Search in a row wise and column wise sorted matrix
Last Updated :
23 Jul, 2025
Given a matrix mat[][] and an integer x, the task is to check if x is present in mat[][] or not. Every row and column of the matrix is sorted in increasing order.
Examples:
Input: x = 62, mat[][] = [[3, 30, 38],
[20, 52, 54],
[35, 60, 69]]
Output: false
Explanation: 62 is not present in the matrix.
Input: x = 55, mat[][] = [[18, 21, 27],
[38, 55, 67]]
Output: true
Explanation: mat[1][1] is equal to 55.
Input: x = 35, mat[][] = [[3, 30, 38],
[20, 52, 54],
[35, 60, 69]]
Output: true
Explanation: mat[2][0] is equal to 35.
[Naive Approach] Comparing with all elements - O(n*m) Time and O(1) Space
The simple idea is to traverse the complete matrix and search for the target element. If the target element is found, return true. Otherwise, return false.
C++
// C++ program to search an element in row-wise
// and column-wise sorted matrix
#include <iostream>
#include <vector>
using namespace std;
bool matSearch(vector<vector<int>> &mat, int x) {
int n = mat.size(), m = mat[0].size();
// Iterate over all the elements to find x
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mat[i][j] == x)
return true;
}
}
// If x was not found, return false
return false;
}
int main() {
vector<vector<int>> mat = {{3, 30, 38},
{20, 52, 54},
{35, 60, 69}};
int x = 35;
if(matSearch(mat, x))
cout << "true";
else
cout << "false";
return 0;
}
// Java program to search an element in row-wise
// and column-wise sorted matrix
class GfG {
static boolean matSearch(int[][] mat, int x) {
int n = mat.length, m = mat[0].length;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mat[i][j] == x)
return true;
}
}
// If x was not found, return false
return false;
}
public static void main(String[] args) {
int[][] mat = {{3, 30, 38},
{20, 52, 54},
{35, 60, 69}};
int x = 35;
if(matSearch(mat, x))
System.out.println("true");
else
System.out.println("false");
}
}
# Python program to search an element in row-wise
# and column-wise sorted matrix
def matSearch(mat, x):
n = len(mat)
m = len(mat[0])
for i in range(n):
for j in range(m):
if mat[i][j] == x:
return True
# If x was not found, return false
return False
if __name__ == "__main__":
mat = [[3, 30, 38],
[20, 52, 54],
[35, 60, 69]]
x = 35
if matSearch(mat, x):
print("true")
else:
print("false")
// C# program to search an element in row-wise
// and column-wise sorted matrix
using System;
class GfG {
static bool matSearch(int[][] mat, int x) {
int n = mat.Length, m = mat[0].Length;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (mat[i][j] == x)
return true;
}
}
// If x was not found, return false
return false;
}
static void Main() {
int[][] mat = new int[][] {
new int[] {3, 30, 38},
new int[] {20, 52, 54},
new int[] {35, 60, 69}
};
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// Java Script program to search an element in row-wise
// and column-wise sorted matrix
function matSearch(mat, x) {
const n = mat.length, m = mat[0].length;
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
if (mat[i][j] === x)
return true;
}
}
// If x was not found, return false
return false;
}
// Driver Code
const mat = [ [ 3, 30, 38 ],
[ 20, 52, 54 ],
[ 35, 60, 69 ] ];
const x = 35;
if (matSearch(mat, x))
console.log("true");
else
console.log("false");
[Better Approach] Binary Search - O(n*logm) Time and O(1) Space:
To optimize the above approach we are going to use the Binary Search algorithm.
The problem specifies that each row in the given matrix is sorted in ascending order. Instead of searching each column sequentially, we can efficiently apply Binary Search on each row to determine if the target is present.
