Print all elements in sorted order from row and column wise sorted matrix - GeeksforGeeks
Given an n x n matrix, where every row and column is sorted in non-decreasing order. Print all elements of matrix in sorted order.
Example:
O(N^3)
We can use Young Tableau to solve the above problem. The idea is to consider given 2D array as Young Tableau and call extract minimum O(N)
O(N2LogN)
The idea is to use a Min Heap of size N which stores elements of first column. The do extract minimum. In extract minimum, replace the minimum element with the next element of the row from which the element is extracted.
http://stackoverflow.com/questions/4279524/how-to-sort-a-m-x-n-matrix-which-has-all-its-m-rows-sorted-and-n-columns-sorted
Given an n x n matrix, where every row and column is sorted in non-decreasing order. Print all elements of matrix in sorted order.
Example:
Input: mat[][] = { {10, 20, 30, 40}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50}, }; Output: Elements of matrix in sorted order 10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50
O(N^3)
We can use Young Tableau to solve the above problem. The idea is to consider given 2D array as Young Tableau and call extract minimum O(N)
// A utility function to youngify a Young Tableau. This is different// from standard youngify. It assumes that the value at mat[0][0] is // infinite.void youngify(int mat[][N], int i, int j){ // Find the values at down and right sides of mat[i][j] int downVal = (i+1 < N)? mat[i+1][j]: INF; int rightVal = (j+1 < N)? mat[i][j+1]: INF; // If mat[i][j] is the down right corner element, return if (downVal==INF && rightVal==INF) return; // Move the smaller of two values (downVal and rightVal) to // mat[i][j] and recur for smaller value if (downVal < rightVal) { mat[i][j] = downVal; mat[i+1][j] = INF; youngify(mat, i+1, j); } else { mat[i][j] = rightVal; mat[i][j+1] = INF; youngify(mat, i, j+1); }}// A utility function to extract minimum element from Young tableauint extractMin(int mat[][N]){ int ret = mat[0][0]; mat[0][0] = INF; youngify(mat, 0, 0); return ret;}// This function uses extractMin() to print elements in sorted ordervoid printSorted(int mat[][N]){ cout << "Elements of matrix in sorted order \n"; for (int i=0; i<N*N; i++) cout << extractMin(mat) << " ";}The idea is to use a Min Heap of size N which stores elements of first column. The do extract minimum. In extract minimum, replace the minimum element with the next element of the row from which the element is extracted.
struct MinHeapNode{ int element; // The element to be stored int i; // index of the row from which the element is taken int j; // index of the next element to be picked from row};void printSorted(int mat[][N]){ // Create a min heap with k heap nodes. Every heap node // has first element of an array MinHeapNode *harr = new MinHeapNode[N]; for (int i = 0; i < N; i++) { harr[i].element = mat[i][0]; // Store the first element harr[i].i = i; // index of row harr[i].j = 1; // Index of next element to be stored from row } MinHeap hp(harr, N); // Create the min heap // Now one by one get the minimum element from min // heap and replace it with next element of its array for (int count = 0; count < N*N; count++) { // Get the minimum element and store it in output MinHeapNode root = hp.getMin(); cout << root.element << " "; // Find the next elelement that will replace current // root of heap. The next element belongs to same // array as the current root. if (root.j < N) { root.element = mat[root.i][root.j]; root.j += 1; } // If root was the last element of its array else root.element = INT_MAX; //INT_MAX is for infinite // Replace root with next element of array hp.replaceMin(root); }}
I don't think you can do it any faster than Ω(m n log(min(m, n)), at least not in the general case.
Suppose (without loss of generality) that m < n. Then your matrix looks like this:
Each circle is a matrix entry and each arrow indicates a known order relation (the entry at the source of the arrow is smaller than the entry at the destination of the arrow).
To sort the matrix, we must resolve all the unknown order relations, some of which are shown in the grey boxes here:
Sorting all of these boxes takes:
Read full article from Print all elements in sorted order from row and column wise sorted matrix - GeeksforGeeks2 Σk < m Ω(k log k) + (n - m + 1) Ω(m log m)= 2 Ω(m² log m) + (n - m + 1) Ω(m log m)= Ω(m n log m)
