Find median in row wise sorted matrix
Last Updated :
23 Jul, 2025
Given a row-wise sorted matrix mat[][] of order n * m, where the number of rows and columns are always odd. The task is to find the median of the matrix.
Note: Median is the middle number in a sorted ascending or descending list of numbers. In case of an even number of elements return the left median.
Examples:
Input: mat[][] = [[1 3 5],
[2 6 9],
[3 6 9]]
Output: 5
Explanation: Elements in sorted order: 1 2 3 3 5 6 6 9 9. There are total 9 elements, thus the median is the element at index 5 (1-based) i.e. 5.
Input: mat[][] = [[1 3 4],
[2 5 6]
[3 7 9]]
Output: 4
Explanation: Elements in sorted order: 1 2 3 3 4 5 6 7 9. There are total 9 elements, thus the median is the element at index 5 (1-based) i.e. 4.
The idea is to store all the elements of the given matrix in a new array of size r x c. Then sort the array and find the median element.
Below is given the implementation:
C++
// C++ program to find median of a matrix
// Using Extra Space
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int median(vector<vector<int>> &mat) {
// Flatten the matrix into a 1D array
vector<int> arr;
for (int i = 0; i < mat.size(); i++) {
for (int j = 0; j < mat[0].size(); j++) {
arr.push_back(mat[i][j]);
}
}
// Sort the array
sort(arr.begin(), arr.end());
// Find the median element
int mid = arr.size() / 2;
return arr[mid];
}
int main() {
vector<vector<int>> matrix = {{1, 3, 5}, {2, 6, 9}, {3, 6, 9}};
cout << median(matrix) << endl;
return 0;
}
// Java program to find median of a matrix
// Using Extra Space
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
class GfG {
static int median(int mat[][]) {
// Flatten the matrix into a 1D array
List<Integer> list = new ArrayList<>();
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[0].length; j++) {
list.add(mat[i][j]);
}
}
// Sort the list
Collections.sort(list);
// Find and return the median element
int mid = list.size() / 2;
return list.get(mid);
}
public static void main(String[] args) {
int[][] matrix = {
{1, 3, 5},
{2, 6, 9},
{3, 6, 9}
};
System.out.println(median(matrix));
}
}
# python3 program to find median of a matrix
# Using Extra Space
def median(mat):
# Flatten the matrix into a 1D array
arr = []
for i in range(len(mat)):
for j in range(len(mat[0])):
arr.append(mat[i][j])
# Sort the array
arr.sort()
# Find the median element
mid = len(arr) // 2
return arr[mid]
if __name__ == "__main__":
mat = [[1, 3, 5],
[2, 6, 9],
[3, 6, 9]]
print(median(mat))
// C# program to find median of a matrix
// Using Extra Space
using System;
using System.Collections.Generic;
using System.Linq;
class GfG {
static int median(int[][] mat) {
// Flatten the matrix into a 1D list
List<int> arr = new List<int>();
for (int i = 0; i < mat.Length; i++) {
for (int j = 0; j < mat[i].Length; j++) {
arr.Add(mat[i][j]);
}
}
// Sort the array
arr.Sort();
// Find the median element
int mid = arr.Count / 2;
return arr[mid];
}
static void Main(string[] args) {
int[][] mat = new int[][] { new int[] { 1, 3, 5 },
new int[] { 2, 6, 9 },
new int[] { 3, 6, 9 } };
Console.WriteLine(median(mat));
}
}
// JavaScript program to find median of a matrix
// Using Extra Space
function median(mat) {
// Flatten the matrix into a 1D array
const arr = mat.reduce((acc, row) => acc.concat(row), []);
// Sort the array
arr.sort((a, b) => a - b);
// Find the median element
const mid = Math.floor(arr.length / 2);
return arr[mid];
}
// Driver Code
const mat = [
[1, 3, 5],
[2, 6, 9],
[3, 6, 9]
];
console.log(median(mat));
Time complexity: O(n * m * log(n * m))
Auxiliary Space: O(n * m)
[Better Approach] - Using Priority Queue - O(n * m * log(n)) Time and O(n) Space
The idea is to use min heap to find the first (n * m) / 2 smallest element and print the element at index (n * m) / 2. To do so, create a min heap using priority queue, where each element is an array that stores, mat[i][j] and its indices i.e. i & j. Now, firstly push elements of 0th column of each row in the queue. Also, declare a variable cnt to store the count of elements selected. Run a while loop until the cnt is less than or equal to (n * m) / 2, and in each iteration take out the top element of queue, storing the smallest element, let it be mat[row][col], increment cnt by 1 and push the mat[row][col+1] in the queue, which is the next greater element. After the loops end, return the last element stored in queue.
