Leetcode 1072: Flip Columns For Maximum Number of Equal Rows
You are given an m x n binary matrix. You can choose any number of columns in the matrix and flip every cell in that column (i.e., change the value of the cell from 0 to 1 or vice versa). Your task is to find the maximum number of rows that can be made equal after performing a number of column flips.
Problem
Approach
Steps
Complexity
Input: The input consists of a binary matrix with m rows and n columns, where each element is either 0 or 1.
Example: Input: matrix = [[1,0],[0,0]]
Constraints:
• 1 <= m, n <= 300
• matrix[i][j] is either 0 or 1.
Output: The output is an integer, which represents the maximum number of rows that can be made equal after performing some number of column flips.
Example: Output: 1
Constraints:
• The output should be an integer value representing the maximum number of rows with identical values after flips.
Goal: The goal is to find the maximum number of rows that can be made identical after flipping columns in the matrix.
Steps:
• 1. Iterate through each row in the matrix.
• 2. For each row, convert it into a binary string where flipped columns are taken into account.
• 3. Track how many times each unique row pattern appears using a hashmap.
• 4. The answer will be the maximum frequency of any row pattern.
Goal: The matrix has a manageable size, so the problem can be solved efficiently using hashing to track row patterns.
Steps:
• 1 <= m, n <= 300
• matrix[i][j] is either 0 or 1.
Assumptions:
• The input matrix is valid and contains only binary values.
• The matrix dimensions are within the specified constraints.
• Input: Input: matrix = [[0, 1], [1, 1]]
• Explanation: In this case, no column flips are needed, as only one row has identical values. The output is 1.
• Input: Input: matrix = [[0, 1], [1, 0]]
• Explanation: Flipping the first column will make both rows identical. The output is 2.
• Input: Input: matrix = [[0, 0, 0], [0, 0, 1], [1, 1, 0]]
• Explanation: Flipping the first two columns will make the second and third rows identical. The output is 2.
Approach: We can solve this problem by tracking the frequency of row patterns. By flipping columns, we can create row patterns that can be stored in a hashmap to count their occurrences. The result is the maximum count of any identical row pattern.
Observations:
• Each row can either stay the same or change after flipping columns.
• The key observation is that rows that are already identical after column flips should be counted together.
• We need a way to count how often each unique row pattern appears after flipping columns, and the maximum count will be our answer.
Steps:
• 1. Initialize a hashmap to store row patterns and their frequencies.
• 2. For each row in the matrix, create a binary string based on the current state of the row (considering flipped columns).
• 3. Track the frequency of each unique binary string.
• 4. Return the maximum frequency from the hashmap as the result.
Empty Inputs:
• The problem does not expect an empty matrix, as there must be at least one row and one column.
Large Inputs:
• Ensure the solution is efficient enough to handle matrices with up to 300 rows and columns.
Special Values:
• If all elements in the matrix are already the same, no flips are needed, and the result is the number of rows.
Constraints:
• The solution must operate efficiently within the given input constraints.
int maxEqualRowsAfterFlips(vector<vector<int>>& mtx) {
unordered_map<string, int> mp;
for(auto x: mtx) {
string s = "";
int top = x[0];
for(int i = 0; i < x.size(); i++) {
if(x[i] == top) s += '1';
else s += '0';
}
mp[s]++;
}
int res = 0;
for(auto [k, v]: mp)
res = max(res, v);
return res;
}
1 : Function Definition
int maxEqualRowsAfterFlips(vector<vector<int>>& mtx) {
This line defines the function `maxEqualRowsAfterFlips`, which takes a 2D vector `mtx` (a binary matrix) and returns the maximum number of rows that can be made equal by flipping some of them.
2 : HashMap Initialization
unordered_map<string, int> mp;
This line initializes an unordered map `mp`, where the keys will be string representations of rows, and the values will be their frequencies (the number of times each row pattern appears).
3 : Matrix Loop
for(auto x: mtx) {
This loop iterates over each row `x` in the matrix `mtx`.
4 : String Initialization
string s = "";
This line initializes an empty string `s`, which will be used to store the binary string representation of the current row.
5 : Top Element Extraction
int top = x[0];
This line stores the first element of the row `x` in the variable `top`. This will be used to normalize the row by flipping the row to make its first element the same for all rows.
6 : Row Loop
for(int i = 0; i < x.size(); i++) {
This loop iterates through each element of the row `x`.
7 : Row Normalization
if(x[i] == top) s += '1';
This line adds '1' to the string `s` if the current element `x[i]` is equal to `top`.
8 : Row Normalization
else s += '0';
This line adds '0' to the string `s` if the current element `x[i]` is not equal to `top`.
9 : Map Update
mp[s]++;
This line updates the map `mp`, incrementing the count of the binary string `s`, which represents the current normalized row.
10 : Result Initialization
int res = 0;
This line initializes the variable `res`, which will store the maximum frequency of any row pattern in the matrix.
11 : Map Loop
for(auto [k, v]: mp)
This loop iterates over the map `mp`, where `k` is the key (row pattern) and `v` is the frequency of that row pattern.
12 : Result Calculation
res = max(res, v);
This line updates the result `res` with the maximum value between the current value of `res` and the frequency `v` of the current row pattern `k`.
13 : Return Statement
return res;
This line returns the value of `res`, which is the maximum frequency of any row pattern in the matrix.
Best Case: O(m * n)
Average Case: O(m * n)
Worst Case: O(m * n)
Description: The time complexity is O(m * n) due to processing each element in the matrix and hashing the row patterns.
Best Case: O(m)
Worst Case: O(m * n)
Description: The space complexity is O(m * n) for storing the row patterns, where m is the number of rows and n is the number of columns in the matrix.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus