Minimize cost to convert a given matrix to another by flipping columns and reordering rows
Given two binary matrices mat[][] and target[][] of dimensions N * M, the task is to find the minimum cost to convert the matrix mat[][] into target[][] using the following operations:
- Flip a particular column in mat[][] such that all 1s become 0s and vice-versa. The cost of this operation is 1.
- Reorder the rows of mat[][]. The cost of this operation is 0.
If its not possible to convert the matrix mat[][] to target[][], then print “-1”.
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Examples:
Input: mat[][] = {{0, 0}, {1, 0}, {1, 1}}, target[][] = {{0, 1}, {1, 0}, {1, 1}}
Output: 1
Explanation:
Step 1: Reorder 2nd and 3rd rows to modify mat[][] to {{0, 0}, {1, 1}, {1, 0}}. Cost = 0
Step 2: Flip the second column . mat[][] modifies to {{0, 1}, {1, 0}, {1, 1}}, which is equal to target[][]. Cost = 1Input: mat[][] = {{0, 0, 0}, {1, 0, 1}, {1, 1, 0}}, target[][] = {{0, 0, 1}, {1, 0, 1}, {1, 1, 1}}
Output: -1
Approach: The idea is to convert the rows of the given matrices into bitsets and then find the minimum cost of operation to make the matrices equal. Below are the steps:
- First, convert the rows of mat[][] to binary numbers using bitset.
- After completing the above step, find all possible rows which can be the first row of the target matrix.
- The number of flips required to convert the row of mat[][] to tar[][] is the count of set bits in the Bitwise XOR value of the bitsets.
- Compute Bitwise XOR of every row with the flip pattern and check if the new matrix, when sorted, is equal to the sorted target[][] matrix or not.
- If they are the same, then store the number of flips.
- Calculate all such count of flips in the above steps and return the minimum among them.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Custom comparator to sort vector// of bitsetsbool static cmp(bitset<105>& p1, bitset<105>& p2){ return p1.to_string() < p2.to_string();} // Function to convert the given// matrix into the target matrixint minCost(vector<vector<int> >& a, vector<vector<int> >& t){ // Number of rows and columns int n = a.size(); int m = a[0].size(); vector<bitset<105> > mat(n), tar(n); // Iterate over rows for (int i = 0; i < n; i++) { string s; for (int j = 0; j < m; j++) { s += char(a[i][j] + '0'); } mat[i] = bitset<105>(s); } // Iterate over rows for (int i = 0; i < n; i++) { string s; for (int j = 0; j < m; j++) { s += char(t[i][j] + '0'); } tar[i] = bitset<105>(s); } // Sort the matrix sort(tar.begin(), tar.end(), cmp); int ans = INT_MAX; // Check all possible rows as the // first row of target for (int i = 0; i < n; i++) { vector<bitset<105> > copy(mat); // Get the flip pattern auto flip = copy[i] ^ tar[0]; for (auto& row : copy) row ^= flip; sort(copy.begin(), copy.end(), cmp); // Number of flip operations // is the count of set bits in flip if (copy == tar) ans = min(ans, (int)flip.count()); } // If it is not possible if (ans == INT_MAX) return -1; // Return the answer return ans;} // Driver Codeint main(){ // Given matrices vector<vector<int> > matrix{ { 0, 0 }, { 1, 0 }, { 1, 1 } }; vector<vector<int> > target{ { 0, 1 }, { 1, 0 }, { 1, 1 } }; // Function Call cout << minCost(matrix, target);} |
1
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)