Below is given the implementation:
C++
// C++ program to find median of a matrix
// using Priority Queue
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int median(vector<vector<int>> &mat) {
int r = mat.size();
int c = mat[0].size();
priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>,
greater<pair<int, pair<int, int>>>> minHeap;
int cnt = 0, res = -1;
// calculate median index
int mid = (r * c) / 2;
// Push the first element of each row into the min-heap
for (int i = 0; i < r; i++) {
minHeap.push({mat[i][0], {i, 0}});
}
// Extract elements from the min-heap to find the median
while (cnt <= mid) {
auto top = minHeap.top();
minHeap.pop();
int row = top.second.first;
int col = top.second.second;
res = top.first;
cnt++;
// If there are more elements in the row, push the next element
if (col + 1 < c) {
minHeap.push({mat[row][col + 1], {row, col + 1}});
}
}
return res;
}
int main() {
vector<vector<int>> matrix = {{1, 3, 5}, {2, 6, 9}, {3, 6, 9}};
cout << median(matrix);
return 0;
}
// Java program to find median of a matrix
// using Priority Queue
import java.util.Comparator;
import java.util.PriorityQueue;
class GfG {
static int median(int[][] mat) {
int r = mat.length;
int c = mat[0].length;
// Min-Heap to store elements
PriorityQueue<int[]> minHeap = new PriorityQueue<>(
(a, b) -> Integer.compare(a[0], b[0])
);
int cnt = 0, res = -1;
// Calculate median index
int mid = (r * c) / 2;
// Push the first element of each row into the min-heap
for (int i = 0; i < r; i++) {
minHeap.offer(new int[]{mat[i][0], i, 0});
}
// Extract elements from the min-heap to find the median
while (cnt <= mid) {
int[] top = minHeap.poll();
int row = top[1];
int col = top[2];
res = top[0];
cnt++;
// If there are more elements in the row, push the next element
if (col + 1 < c) {
minHeap.offer(new int[]{mat[row][col + 1], row, col + 1});
}
}
return res;
}
public static void main(String[] args) {
int[][] mat = {{1, 3, 5}, {2, 6, 9}, {3, 6, 9}};
System.out.println(median(mat));
}
}
# python3 program to find median of a matrix
# Using Priority queue
import heapq
def median(mat):
r = len(mat)
c = len(mat[0])
# Min-Heap to store elements
minHeap = []
cnt = 0
res = -1
# Calculate median index
mid = (r * c) // 2
# Push the first element of each row into the min-heap
for i in range(r):
heapq.heappush(minHeap, (mat[i][0], i, 0))
# Extract elements from the min-heap to find the median
while cnt <= mid:
val, row, col = heapq.heappop(minHeap)
res = val
cnt += 1
# If there are more elements in the row, push the next element
if col + 1 < c:
heapq.heappush(minHeap, (mat[row][col + 1], row, col + 1))
return res
if __name__ == "__main__":
matrix = [[1, 3, 5], [2, 6, 9], [3, 6, 9]]
print(median(matrix))
// Javascript program to find median of a matrix
// Using Priority queue
class PriorityQueue {
constructor() {
this.heap = [];
}
enqueue(value, row, col) {
this.heap.push({ value, row, col });
// Min-heap based on value
this.heap.sort((a, b) => a.value - b.value);
}
dequeue() {
// Remove and return the smallest element
return this.heap.shift();
}
isEmpty() {
return this.heap.length === 0;
}
}
function median(mat) {
let r = mat.length;
let c = mat[0].length;
let minHeap = new PriorityQueue();
let cnt = 0;
let res = -1;
// Calculate the median index
let mid = Math.floor((r * c) / 2);
// Push the first element of each row into the priority queue
for (let i = 0; i < r; i++) {
minHeap.enqueue(mat[i][0], i, 0);
}
// Extract elements from the priority queue to find the median
while (cnt <= mid) {
let { value, row, col } = minHeap.dequeue();
res = value;
cnt++;
// If there are more elements in the row, push the next element
if (col + 1 < c) {
minHeap.enqueue(mat[row][col + 1], row, col + 1);
}
}
return res;
}
// Driver code
let mat = [
[1, 3, 5],
[2, 6, 9],
[3, 6, 9]
];
console.log(median(mat));
Time complexity: O(( n * m)/2 * log(n) ), since we are going to insert and remove the top element in priority queue (n * m)/2 times and every time the priority queue will be having elements in it.
