Search in row wise sorted matrix
Last Updated :
23 Jul, 2025
Given a row-wise sorted matrix mat[][] and an integer x, the task is to check if x is present in mat[][] or not. Note that there is no ordering among column elements.
Examples:
Input: x = 9, mat[][] = [[3, 4, 9],
[2, 5, 6],
[9, 25, 27]]
Output: true
Explanation: mat[0][2] is equal to 9.
Input: x = 56, mat[][] = [[19, 22, 27, 38, 55, 67]]
Output: false
Explanation: 56 is not present in mat[][].
[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. Please refer to Searching Algorithms for 2D Arrays (Matrix) to know more about the implementation.
[Expected Approach] Using Binary Search - O(n * log(m)) Time and O(1) Space
The idea is simple, we traverse through the matrix and apply binary search on each row to find the element.
C++
// C++ program to search an element in row-wise
// sorted matrix
#include <iostream>
#include <vector>
using namespace std;
bool search(vector<int> &arr, int x) {
int lo = 0, hi = arr.size();
while(lo <= hi) {
int mid = (lo + hi) / 2;
// If x = mid, return true
if(x == arr[mid])
return true;
// If x < arr[mid], search in left half
if(x < arr[mid])
hi = mid - 1;
// If x > arr[mid], search in right half
else
lo = mid + 1;
}
return false;
}
bool searchRowMatrix(vector<vector<int>> &mat, int x) {
int n = mat.size(), m = mat[0].size();
// Iterate over all the rows to find x
for(int i = 0; i < n; i++) {
// binary search on ith row
if(search(mat[i], x))
return true;
}
// If x was not found, return false
return false;
}
int main() {
vector<vector<int>> mat = {{3, 4, 9},
{2, 5, 6},
{9, 25, 27}};
int x = 9;
if(searchRowMatrix(mat, x))
cout << "true";
else
cout << "false";
return 0;
}
// Java program to search an element in row-wise
// sorted matrix
import java.util.*;
class GfG {
static boolean search(int[] arr, int x) {
int lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
// If x == mid, return true
if (x == arr[mid])
return true;
// If x < arr[mid], search in the left half
if (x < arr[mid])
hi = mid - 1;
// If x > arr[mid], search in the right half
else
lo = mid + 1;
}
return false;
}
static boolean searchRowMatrix(int[][] mat, int x) {
int n = mat.length, m = mat[0].length;
// Iterate over all the rows to find x
for (int i = 0; i < n; i++) {
// Perform binary search on the ith row
if (search(mat[i], x))
return true;
}
// If x was not found, return false
return false;
}
public static void main(String[] args) {
int[][] mat = {
{3, 4, 9},
{2, 5, 6},
{9, 25, 27}
};
int x = 9;
if (searchRowMatrix(mat, x))
System.out.println("true");
else
System.out.println("false");
}
}
# Python program to search an element in row-wise
# sorted matrix
def search(arr, x):
lo, hi = 0, len(arr) - 1
while lo <= hi:
mid = (lo + hi) // 2
# If x == mid, return true
if x == arr[mid]:
return True
# If x < arr[mid], search in the left half
if x < arr[mid]:
hi = mid - 1
# If x > arr[mid], search in the right half
else:
lo = mid + 1
return False
def searchRowMatrix(mat, x):
n, m = len(mat), len(mat[0])
# Iterate over all the rows to find x
for i in range(n):
# Perform binary search on the ith row
if search(mat[i], x):
return True
# If x was not found, return false
return False
if __name__ == "__main__":
mat = [[3, 4, 9],
[2, 5, 6],
[9, 25, 27]]
x = 9
if searchRowMatrix(mat, x):
print("true")
else:
print("false")
// C# program to search an element in row-wise
// sorted matrix
using System;
class GfG {
static bool Search(int[] arr, int x) {
int lo = 0, hi = arr.Length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
// If x == mid, return true
if (x == arr[mid])
return true;
// If x < arr[mid], search in the left half
if (x < arr[mid])
hi = mid - 1;
// If x > arr[mid], search in the right half
else
lo = mid + 1;
}
return false;
}
// Function to search for element x in the matrix
static bool SearchRowMatrix(int[][] mat, int x) {
int n = mat.Length, m = mat[0].Length;
// Iterate over all the rows to find x
for (int i = 0; i < n; i++) {
// Perform binary search on the ith row
if (Search(mat[i], x))
return true;
}
// If x was not found, return false
return false;
}
static void Main() {
int[][] mat = {
new int[] { 3, 4, 9 },
new int[] { 2, 5, 6 },
new int[] { 9, 25, 27 }
};
int x = 9;
if (SearchRowMatrix(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// JavaScript program to search an element in row-wise
// sorted matrix
function search(arr, x) {
let lo = 0, hi = arr.length - 1;
while (lo <= hi) {
let mid = Math.floor((lo + hi) / 2);
// If x == mid, return true
if (x === arr[mid])
return true;
// If x < arr[mid], search in the left half
if (x < arr[mid])
hi = mid - 1;
// If x > arr[mid], search in the right half
else
lo = mid + 1;
}
return false;
}
function searchRowMatrix(mat, x) {
let n = mat.length, m = mat[0].length;
// Iterate over all the rows to find x
for (let i = 0; i < n; i++) {
// Perform binary search on the ith row
if (search(mat[i], x))
return true;
}
// If x was not found, return false
return false;
}
// Driver Code
let mat = [
[3, 4, 9],
[2, 5, 6],
[9, 25, 27]
];
let x = 9;
if (searchRowMatrix(mat, x))
console.log("true");
else
console.log("false");
Time Complexity: O(n*logm), where n is the number of rows and m is the number of columns.
Auxiliary Space: O(1)
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