Kth smallest element in a row-wise and column-wise sorted 2D array
Last Updated :
13 Aug, 2025
Given an n × n matrix mat[][] where each row and column is sorted in non-decreasing order, find the kth smallest element, where k lies in the range [1, n²].
Example:
Input: mat[][] = [[10, 20, 30, 40], k = 3
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]]
Output: 20
Explanation: The sorted order is [10, 15, 20, ...]; the 3rd element is 20.
Input: mat[][] = [[10, 20, 30, 40], k = 7
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]]
Output: 30
Explanation: The sorted order is [10, 15, 20, 24, 25, 29, 30, ...]; the 7th element is 30.
[Naive Approach] Using Sorting - O(n2 × log n2) Time and O(n) Space
Initialize a 1-dimensional array of size n*n to store all the elements of the mat[][] , we will get our kth minimum element by sorting the 1-dimensional array in non-decreasing order.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int kthSmallest(vector<vector<int>>& mat, int k) {
int n = mat.size();
// create a vector to store all elements
vector<int> arr;
// store all elements of the mat
// into the array
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
arr.push_back(mat[i][j]);
}
}
// sort the array
sort(arr.begin(), arr.end());
// return the kth smallest element
// (0-based index, hence k-1)
return arr[k - 1];
}
int main() {
vector<vector<int>> mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50 }};
int k = 3;
int result = kthSmallest(mat, k);
cout << result << endl;
return 0;
}
import java.util.ArrayList;
import java.util.Collections;
class GfG {
static int kthSmallest(int[][] mat, int k) {
int n = mat.length;
// create an ArrayList to store all elements
ArrayList<Integer> arr = new ArrayList<Integer>();
// store all elements of the mat into the array
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
arr.add(mat[i][j]);
}
}
// sort the array
Collections.sort(arr);
// return the kth smallest element
// (0-based index, hence k-1)
return arr.get(k - 1);
}
public static void main(String[] args) {
int[][] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
System.out.println(result);
}
}
def kthSmallest(mat, k):
n = len(mat)
# create a list to store all elements
arr = []
# store all elements of the mat into the array
for i in range(n):
for j in range(n):
arr.append(mat[i][j])
# sort the array
arr.sort()
# return the kth smallest element
# (0-based index, hence k-1)
return arr[k - 1]
if __name__ == "__main__":
mat = [
[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]
]
k = 3
result = kthSmallest(mat, k)
print(result)
using System;
using System.Collections.Generic;
class GfG {
// function to find the kth smallest
// element in a sorted 2D mat
static int kthSmallest(int[,] mat, int k) {
int n = mat.GetLength(0);
// create a List to store all elements
List<int> arr = new List<int>();
// store all elements of the mat
// into the array
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
arr.Add(mat[i,j]);
}
}
// sort the array
arr.Sort();
// return the kth smallest element
// (0-based index, hence k-1)
return arr[k - 1];
}
public static void Main(string[] args) {
int[,] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
Console.WriteLine(result);
}
}
function kthSmallest(mat, k) {
let n = mat.length;
// create an array to store all elements
let arr = [];
// store all elements of the mat
// into the array
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
arr.push(mat[i][j]);
}
}
// sort the array
arr.sort((a, b) => a - b);
// return the kth smallest element
// (0-based index, hence k-1)
return arr[k - 1];
}
// Driver Code
let mat = [[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]];
let k = 3;
let result = kthSmallest(mat, k);
console.log(result);
[Better Approach] Using Priority Queue - O(n2 × log k) Time and O(k) Space
The idea is to use a max-heap to store and maintain the track of k smallest elements in the heap. If the size of the heap exceeds more than k while inserting the elements , we will pop the top element from max-heap so as to maintain the size of k elements. After successful traversal in mat[][], the top element of the max-heap will be the kth minimum element.
C++
#include <iostream>
#include <vector>
#include <queue> // for priority_queue
using namespace std;
int kthSmallest(vector<vector<int>>& mat, int k) {
int n = mat.size();
priority_queue<int> pq; // max-heap
// traverse all elements in the mat
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int curr = mat[i][j];
// push the current element into the max-heap
pq.push(curr);
// if size exceeds k, remove the largest
if (pq.size() > k) {
pq.pop();
}
}
}
// top of the heap is the kth smallest element
return pq.top();
}
int main() {
vector<vector<int>> mat = {
{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}
};
int k = 3;
int result = kthSmallest(mat, k);
cout << result << endl;
return 0;
}
import java.util.*;
class GfG {
static int kthSmallest(int[][] mat, int k) {
int n = mat.length;
PriorityQueue<Integer> pq =
new PriorityQueue<>(Collections.reverseOrder());
// traverse all elements in the mat
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int curr = mat[i][j];
// push the current element into the max-heap
pq.offer(curr);
// if the size of the max-heap exceeds k,
// remove the largest element
if (pq.size() > k) {
pq.poll();
}
}
}
// the top element of the max-heap
// is the kth smallest element
return pq.peek();
}
public static void main(String[] args) {
int[][] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
System.out.println(result);
}
}
import heapq
def kthSmallest(mat, k):
n = len(mat)
pq = []
# traverse all elements in the mat
for i in range(n):
for j in range(n):
curr = mat[i][j]
# push the current element into the max-heap
heapq.heappush(pq, -curr)
# if the size of the max-heap exceeds k,
# remove the largest element
if len(pq) > k:
heapq.heappop(pq)
# the top element of the max-heap
# is the kth smallest element
return -pq[0]
if __name__ == "__main__":
mat = [
[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]]
k = 3
result = kthSmallest(mat, k)
print(result)
using System;
using System.Collections.Generic;
class GfG {
static int kthSmallest(int[,] mat, int k) {
int n = mat.GetLength(0);
PriorityQueue<int> pq =
new PriorityQueue<int>
(Comparer<int>.Create((a, b) => b.CompareTo(a)));
// traverse all elements in the mat
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int curr = mat[i,j];
// push the current element into the max-heap
pq.Enqueue(curr);
// if the size of the max-heap exceeds k,
// remove the largest element
if (pq.Count > k) {
pq.Dequeue();
}
}
}
// the top element of the max-heap
// is the kth smallest element
return pq.Peek();
}
public static void Main(string[] args) {
int[,] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
Console.WriteLine(result);
}
}
public class PriorityQueue<T> {
private List<T> data;
private IComparer<T> comparer;
public PriorityQueue(IComparer<T> comparer) {
this.data = new List<T>();
this.comparer = comparer;
}
public void Enqueue(T item) {
data.Add(item);
int ci = data.Count - 1;
while (ci > 0) {
int pi = (ci - 1) / 2;
if (comparer.Compare(data[ci], data[pi]) >= 0) break;
T tmp = data[ci]; data[ci] = data[pi]; data[pi] = tmp;
ci = pi;
}
}
public T Dequeue() {
int li = data.Count - 1;
T frontItem = data[0];
data[0] = data[li];
data.RemoveAt(li);
--li;
int pi = 0;
while (true) {
int ci = pi * 2 + 1;
if (ci > li) break;
int rc = ci + 1;
if (rc <= li && comparer.Compare(data[rc], data[ci]) < 0)
ci = rc;
if (comparer.Compare(data[pi], data[ci]) <= 0) break;
T tmp = data[pi]; data[pi] = data[ci]; data[ci] = tmp;
pi = ci;
}
return frontItem;
}
public T Peek() {
return data[0];
}
public int Count {
get { return data.Count; }
}
}
class MaxHeap {
constructor() {
this.heap = [];
}
push(val) {
this.heap.push(val);
this.bubbleUp(this.heap.length - 1);
}
pop() {
const max = this.heap[0];
const end = this.heap.pop();
if (this.heap.length > 0) {
this.heap[0] = end;
this.bubbleDown(0);
}
return max;
}
peek() {
return this.heap[0];
}
size() {
return this.heap.length;
}
bubbleUp(idx) {
const element = this.heap[idx];
while (idx > 0) {
const parentIdx = Math.floor((idx - 1) / 2);
const parent = this.heap[parentIdx];
if (element <= parent) break;
this.heap[idx] = parent;
this.heap[parentIdx] = element;
idx = parentIdx;
}
}
bubbleDown(idx) {
const length = this.heap.length;
const element = this.heap[idx];
while (true) {
const leftChildIdx = 2 * idx + 1;
const rightChildIdx = 2 * idx + 2;
let leftChild, rightChild;
let swap = null;
if (leftChildIdx < length) {
leftChild = this.heap[leftChildIdx];
if (leftChild > element) {
swap = leftChildIdx;
}
}
if (rightChildIdx < length) {
rightChild = this.heap[rightChildIdx];
if (
(swap === null && rightChild > element) ||
(swap !== null && rightChild > leftChild)
) {
swap = rightChildIdx;
}
}
if (swap === null) break;
this.heap[idx] = this.heap[swap];
this.heap[swap] = element;
idx = swap;
}
}
}
// function to find the kth smallest
// element in a sorted 2D mat
function kthSmallest(mat, k) {
const n = mat.length;
const pq = new MaxHeap();
// traverse all elements in the mat
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
const curr = mat[i][j];
// push the current element into the max-heap
pq.push(curr);
// if the size of the max-heap exceeds k,
// remove the largest element
if (pq.size() > k) {
pq.pop();
}
}
}
// the top element of the max-heap
// is the kth smallest element
return pq.peek();
}
const mat =
[[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]];
const k = 3;
const result = kthSmallest(mat, k);
console.log(result);
[Efficient Approach] Binary Search on Answer
This approach uses binary search to iterate over possible solutions. As answer lies in the range from mat[0][0] to mat[n-1][n-1], So we do a binary search on this range and in each iteration determine the no of elements smaller than or equal to our current middle element.
Step by Step Implementation:
- Initialize a variable, say low equals to the mat[0][0] (minimum value of matrix).
- Initialize a variable, say high equals to the mat[n-1][n-1] (maximum value of matrix).
- Initialize ans to 0.
- Perform Binary Search on the range from low to high:
=> Calculate the midpoint in the range say mid.
=> If the countSmallerEqual(function which will return the count of elements less than or equal to mid) is less than k, update low to mid+ 1.
=> if the returned value is greater or equal to k , this can be our possible ans. So, update ans to mid and narrow the search range by setting high to mid - 1. - countSmallerEqual (helper function that counts the number of elements in the matrix less than or equal to the given mid.)
=> initialize a pointer, say row and col points to 0 and n-1 respectively. And a variable count = 0.
=> if the mat[row][col] is greater than mid, move left in the matrix by decrementing col.
=> if the mat[row][col] is less than or equal to mid, increment the count as by col + 1 and move down in the matrix by incrementing row.
C++
#include <iostream>
#include <vector>
using namespace std;
// function to count the number of elements
// less than or equal to x
int countSmallerEqual(vector<vector<int>>& mat, int x) {
int n = mat.size();
int count = 0;
int row = 0;
int col = n - 1;
// traverse from the top-right corner
while (row < n && col >= 0) {
if (mat[row][col] <= x) {
// if current element is less than
// or equal to x, all elements in this
// row up to the current column are <= x
count += (col + 1);
row++;
}
else{
// move left in the mat
col--;
}
}
return count;
}
// function to find the kth smallest
// element in a sorted 2D mat
int kthSmallest(vector<vector<int>>& mat, int k) {
int n = mat.size();
int low = mat[0][0];
int high = mat[n - 1][n - 1];
int ans = 0;
while (low <= high) {
int mid = low + (high - low) / 2;
// count elements less than or equal to mid
int count = countSmallerEqual(mat, mid);
if (count < k) {
// if there are less than k elements
// <= mid, the kth smallest is larger
low = mid + 1;
} else {
// otherwise, mid might be the answer,
// but we need to check for smaller values
ans = mid;
high = mid - 1;
}
}
return ans;
}
int main() {
vector<vector<int>> mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50 }};
int k = 3;
int result = kthSmallest(mat, k);
cout << result << endl;
return 0;
}
class GfG {
// function to count the number of elements
// less than or equal to x
static int countSmallerEqual(int[][] mat, int x) {
int n = mat.length;
int count = 0;
int row = 0;
int col = n - 1;
// traverse from the top-right corner
while (row < n && col >= 0) {
if (mat[row][col] <= x) {
// if current element is less than
// or equal to x, all elements in this
// row up to the current column are <= x
count += (col + 1);
row++;
}
else{
// move left in the mat
col--;
}
}
return count;
}
// function to find the kth smallest
// element in a sorted 2D mat
static int kthSmallest(int[][] mat, int k) {
int n = mat.length;
int low = mat[0][0];
int high = mat[n - 1][n - 1];
int ans = 0;
while (low <= high) {
int mid = low + (high - low) / 2;
// count elements less than or equal to mid
int count = countSmallerEqual(mat, mid);
if (count < k) {
// if there are less than k elements
// <= mid, the kth smallest is larger
low = mid + 1;
}
else {
// otherwise, mid might be the answer,
// but we need to check for smaller values
ans = mid;
high = mid - 1;
}
}
return ans;
}
public static void main(String[] args) {
int[][] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
System.out.println(result);
}
}
# function to count the number of elements
# less than or equal to x
def countSmallerEqual(mat, x):
n = len(mat)
count = 0
row = 0
col = n - 1
# traverse from the top-right corner
while row < n and col >= 0:
if mat[row][col] <= x:
