Given a n x n matrix. The problem is to sort the given matrix in strict order. Here strict order means that matrix is sorted in a way such that all elements in a row are sorted in increasing order and for row ‘i’, where 1 <= i <= n-1, first element of row ‘i’ is greater than or equal to the last element of row ‘i-1’.
Examples:
Input : mat[][] = { {5, 4, 7}, {1, 3, 8}, {2, 9, 6} } Output : 1 2 3 4 5 6 7 8 9
Approach: Create a temp[] array of size n^2. Starting with the first row one by one copy the elements of the given matrix into temp[]. Sort temp[]. Now one by one copy the elements of temp[] back to the given matrix.
C++
// C++ implementation to sort the given matrix #include <bits/stdc++.h> using namespace std; #define SIZE 10 // function to sort the given matrix void sortMat( int mat[SIZE][SIZE], int n) { // temporary matrix of size n^2 int temp[n * n]; int k = 0; // copy the elements of matrix one by one // into temp[] for ( int i = 0; i < n; i++) for ( int j = 0; j < n; j++) temp[k++] = mat[i][j]; // sort temp[] sort(temp, temp + k); // copy the elements of temp[] one by one // in mat[][] k = 0; for ( int i = 0; i < n; i++) for ( int j = 0; j < n; j++) mat[i][j] = temp[k++]; } // function to print the given matrix void printMat( int mat[SIZE][SIZE], int n) { for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) cout << mat[i][j] << " " ; cout << endl; } } // Driver program to test above int main() { int mat[SIZE][SIZE] = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } }; int n = 3; cout << "Original Matrix: "; printMat(mat, n); sortMat(mat, n); cout << " Matrix After Sorting: "; printMat(mat, n); return 0; } |
Output:
Original Matrix: 5 4 7 1 3 8 2 9 6 Matrix After Sorting: 1 2 3 4 5 6 7 8 9
Time Complexity: O(n2log2n).
Auxiliary Space: O(n2).
Please refer complete article on Sort the given matrix for more details!
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