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Find median in row wise sorted matrix

• Difficulty Level : Hard
• Last Updated : 02 Jan, 2023

We are given a row-wise sorted matrix of size r*c, we need to find the median of the matrix given. It is assumed that r*c is always odd.

Examples:

Input: 1 3 5
2 6 9
3 6 9
Output: Median is 5
If we put all the values in a sorted
array A[] = 1 2 3 3 5 6 6 9 9)

Input: 1 3 4
2 5 6
7 8 9
Output: Median is 5

Simple Method: The simplest method to solve this problem is to store all the elements of the given matrix in an array of size r*c. Then we can either sort the array and find the median element in O(r*clog(r*c)) or we can use the approach discussed here to find the median in O(r*c). Auxiliary space required will be O(r*c) in both cases.

An efficient approach for this problem is to use a binary search algorithm. The idea is that for a number to be median there should be exactly (n/2) numbers that are less than this number. So, we try to find the count of numbers less than all the numbers. Below is the step-by-step algorithm for this approach:

Algorithm:

1. First, we find the minimum and maximum elements in the matrix. The minimum element can be easily found by comparing the first element of each row, and similarly, the maximum element can be found by comparing the last element of each row.
2. Then we use binary search on our range of numbers from minimum to maximum, we find the mid of the min and max and get a count of numbers less than or equal to our mid. And accordingly change the min or max.
3. For a number to be median, there should be (r*c)/2 numbers smaller than that number. So for every number, we get the count of numbers less than or equal to that by using upper_bound() in each row of the matrix, if it is less than the required count, the median must be greater than the selected number, else the median must be less than or equal to the selected number.

Below is the implementation of the above approach:

C++

 // C++ program to find median of a matrix// sorted row wise#includeusing namespace std; const int MAX = 100; // function to find median in the matrixint binaryMedian(int m[][MAX], int r ,int c){    int min = INT_MAX, max = INT_MIN;    for (int i=0; i max)            max = m[i][c-1];    }     int desired = (r * c + 1) / 2;    while (min < max)    {        int mid = min + (max - min) / 2;        int place = 0;         // Find count of elements smaller than or equal to mid        for (int i = 0; i < r; ++i)            place += upper_bound(m[i], m[i]+c, mid) - m[i];        if (place < desired)            min = mid + 1;        else            max = mid;    }    return min;} // driver program to check above functionsint main(){    int r = 3, c = 3;    int m[][MAX]= { {1,3,5}, {2,6,9}, {3,6,9} };    cout << "Median is " << binaryMedian(m, r, c) << endl;    return 0;}

Java

 // Java program to find median of a matrix// sorted row wiseimport java.util.Arrays; public class MedianInRowSorted {    // function to find median in the matrix    static int binaryMedian(int m[][], int r, int c)    {        int max = Integer.MIN_VALUE;        int min = Integer.MAX_VALUE;         for (int i = 0; i < r; i++) {             // Finding the minimum element            if (m[i][0] < min)                min = m[i][0];             // Finding the maximum element            if (m[i] > max)                max = m[i];        }         int desired = (r * c + 1) / 2;        while (min < max) {            int mid = min + (max - min) / 2;            int place = 0;            int get = 0;             // Find count of elements smaller than or equal            // to mid            for (int i = 0; i < r; ++i) {                 get = Arrays.binarySearch(m[i], mid);                 // If element is not found in the array the                // binarySearch() method returns                // (-(insertion_point) - 1). So once we know                // the insertion point we can find elements                // Smaller than the searched element by the                // following calculation                if (get < 0)                    get = Math.abs(get) - 1;                 // If element is found in the array it                // returns the index(any index in case of                // duplicate). So we go to last index of                // element which will give  the number of                // elements smaller than the number                // including the searched element.                else {                    while (get < m[i].length                           && m[i][get] == mid)                        get += 1;                }                 place = place + get;            }             if (place < desired)                min = mid + 1;            else                max = mid;        }        return min;    }     // Driver Program to test above method.    public static void main(String[] args)    {        int r = 3, c = 3;        int m[][]            = { { 1, 3, 5 }, { 2, 6, 9 }, { 3, 6, 9 } };         System.out.println("Median is "                           + binaryMedian(m, r, c));    }} // This code is contributed by Sumit Ghosh