Auxiliary Space: O(n), the min-heap stores at most one element per row of the matrix at any point in time.
[Expected Approach] - Using Binary Search - O(n * log(m) * log(maxVal - minVal)) Time and O(1) Space
An efficient approach for this problem is to use binary search algorithm. The idea is that for a number x to be median there should be exactly r * c / 2 numbers that are less than this number. So, we apply binary search on search space [minimum Element to maximum Element] and try to find the count of numbers less than mid. If this count is less than r * c / 2 then search in upper half to increase the count, otherwise search in lower half to decrease the count.
Below is given the implementation:
C++
// C++ program to find median of a matrix
// using binary search
#include <algorithm>
#include <iostream>
#include <climits>
#include <vector>
using namespace std;
// Function to find the median in the matrix
int median(vector<vector<int>> &mat) {
int r = mat.size();
int c = mat[0].size();
int minVal = INT_MAX, maxVal = INT_MIN;
// Finding the minimum and maximum elements in the matrix
for (int i = 0; i < r; i++) {
if (mat[i][0] < minVal)
minVal = mat[i][0];
if (mat[i][c - 1] > maxVal)
maxVal = mat[i][c - 1];
}
int desired = (r * c + 1) / 2;
int lo = minVal, hi = maxVal;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
int place = 0;
// Count elements smaller than or equal to mid
for (int i = 0; i < r; ++i)
place += upper_bound(mat[i].begin(), mat[i].end(), mid)
- mat[i].begin();
// Adjust the range based on the count of elements found
if (place < desired)
lo = mid + 1;
else
hi = mid;
}
return lo;
}
int main() {
vector<vector<int>> mat = {{1, 3, 5}, {2, 6, 9}, {3, 6, 9}};
cout << median(mat) << endl;
return 0;
}
// Java program to find median of a matrix
// using binary search
import java.util.Arrays;
class GfG {
// Function to find median in the matrix using binary search
static int median(int mat[][]) {
int n = mat.length;
int m = mat[0].length;
// Initializing the minimum and maximum values
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
// Iterating through each row of the matrix
for (int i = 0; i < n; i++) {
// Updating the minimum value if current element is smaller
if (mat[i][0] < min)
min = mat[i][0];
// Updating the maximum value if current element is larger
if (mat[i][m - 1] > max)
max = mat[i][m - 1];
}
// Calculating the desired position of the median
int desired = (n * m + 1) / 2;
// Using binary search to find the median value
while (min < max) {
// Calculating the middle value
int mid = min + (max - min) / 2;
// Counting the number of elements less than or equal to mid
int place = 0;
for (int i = 0; i < n; i++) {
place += upperBound(mat[i], mid);
}
// Updating the search range based on the count
if (place < desired) {
min = mid + 1;
} else {
max = mid;
}
}
// Returning the median value
return min;
}
// Helper function to find the upper bound of a number in a row
static int upperBound(int[] row, int num) {
int low = 0, high = row.length;
while (low < high) {
int mid = low + (high - low) / 2;
if (row[mid] <= num) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
public static void main(String[] args) {
int mat[][] = { { 1, 3, 5 },
{ 2, 6, 9 },
{ 3, 6, 9 }};
System.out.println(median(mat));
}
}
# python3 program to find median of a matrix
# using binary search
from bisect import bisect_right
def median(mat):
r = len(mat)
c = len(mat[0])
minVal = float('inf')
maxVal = float('-inf')
# Finding the minimum and maximum elements in the matrix
for i in range(r):
minVal = min(minVal, mat[i][0])
maxVal = max(maxVal, mat[i][c - 1])
desired = (r * c + 1) // 2
lo = minVal