C++
// C++ program to search an element in row-wise
// and column-wise sorted matrix
#include <iostream>
#include <vector>
using namespace std;
bool binarySearch(vector<int> &mat, int target) {
int n = mat.size();
int low = 0, high = n - 1;
// Standard binary search algorithm
while (low <= high) {
int mid = (low + high) / 2;
if (mat[mid] == target)
return true; // Element found
else if (target > mat[mid])
low = mid + 1; // Search in the right half
else
high = mid - 1; // Search in the left half
}
return false; // Element not found
}
bool matSearch(vector<vector<int>> &mat, int x) {
int n = mat.size();
// Iterate over each row and perform binary search
for (int i = 0; i < n; i++) {
if (binarySearch(mat[i], x))
return true; // Element found in one of the rows
}
return false; // Element not found in any row
}
int main() {
vector<vector<int>> mat = {{3, 30, 38},
{20, 52, 54},
{35, 60, 69}};
int x = 35;
if(matSearch(mat, x))
cout << "true";
else
cout << "false";
return 0;
}
// Java program to search an element in row-wise
// and column-wise sorted matrix
class GfG {
public static boolean binarySearch(int[] mat, int target) {
int low = 0, high = mat.length - 1;
// Standard binary search algorithm
while (low <= high) {
int mid = (low + high) / 2;
if (mat[mid] == target)
return true; // Element found
else if (target > mat[mid])
low = mid + 1; // Search in the right half
else
high = mid - 1; // Search in the left half
}
return false; // Element not found
}
static boolean matSearch(int[][] mat, int x) {
int n = mat.length; // Number of rows
// Iterate over each row and perform binary search
for (int i = 0; i < n; i++) {
if (binarySearch(mat[i], x))
return true; // Element found in one of the rows
}
return false; // Element not found in any row
}
public static void main(String[] args) {
int[][] mat = {{3, 30, 38},
{20, 52, 54},
{35, 60, 69}};
int x = 35;
if(matSearch(mat, x))
System.out.println("true");
else
System.out.println("false");
}
}
# Python program to search an element in row-wise
# and column-wise sorted matrix
def binarySearch(mat, target):
n = len(mat)
low, high = 0, n - 1
# Standard binary search algorithm
while low <= high:
mid = (low + high) // 2 # Midpoint index
if mat[mid] == target:
return True # Element found
elif target > mat[mid]:
low = mid + 1 # Search in the right half
else:
high = mid - 1 # Search in the left half
return False # Element not found
def matSearch(mat, x):
n = len(mat)
m = len(mat[0])
# Iterate over each row and perform binary search
for i in range(n):
if binarySearch(mat[i], x):
return True # Element found in one of the rows
return False # Element not found in any row
if __name__ == "__main__":
mat = [
[3, 30, 38],
[20, 52, 54],
[35, 60, 69]
]
x = 35
if matSearch(mat, x):
print("true")
else:
print("false")
// C# program to search an element in row-wise
// and column-wise sorted matrix
using System;
class GfG {
// Function to perform binary search on a sorted row (1D array)
static bool BinarySearch(int[] mat, int target)
{
int low = 0, high = mat.Length - 1;
// Standard binary search algorithm
while (low <= high)
{
int mid = (low + high) / 2;
if (mat[mid] == target)
return true; // Element found
else if (target > mat[mid])
low = mid + 1; // Search in the right half
else
high = mid - 1; // Search in the left half
}
return false; // Element not found
}
// Function to search an element in a row-wise sorted matrix
static bool matSearch(int[][] mat, int x) {
int n = mat.Length;
// Iterate over each row and perform binary search
for (int i = 0; i < n; i++)
{
if (BinarySearch(mat[i], x))
return true; // Element found in one of the rows
}
return false; // Element not found in any row
}
static void Main() {
int[][] mat = new int[][] {
new int[] {3, 30, 38},
new int[] {20, 52, 54},
new int[] {35, 60, 69}
};
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// JavaScript program to search an element in row-wise
// and column-wise sorted matrix
function binarySearch(mat, target) {
let low = 0, high = mat.length - 1;
// Standard binary search algorithm
while (low <= high) {
let mid = Math.floor((low + high) / 2); // Use Math.floor() to get an integer index
if (mat[mid] === target)
return true; // Element found
else if (target > mat[mid])
low = mid + 1; // Search in the right half
else
high = mid - 1; // Search in the left half
}
return false; // Element not found
}
function matSearch(mat, x) {
let n = mat.length;
// Iterate over each row and perform binary search
for (let i = 0; i < n; i++) {
if (binarySearch(mat[i], x))
return true; // Element found in one of the rows
}
return false; // Element not found in any row
}
// Driver Code
let mat = [
[3, 30, 38],
[20, 52, 54],
[35, 60, 69]
];
let x = 35;
if (matSearch(mat, x))
console.log("true");
else
console.log("false");
[Expected Approach] Eliminating rows or columns - O(n + m) Time and O(1) Space:
The idea is to remove a row or column in each comparison until an element is found. Start searching from the top-right corner of the matrix. There are 3 possible cases:
- x is greater than the current element: This ensures that all the elements in the current row are smaller than the given number as the pointer is already at the right-most element and the row is sorted. Thus, the entire row gets eliminated and continues the search from the next row.
- x is smaller than the current element: This ensures that all the elements in the current column are greater than the given number. Thus, the entire column gets eliminated and continues the search from the previous column, i.e. the column on the immediate left.
- The given number is equal to the current number: This will end the search.
Illustration:
C++
// C++ program to search an element in row-wise
// and column-wise sorted matrix
#include <iostream>
#include <vector>
using namespace std;
bool matSearch(vector<vector<int>> &mat, int x) {
int n = mat.size(), m = mat[0].size();
int i = 0, j = m - 1;
while(i < n && j >= 0) {
// If x > mat[i][j], then x will be greater
// than all elements to the left of
// mat[i][j] in row i, so increment i
if(x > mat[i][j]) {
i++;
}
// If x < mat[i][j], then x will be smaller
// than all elements to the bottom of
// mat[i][j] in column j, so decrement j
else if(x < mat[i][j]) {
j--;
}
// If x = mat[i][j], return true
else {
return true;
}
}
// If x was not found, return false
return false;
}
int main() {
vector<vector<int>> mat = {{3, 30, 38},
{20, 52, 54},
{35, 60, 69}};
int x = 35;
if(matSearch(mat, x))
cout << "true";
else
cout << "false";
return 0;
}
// Java program to search an element in row-wise
// and column-wise sorted matrix
import java.util.*;
class GfG {
static boolean matSearch(int[][] mat, int x) {
int n = mat.length, m = mat[0].length;
int i = 0, j = m - 1;
while (i < n && j >= 0) {
// If x > mat[i][j], then x will be greater
// than all elements to the left of
// mat[i][j] in row i, so increment i
if (x > mat[i][j]) {
i++;
}
// If x < mat[i][j], then x will be smaller
// than all elements to the bottom of
// mat[i][j] in column j, so decrement j
else if (x < mat[i][j]) {
j--;
}
// If x = mat[i][j], return true
else {
return true;
}
}
// If x was not found, return false
return false;
}
public static void main(String[] args) {
int[][] mat = {
{3, 30, 38},
{20, 52, 54},
{35, 60, 69}
};
int x = 35;
if (matSearch(mat, x))
System.out.println("true");
else
System.out.println("false");
}
}
# Python program to search an element in row-wise
# and column-wise sorted matrix
def matSearch(mat, x):
n = len(mat)
m = len(mat[0])
i = 0
j = m - 1
while i < n and j >= 0:
# If x > mat[i][j], then x will be greater
# than all elements to the left of
# mat[i][j] in row i, so increment i
if x > mat[i][j]:
i += 1
# If x < mat[i][j], then x will be smaller
# than all elements to the bottom of
# mat[i][j] in column j, so decrement j
elif x < mat[i][j]:
j -= 1
# If x = mat[i][j], return true
else:
return True
# If x was not found, return false
return False
if __name__ == "__main__":
mat = [
[3, 30, 38],
[20, 52, 54],
[35, 60, 69]
]
x = 35
if matSearch(mat, x):
print("true")
else:
print("false")
// C# program to search an element in row-wise
// and column-wise sorted matrix
using System;
class GfG {
static bool matSearch(int[][] mat, int x) {
int n = mat.Length, m = mat[0].Length;
int i = 0, j = m - 1;
while (i < n && j >= 0) {
// If x > mat[i][j], then x will be greater
// than all elements to the left of
// mat[i][j] in row i, so increment i
if (x > mat[i][j]) {
i++;
}
// If x < mat[i][j], then x will be smaller
// than all elements to the bottom of
// mat[i][j] in column j, so decrement j
else if (x < mat[i][j]) {
j--;
}
// If x = mat[i][j], return true
else {
return true;
}
}
// If x was not found, return false
return false;
}
static void Main() {
int[][] mat = new int[][] {
new int[] {3, 30, 38},
new int[] {20, 52, 54},
new int[] {35, 60, 69}
};
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// JavaScript program to search an element in row-wise
// and column-wise sorted matrix
function matSearch(mat, x) {
let n = mat.length, m = mat[0].length;
let i = 0, j = m - 1;
while (i < n && j >= 0) {
// If x > mat[i][j], then x will be greater
// than all elements to the left of
// mat[i][j] in row i, so increment i
if (x > mat[i][j]) {
i++;
}
// If x < mat[i][j], then x will be smaller
// than all elements to the bottom of
// mat[i][j] in column j, so decrement j
else if (x < mat[i][j]) {
j--;
}
// If x = mat[i][j], return true
else {
return true;
}
}
// If x was not found, return false
return false;
}
// Driver Code
let mat = [
[3, 30, 38],
[20, 52, 54],
[35, 60, 69]
];
let x = 35;
if (matSearch(mat, x))
console.log("true");
else
console.log("false");
Related Article: Search element in a sorted matrix
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