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

We strongly recommend to minimize the browser and try this yourself first.

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 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;
}

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





  • Debabrata

    why we cant apply simple merge technique?

    #define ROW 4
    #define COL 4
    void merge(int arr[ROW][COL],int *sa){
    int tmpArr[ROW*COL];
    int i_end = COL;
    int j_end = COL;
    int i,ith_row ;
    int j ,k;
    for(ith_row=0;ith_row<COL;ith_row++){
    sa[ith_row] = arr[0][ith_row];
    }
    for(ith_row =1 ; ith_row<ROW;ith_row++){
    i=0;j=0;
    for(k=0;k<i_end+j_end;k++){
    if((i=j_end)|| sa[i] < arr[ith_row][j])){
    tmpArr[k] = sa[i++];
    }else{
    tmpArr[k] = arr[ith_row][j++];
    }

    }
    for(k=0;k<i_end+j_end;k++){
    sa[k] = tmpArr[k];
    }
    i_end += COL;

    }

    }
    void print_1D_array(int *arr,size_t size){
    for(size_t i=0;i<size-1;i++){
    std::cout<< arr[i]<<"-";
    }
    }
    int main(){

    int arr[ROW][COL] ={ {1,6,9,15},
    {12,14,20,30},
    {3,6,9,10},
    {40,44,46,50}};

    int sorted_array[ROW * COL];
    merge(arr,sorted_array);
    print_1D_array(sorted_array, ROW * COL);
    return 0;
    }

  • Adarsh Singh

    Using PriorityQueue Library in Java

    import java.util.*;

    class pair
    {
    X _data ;
    Y _row;
    Z _column;

    pair(X x , Y y , Z z){_data=x; _row=y; _column=z;}

    X data(){return _data;}
    Y row(){return _row;}
    Z column(){return _column;}

    void setData(X x){_data=x;}
    void setRow(Y y){_row=y;}
    void setColumn(Z z){_column=z;}
    }

    public class KthSmallestNumberInRow_ColumnWiseSortedMatrix {

    public static void main(String[]args)
    {
    int a[][]={ {10, 20, 30, 40},{15, 25, 35, 45},{24, 29, 37, 48},{32, 33, 39, 50} };
    getKthSmallestElement(a);

    }

    private static void getKthSmallestElement(int[][] a) {

    PriorityQueue<pair> pq=new PriorityQueue<pair>(1,
    new Comparator<pair>(){
    public int compare(pair t, pair t1) {
    return t.data()-t1.data();
    }
    }
    );

    for(int i=0;i<4;i++) // number of column
    {
    pq.offer(new pair(a[0][i],0,i));
    }

    int count=4; // 4th smallest
    int answer=0;
    for(int c=1;c<=count;c++)
    {
    pair res = pq.poll();
    answer=res.data();
    // System.out.println(answer);
    int row=res.row();
    int col=res.column();
    if(row+1<4)pq.offer(new pair(a[row+1][col],row+1,col));

    }

    System.out.println(answer);
    }
    }

  • Adarsh Singh

    Using PriorityQueue Library in Java

    import java.util.*;

    class pair
    {
    X _data ;
    Y _row;
    Z _column;

    pair(X x , Y y , Z z){_data=x; _row=y; _column=z;}

    X data(){return _data;}
    Y row(){return _row;}
    Z column(){return _column;}

    void setData(X x){_data=x;}
    void setRow(Y y){_row=y;}
    void setColumn(Z z){_column=z;}
    }

    public class KthSmallestNumberInRow_ColumnWiseSortedMatrix {

    public static void main(String[]args)
    {
    int a[][]={ {10, 20, 30, 40},{15, 25, 35, 45},{24, 29, 37, 48},{32, 33, 39, 50} };
    getKthSmallestElement(a);

    }

    private static void getKthSmallestElement(int[][] a) {

    PriorityQueue<pair> pq=new PriorityQueue<pair>(1,
    new Comparator<pair>(){
    public int compare(pair t, pair t1) {
    return t.data()-t1.data();
    }
    }
    );

    for(int i=0;i<4;i++) // number of column
    {
    pq.offer(new pair(a[0][i],0,i));
    }

    int count=4; // 4th smallest
    int answer=0;
    for(int c=1;c<=count;c++)
    {
    pair res = pq.poll();
    answer=res.data();
    // System.out.println(answer);
    int row=res.row();
    int col=res.column();
    if(row+1<4)pq.offer(new pair(a[row+1][col],row+1,col));

    }

    System.out.println(answer);
    }
    }

  • novice

    Do we really need a heap ? Why not

    row = 1
    col = 1
    count = 0

    while (count <= k ) {
    // checks to ensure row , col < n omitted for simplicity
    if (a[row][col+1] n)
    ++row
    }
    else {
    ++row
    if ( row > n)
    ++col
    }

    ++count
    }

    return a[row][col]

    o(k) complexity

    • Mearaj

      I guess you have a bug here, you are checking only right side and down , rather you should check all the corners. Here after 24 , you shd move to 25 not 29.

      10, 20, 30, 40
      |
      15, 25, 35, 45
      |
      24,–> 29 –> 37, 48

      32, 33, 39, 50

      • novice

        You are right. But my main point is that a heap might not be necessary.

    • Maulish Soni

      How about using RANDOM-SELECTION algo?

  • sunil

    But this is as good as sorting n sorted arrays i.e you are not making use of the fact that they are sorted column wise as well.

  • gg

    can we do it by using k-way merge?

    • pulkit mehra

      We can, but the time complexity would be more I guess, as we would have to process all the rows( we would have to merge the row elements) instead of processing n+k elements in the worst case for the above given method.

  • Chenzhan Liu

    Instead of adding items into heap row-wisely, I suppose the following strategy might be more efficient:
    (1)pop out the top value and get its row and col
    (2)add matrix[row+1][col] and matrix[row][col+1] into heap, if they exist and have not been in heap

    • pulkit mehra

      Can you explain as to how will it be more effective ?

    • Ankul

      +1 on above soln. We can start the heap with empty heap each time pushing these 2 elements into it after the min is popped. This removes the extra overhead of O(n) or O(k).

    • https://www.facebook.com/aditya.pearl7 Aditya Gaurav

      in this method the maximum size of heap at any point of time can O(n-1 + n-1) & insertion and extraction complexity is O(log (size of heap)) and hence the worst case complexity is O(k*log(n));

  • Ahmed Saleh

    I would just converted the 2D Array to 1D, sort it, then get kth element.

    • omar salem

      that’s N^2 right there

    • Shivam Goel

      n^2 (logn)

      • Ahmed Saleh

        Converting 2D array to 1D is O(N^2), sort it is O(NLogN).