hi = maxVal
# Binary search to find the median
while lo < hi:
mid = lo + (hi - lo) // 2
place = 0
# Count elements smaller than or equal to mid using bisect_right
for i in range(r):
place += bisect_right(mat[i], mid)
if place < desired:
lo = mid + 1
else:
hi = mid
return lo
if __name__ == "__main__":
mat = [[1, 3, 5], [2, 6, 9], [3, 6, 9]]
print(median(mat))
// C# program to find median of a matrix
// using binary search
using System;
class GfG {
// Function to find median in the matrix using binary search
static int median(int[][] mat) {
int n = mat.Length;
int m = mat[0].Length;
// Initializing the minimum and maximum values
int min = int.MaxValue, max = int.MinValue;
// Iterating through each row of the matrix
for (int i = 0; i < n; i++) {
// Updating the minimum value if current element is smaller
if (mat[i][0] < min)
min = mat[i][0];
// Updating the maximum value if current element is larger
if (mat[i][m - 1] > max)
max = mat[i][m - 1];
}
// Calculating the desired position of the median
int desired = (n * m + 1) / 2;
// Using binary search to find the median value
int lo = min, hi = max;
while (lo < hi)
{
// Calculating the middle value
int mid = lo + (hi - lo) / 2;
// Counting the number of elements less than or equal to mid
int place = 0;
for (int i = 0; i < n; i++)
place += upperBound(mat[i], mid);
// Updating the search range based on the count
if (place < desired)
lo = mid + 1;
else
hi = mid;
}
// Returning the median value
return lo;
}
// Helper function to find the upper bound of a number in a row
static int upperBound(int[] row, int num) {
int lo = 0, hi = row.Length;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
if (row[mid] <= num)
lo = mid + 1;
else
hi = mid;
}
return lo;
}
static void Main(string[] args) {
int[][] mat = new int[][] {
new int[] { 1, 3, 5 },
new int[] { 2, 6, 9 },
new int[] { 3, 6, 9 }
};
Console.WriteLine(median(mat));
}
}
// JavaScript program to find median of a matrix
// using binary search
// Function to find median in the matrix using binary search
function median(mat) {
const n = mat.length;
const m = mat[0].length;
// Initializing the minimum and maximum values
let min = Number.MAX_VALUE, max = Number.MIN_VALUE;
// Iterating through each row of the matrix
for (let i = 0; i < n; i++) {
// Updating the minimum value if current element is smaller
if (mat[i][0] < min)
min = mat[i][0];
// Updating the maximum value if current element is larger
if (mat[i][m - 1] > max)
max = mat[i][m - 1];
}
// Calculating the desired position of the median
const desired = Math.floor((n * m + 1) / 2);
// Using binary search to find the median value
let lo = min, hi = max;
while (lo < hi) {
// Calculating the middle value
const mid = Math.floor(lo + (hi - lo) / 2);
// Counting the number of elements less than or equal to mid
let place = 0;
for (let i = 0; i < n; i++) {
place += upperBound(mat[i], mid);
}
// Updating the search range based on the count
if (place < desired) {
lo = mid + 1;
} else {
hi = mid;
}
}
// Returning the median value
return lo;
}
// Helper function to find the upper bound of a number in a row
function upperBound(row, num) {
let lo = 0, hi = row.length;
while (lo < hi) {
const mid = Math.floor(lo + (hi - lo) / 2);
if (row[mid] <= num) {
lo = mid + 1;
} else {
hi = mid;
}
}
return lo;
}
// Driver Code
const mat = [
[1, 3, 5],
[2, 6, 9],
[3, 6, 9]
];
console.log(median(mat));
Time Complexity: O(n * log(m) * log(maxVal - minVal)), the upper bound function will take log(m) time and is performed for each row. And binary search is performed from minVal to maxVal.
Auxiliary Space: O(1)
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