Given a 2D integer matrix mat[][] of size n x m, where every row and column is sorted in increasing order and a number x, return true if the element x is present in the matrix. Otherwise, return false.
Examples:Â
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.
Try It Yourself
Table of Content
[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.
#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;
}
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");
}
}
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")
using System;
class GfG {
static bool matSearch(int[, ] mat, int x)
{
int n = mat.GetLength(0), m = mat.GetLength(1);
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 = { { 3, 30, 38 },
{ 20, 52, 54 },
{ 35, 60, 69 } };
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
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");
Output
true
[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.
#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;
// Element found
if (mat[mid] == target)
return true;
// Search in the right half
else if (target > mat[mid])
low = mid + 1;
// Search in the left half
else
high = mid - 1;
}
// Element not found
return false;
}
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))
// Element found in one of the rows
return true;
}
// Element not found in any row
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;
}
public class Main {
public static boolean binarySearch(int[] mat,
int target)
{
int n = mat.length;
int low = 0, high = n - 1;
// Standard binary search algorithm
while (low <= high) {
int mid = (low + high) / 2;
// Element found
if (mat[mid] == target)
return true;
// Search in the right half
else if (target > mat[mid])
low = mid + 1;
// Search in the left half
else
high = mid - 1;
}
// Element not found
return false;
}
public static boolean 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))
// Element found in one of the rows
return true;
}
// Element not found in any row
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");
}
}
def binarySearch(mat, target):
n = len(mat)
low, high = 0, n - 1
# Standard binary search algorithm
while low <= high:
mid = (low + high) // 2
# Element found
if mat[mid] == target:
return True
# Search in the right half
elif target > mat[mid]:
low = mid + 1
# Search in the left half
else:
high = mid - 1
# Element not found
return False
def matSearch(mat, x):
n = len(mat)
# Iterate over each row and perform binary search
for i in range(n):
if binarySearch(mat[i], x):
# Element found in one of the rows
return True
# Element not found in any row
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")
using System;
public class MainClass {
public static bool binarySearch(int[,] mat, int row, int target)
{
int n = mat.GetLength(1);
int low = 0, high = n - 1;
// Standard binary search algorithm
while (low <= high) {
int mid = (low + high) / 2;
// Element found
if (mat[row, mid] == target)
return true;
// Search in the right half
else if (target > mat[row, mid])
low = mid + 1;
// Search in the left half
else
high = mid - 1;
}
// Element not found
return false;
}
public static bool matSearch(int[,] mat, int x)
{
int n = mat.GetLength(0);
// Iterate over each row and perform binary search
for (int i = 0; i < n; i++) {
if (binarySearch(mat, i, x))
// Element found in one of the rows
return true;
}
// Element not found in any row
return false;
}
public static void Main(string[] args)
{
int[,] mat = new int[,] { { 3, 30, 38 },
{ 20, 52, 54 },
{ 35, 60, 69 } };
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
function binarySearch(mat, target)
{
let n = mat.length;
let low = 0, high = n - 1;
// Standard binary search algorithm
while (low <= high) {
let mid = Math.floor((low + high) / 2);
// Element found
if (mat[mid] === target)
return true;
// Search in the right half
else if (target > mat[mid])
low = mid + 1;
// Search in the left half
else
high = mid - 1;
}
// Element not found
return false;
}
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))
// Element found in one of the rows
return true;
}
// Element not found in any row
return false;
}
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");
Output
true
[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.
#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;
}
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");
}
}
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")
using System;
class GfG {
static bool matSearch(int[, ] mat, int x)
{
int n = mat.GetLength(0), m = mat.GetLength(1);
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 = { { 3, 30, 38 },
{ 20, 52, 54 },
{ 35, 60, 69 } };
int x = 35;
if (matSearch(mat, x))
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
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");
Output
true