Python3

 # Python program to find median of matrix# sorted row wise from bisect import bisect_right as upper_bound MAX = 100; # Function to find median in the matrixdef binaryMedian(m, r, d):    mi = m[0][0]    mx = 0    for i in range(r):        if m[i][0] < mi:            mi = m[i][0]        if m[i][d-1] > mx :            mx =  m[i][d-1]         desired = (r * d + 1) // 2         while (mi < mx):        mid = mi + (mx - mi) // 2        place = [0];                 # Find count of elements smaller than or equal to mid        for i in range(r):             j = upper_bound(m[i], mid)             place[0] = place[0] + j        if place[0] < desired:            mi = mid + 1        else:            mx = mid    print ("Median is", mi)    return        # Driver coder, d = 3, 3 m = [ [1, 3, 5], [2, 6, 9], [3, 6, 9]]binaryMedian(m, r, d) # This code is contributed by Sachin BIsht

C#

 // C# program to find median// of a matrix sorted row wiseusing System;class MedianInRowSorted{ // Function to find median// in the matrixstatic int binaryMedian(int [,]m,                        int r, int c){  int max = int.MinValue;  int min = int.MaxValue;   for(int i = 0; i < r; i++)  {    // Finding the minimum    // element    if(m[i, 0] < min)      min = m[i, 0];     // Finding the maximum    // element    if(m[i, c - 1] > max)      max = m[i, c - 1];  }   int desired = (r * c + 1) / 2;  while(min < max)  {    int mid = min + (max - min) / 2;    int place = 0;    int get = 0;     // Find count of elements    // smaller than or equal to mid    for(int i = 0; i < r; ++i)    {      get = Array.BinarySearch(            GetRow(m, i), mid);       // If element is not found      // in the array the binarySearch()      // method returns (-(insertion_      // point) - 1). So once we know      // the insertion point we can      // find elements Smaller than      // the searched element by the      // following calculation      if(get < 0)        get = Math.Abs(get) - 1;       // If element is found in the      // array it returns the index(any      // index in case of duplicate). So      // we go to last index of element      // which will give  the number of      // elements smaller than the number      // including the searched element.      else      {        while(get < GetRow(m, i).GetLength(0) &&              m[i, get] == mid)          get += 1;      }       place = place + get;    }     if (place < desired)      min = mid + 1;    else      max = mid;  }  return min;} public static int[] GetRow(int[,] matrix,                           int row){  var rowLength = matrix.GetLength(1);  var rowVector = new int[rowLength];   for (var i = 0; i < rowLength; i++)    rowVector[i] = matrix[row, i];   return rowVector;}   // Driver codepublic static void Main(String[] args){  int r = 3, c = 3;  int [,]m = {{1,3,5},              {2,6,9},              {3,6,9} };   Console.WriteLine("Median is " +                     binaryMedian(m, r, c));}} // This code is contributed by Princi Singh

Javascript



Output

Median is 5

Time Complexity: O(32 * r * log(c)). The upper bound function will take log(c) time and is performed for each row. And since the numbers will be max of 32 bit, so binary search of numbers from min to max will be performed in at most 32 ( log2(2^32) = 32 ) operations.
Auxiliary Space: O(1)

This article is contributed by Akshit Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Method 2 – Using a min heap – Priority Queue
When we need to find the nth smallest element in a row or a column sorted matrix, we can use a min heap and a class node{val,row,col},
Since the matrix is row sorted the smallest element will be one among the first elements of all row ,
so push all first element of rows in the min heap,  then for top element in heap push the second column element in that row ,if it is smallest it will be on top, or the other smallest element will be on top.
Repeat this process of popping the top elements till (r*c)/2.

In this problem this concept is used to find the median of the row sorted matrix.

C++

 #includeusing namespace std; class node{    public:        int data;        int row;        int col;                 node(int d, int r, int c){            row = r;            col = c;            data = d;        }}; class compare{    public:    bool operator()(node* a, node* b){        return a->data > b->data;    }}; class Solution{  public:    int median(vector> &matrix, int R, int C){        priority_queue, compare>minheap;        int count = 0, median = -1;        int medianindex = (R*C)/2;                 for(int i = 0; irow;            int col = top->col;            median = top->data;                         count++;            if(col+1 < C){                col++;                 node* temp = new node(matrix[row][col], row, col);                 minheap.push(temp);            }        }        return median;    } }; int main(){             int r=3;        int c=3;                 vector> matrix={{1, 3, 5},                                    {2, 6, 9},                                    {3, 6, 9}};                Solution obj;        cout<

Time Complexity =  O(log(r*c))- for insertion and deletion in min heap
Auxiliary Space = O((r*c)log(r*c)) – For creating a min heap, with the maximum number of elements equal to r*c/2

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