有序矩阵中查找第k小的元素 Kth smallest element in a row-wise and column-wise sorted 2D array

相关问题：[LeetCode] Merge K Sorted Linked Lists

有序矩阵中查找第k小的元素 Kth smallest element in a row-wise and column-wise sorted 2D array

http://www.geeksforgeeks.org/kth-smallest-element-in-a-row-wise-and-column-wise-sorted-2d-array-set-1/

Given an n x n matrix, where every row and column is sorted in non-decreasing order. Find the kth smallest element in the given 2D array.

For example, consider the following 2D array.

        10, 20, 30, 40        15, 25, 35, 45        24, 29, 37, 48        32, 33, 39, 50The 3rd smallest element is 20 and 7th smallest element is 30

The idea is to use min heap. Following are detailed step.1) Build a min heap of elements from first row. A heap entry also stores row number and column number.2) Do following k times.…a) Get minimum element (or root) from min heap.…b) Find row number and column number of the minimum element.…c) Replace root with the next element from same column and min-heapify the root.3) Return the last extracted root.
Following is C++ implementation of above algorithm.
// kth largest element in a 2d array sorted row-wise and column-wise#include<iostream>#include<climits>using namespace std; // A structure to store an entry of heap.  The entry contains// a value from 2D array, row and column numbers of the valuestruct HeapNode {    int val;  // value to be stored    int r;    // Row number of value in 2D array    int c;    // Column number of value in 2D array}; // A utility function to swap two HeapNode items.void swap(HeapNode *x, HeapNode *y) {    HeapNode z = *x;    *x = *y;    *y = z;} // A utility function to minheapify the node harr[i] of a heap// stored in harr[]void minHeapify(HeapNode harr[], int i, int heap_size){    int l = i*2 + 1;    int r = i*2 + 2;    int smallest = i;    if (l < heap_size && harr[l].val < harr[i].val)        smallest = l;    if (r < heap_size && harr[r].val < harr[smallest].val)        smallest = r;    if (smallest != i)    {        swap(&harr[i], &harr[smallest]);        minHeapify(harr, smallest, heap_size);    }} // A utility function to convert harr[] to a max heapvoid buildHeap(HeapNode harr[], int n){    int i = (n - 1)/2;    while (i >= 0)    {        minHeapify(harr, i, n);        i--;    }} // This function returns kth smallest element in a 2D array mat[][]int kthSmallest(int mat[4][4], int n, int k){    // k must be greater than 0 and smaller than n*n    if (k <= 0 || k > n*n)       return INT_MAX;     // Create a min heap of elements from first row of 2D array    HeapNode harr[n];    for (int i = 0; i < n; i++)        harr[i] =  {mat[0][i], 0, i};    buildHeap(harr, n);     HeapNode hr;    for (int i = 0; i < k; i++)    {       // Get current heap root       hr = harr[0];        // Get next value from column of root's value. If the       // value stored at root was last value in its column,       // then assign INFINITE as next value       int nextval = (hr.r < (n-1))? mat[hr.r + 1][hr.c]: INT_MAX;        // Update heap root with next value       harr[0] =  {nextval, (hr.r) + 1, hr.c};        // Heapify root       minHeapify(harr, 0, n);    }     // Return the value at last extracted root    return hr.val;} // driver program to test above functionint main(){  int mat[4][4] = { {10, 20, 30, 40},                    {15, 25, 35, 45},                    {25, 29, 37, 48},                    {32, 33, 39, 50},                  };  cout << "7th smallest element is " << kthSmallest(mat, 4, 7);  return 0;}
Output:
7th smallest element is 30
Time Complexity: The above solution involves following steps.
1) Build a min heap which takes O(n) time
2) Heapify k times which takes O(kLogn) time.
Therefore, overall time complexity is O(n + kLogn) time.
The above code can be optimized to build a heap of size k when k is smaller than n. In that case, the kth smallest element must be in first k rows and k columns.