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
We strongly recommend to minimize the browser and try this yourself first.
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 C++ program to Print all elements in sorted order from row and// column wise sorted matrix#include<iostream>#include<climits>using namespace std;#define INF INT_MAX#define N 4// 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) << " ";}// driver program to test above functionint main(){ int mat[N][N] = { {10, 20, 30, 40}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50}, }; printSorted(mat); return 0;} |
Output:
Elements of matrix in sorted order 10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50
Time complexity of extract minimum is O(N) and it is called O(N2) times. Therefore the overall time complexity is O(N3).
A better solution is to use the approach used for merging k sorted arrays. 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. Time complexity of this solution is O(N2LogN).
// C++ program to merge k sorted arrays of size n each.#include<iostream>#include<climits>using namespace std;#define N 4// A min heap nodestruct 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};// Prototype of a utility function to swap two min heap nodesvoid swap(MinHeapNode *x, MinHeapNode *y);// A class for Min Heapclass MinHeap{ MinHeapNode *harr; // pointer to array of elements in heap int heap_size; // size of min heappublic: // Constructor: creates a min heap of given size MinHeap(MinHeapNode a[], int size); // to heapify a subtree with root at given index void MinHeapify(int ); // to get index of left child of node at index i int left(int i) { return (2*i + 1); } // to get index of right child of node at index i int right(int i) { return (2*i + 2); } // to get the root MinHeapNode getMin() { return harr[0]; } // to replace root with new node x and heapify() new root void replaceMin(MinHeapNode x) { harr[0] = x; MinHeapify(0); }};// This function prints elements of a given matrix in non-decreasing// order. It assumes that ma[][] is sorted row wise sorted.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); }}// FOLLOWING ARE IMPLEMENTATIONS OF STANDARD MIN HEAP METHODS// FROM CORMEN BOOK// Constructor: Builds a heap from a given array a[] of given sizeMinHeap::MinHeap(MinHeapNode a[], int size){ heap_size = size; harr = a; // store address of array int i = (heap_size - 1)/2; while (i >= 0) { MinHeapify(i); i--; }}// A recursive method to heapify a subtree with root at given index// This method assumes that the subtrees are already heapifiedvoid MinHeap::MinHeapify(int i){ int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l].element < harr[i].element) smallest = l; if (r < heap_size && harr[r].element < harr[smallest].element) smallest = r; if (smallest != i) { swap(&harr[i], &harr[smallest]); MinHeapify(smallest); }}// A utility function to swap two elementsvoid swap(MinHeapNode *x, MinHeapNode *y){ MinHeapNode temp = *x; *x = *y; *y = temp;}// driver program to test above functionint main(){ int mat[N][N] = { {10, 20, 30, 40}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50}, }; printSorted(mat); return 0;} |
Output:
10 15 20 25 27 29 30 32 33 35 37 39 40 45 48 50
No comments:
Post a Comment