# if current element is less than
# or equal to x, all elements in this
# row up to the current column are <= x
count += (col + 1)
row += 1
else:
# move left in the mat
col -= 1
return count
# function to find the kth smallest
# element in a sorted 2D mat
def kthSmallest(mat, k):
n = len(mat)
low = mat[0][0]
high = mat[n - 1][n - 1]
ans = 0
while low <= high:
mid = low + (high - low) // 2
# count elements less than or equal to mid
count = countSmallerEqual(mat, mid)
if count < k:
# if there are less than k elements
# <= mid, the kth smallest is larger
low = mid + 1
else:
# otherwise, mid might be the answer,
# but we need to check for smaller values
ans = mid
high = mid - 1
return ans
if __name__ == "__main__":
mat = [
[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]]
k = 3
result = kthSmallest(mat, k)
print(result)
using System;
class GfG {
// function to count the number of elements
// less than or equal to x
static int countSmallerEqual(int[,] mat, int x) {
int n = mat.GetLength(0);
int count = 0;
int row = 0;
int col = n - 1;
// traverse from the top-right corner
while (row < n && col >= 0) {
if (mat[row,col] <= x) {
// if current element is less than
// or equal to x, all elements in this
// row up to the current column are <= x
count += (col + 1);
row++;
}
else {
// Move left in the mat
col--;
}
}
return count;
}
// function to find the kth smallest
// element in a sorted 2D mat
static int kthSmallest(int[,] mat, int k) {
int n = mat.GetLength(0);
int low = mat[0,0];
int high = mat[n - 1,n - 1];
int ans = 0;
while (low <= high) {
int mid = low + (high - low) / 2;
// count elements less than or equal to mid
int count = countSmallerEqual(mat, mid);
if (count < k) {
// if there are less than k elements
// <= mid, the kth smallest is larger
low = mid + 1;
} else {
// otherwise, mid might be the answer,
// but we need to check for smaller values
ans = mid;
high = mid - 1;
}
}
return ans;
}
public static void Main(string[] args) {
int[,] mat =
{{10, 20, 30, 40},
{15, 25, 35, 45},
{24, 29, 37, 48},
{32, 33, 39, 50}};
int k = 3;
int result = kthSmallest(mat, k);
Console.WriteLine(result);
}
}
// function to count the number of elements
// less than or equal to x
function countSmallerEqual(mat, x) {
const n = mat.length;
let count = 0;
let row = 0;
let col = n - 1;
// traverse from the top-right corner
while (row < n && col >= 0) {
if (mat[row][col] <= x) {
// if current element is less than
// or equal to x, all elements in this
// row up to the current column are <= x
count += (col + 1);
row++;
}
else {
// move left in the mat
col--;
}
}
return count;
}
// function to find the kth smallest
// element in a sorted 2D mat
function kthSmallest(mat, k) {
const n = mat.length;
let low = mat[0][0];
let high = mat[n - 1][n - 1];
let ans = 0;
while (low <= high) {
const mid = low + Math.floor((high - low) / 2);
// count elements less than or equal to mid
const count = countSmallerEqual(mat, mid);
if (count < k) {
// if there are less than k elements
// <= mid, the kth smallest is larger
low = mid + 1;
} else {
// otherwise, mid might be the answer,
// but we need to check for smaller values
ans = mid;
high = mid - 1;
}
}
return ans;
}
// Driver Code
const mat =
[[10, 20, 30, 40],
[15, 25, 35, 45],
[24, 29, 37, 48],
[32, 33, 39, 50]];
const k = 3;
const result = kthSmallest(mat, k);
console.log(result);
Time Complexity: O(n × log(max(mat) − min(mat))), Binary search is applied over the value range of the matrix, which takes log(max(mat) − min(mat)) steps. Each step calls a counting function that runs in O(n) time by scanning from top-right to bottom-left.
Auxiliary Space: O(1), The algorithm uses only fixed variables and no extra data structures, so space usage is constant.
SDE Sheet - Kth Smallest Element in a Row-Column Wise Sorted Matrix
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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