose middle
010111
011011
011011 0/ 2 + 3 + 1
1/ 3 + 2 + 1
010011
010000 2 + 2 + 1
101111 3
011111 3 + 2
111111 3 + 2 + 1 = 6
010000 3 = 3
100000 3 + 2 = 5
000000 3 + 2 + 1 = 6
*/
int n = s.size();
if(n == 1) return 0;
long long res = 0;
char k = s[n / 2];
int i, j;
if(n % 2)
i = n/2 + 1, j = n/2 - 1;
else i = n/2, j = n / 2 - 1;
bool ltog = true, rtog = true;
while(i < n && j >= 0) {
if(s[i] == k && rtog || s[i] != k && !rtog) {
} else {
rtog = !rtog;
res += n - i;
}
i++;
if((s[j] == k && ltog) || (s[j] != k && !ltog)) {
} else {
ltog = !ltog;
res += j + 1;
}
j--;
}
return res;
}
1 : String Manipulations
long long minimumCost(string s) {
Defines the function to calculate the minimum cost of a string transformation.
2 : String Operations
int n = s.size();
Finds the size of the input string `s`.
3 : Conditional Checks
if(n == 1) return 0;
Handles the edge case where the string has only one character.
4 : Variable Initialization
long long res = 0;
Initializes the result variable `res` to store the cost.
5 : Character Operations
char k = s[n / 2];
Determines the target character from the middle of the string.
6 : Variable Initialization
int i, j;
Declares indices for traversal from the middle outwards.
7 : Conditional Checks
if(n % 2)
Checks whether the length of the string is odd.
8 : Index Calculation
i = n/2 + 1, j = n/2 - 1;
Initializes traversal indices for an odd-length string.
9 : Index Calculation
else i = n/2, j = n / 2 - 1;
Initializes traversal indices for an even-length string.
10 : Boolean Variables
bool ltog = true, rtog = true;
Initializes toggle variables for tracking left and right flips.
11 : Loop Constructs
while(i < n && j >= 0) {
Iterates over the string from the middle to the edges.
12 : Conditional Checks
if(s[i] == k && rtog || s[i] != k && !rtog) {
Checks whether the current right character matches the target.
13 : No Operation
} else {
Handles cases where no action is required.
14 : Boolean Operations
rtog = !rtog;
Toggles the right flip state.
15 : Sum Calculation
res += n - i;
Adds the cost of flipping the right character.
16 : Index Update
i++;
Moves to the next right character.
17 : Conditional Checks
if((s[j] == k && ltog) || (s[j] != k && !ltog)) {
Checks whether the current left character matches the target.
18 : No Operation
} else {
Handles cases where no action is required.
19 : Boolean Operations
ltog = !ltog;
Toggles the left flip state.
20 : Sum Calculation
res += j + 1;
Adds the cost of flipping the left character.
21 : Index Update
j--;
Moves to the next left character.
22 : Return Statement
return res;
Returns the total calculated cost.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The solution runs in linear time since we need to check each character at most once.
Best Case: O(1)
Worst Case: O(1)
Description: The solution uses constant space as we only store a few variables to track the current state.
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