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, 50
The 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 the implementation of above algorithm.

C++

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// 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 value
struct 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 heap
void 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 function
int 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;
}

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Python3

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# Program for kth largest element in a 2d array 
# sorted row-wise and column-wise
from sys import maxsize
  
# A structure to store an entry of heap. 
# The entry contains a value from 2D array,
# row and column numbers of the value
class HeapNode:
    def __init__(self, val, r, c):
        self.val = val # value to be stored
        self.r = r # Row number of value in 2D array
        self.c = c # Column number of value in 2D array
  
# A utility function to minheapify the node harr[i] 
# of a heap stored in harr[]
def minHeapify(harr, i, heap_size):
    l = i * 2 + 1
    r = i * 2 + 2
    smallest = i
    if l < heap_size and harr[l].val < harr[i].val:
        smallest = l
    if r < heap_size and harr[r].val < harr[smallest].val:
        smallest = r
  
    if smallest != i:
        harr[i], harr[smallest] = harr[smallest], harr[i]
        minHeapify(harr, smallest, heap_size)
  
# A utility function to convert harr[] to a max heap
def buildHeap(harr, n):
    i = (n - 1) // 2
    while i >= 0:
        minHeapify(harr, i, n)
        i -= 1
  
# This function returns kth smallest element
# in a 2D array mat[][]
def kthSmallest(mat, n, k):
  
    # k must be greater than 0 and smaller than n*n
    if k <= 0 or k > n * n:
        return maxsize
  
    # Create a min heap of elements from
    # first row of 2D array
    harr = [0] * n
    for i in range(n):
        harr[i] = HeapNode(mat[0][i], 0, i)
    buildHeap(harr, n)
  
    hr = HeapNode(0, 0, 0)
    for i in range(k):
  
        # 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
        nextval = mat[hr.r + 1][hr.c] if (hr.r < n - 1) else maxsize
  
        # Update heap root with next value
        harr[0] = HeapNode(nextval, hr.r + 1, hr.c)
  
        # Heapify root
        minHeapify(harr, 0, n)
  
    # Return the value at last extracted root
    return hr.val
  
# Driver Code
if __name__ == "__main__":
    mat = [[10, 20, 30, 40], 
           [15, 25, 35, 45], 
           [25, 29, 37, 48],
           [32, 33, 39, 50]]
    print("7th smallest element is"
             kthSmallest(mat, 4, 7))
  
# This code is contributed by
# sanjeev2552

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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.

We will soon be publishing more efficient algorithms for finding the kth smallest element.

This article is compiled by Ravi Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : sanjeev2552