Given an n x n matrix and a number x, find the position of x in the matrix if it is present in it. Otherwise, print “Not Found”. In the given matrix, every row and column is sorted in increasing order. The designed algorithm should have linear time complexity.
Example:
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Input: mat[4][4] = { {10, 20, 30, 40},
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}};
x = 29
Output: Found at (2, 1)
Explanation: Element at (2,1) is 29
Input : mat[4][4] = { {10, 20, 30, 40},
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}};
x = 100
Output : Element not found
Explanation: Element 100 is not foundSimple Solution
- Approach: The simple idea is to traverse the array and to search elements one by one.
- Algorithm:
- Run a nested loop, outer loop for row and inner loop for the column
- Check every element with x and if the element is found then print “element found”
- If the element is not found, then print “element not found”.
- Implementation:
CPP14
#include <bits/stdc++.h>
using namespace std;
int search(int mat[4][4], int n, int x)
{
if (n == 0)
return -1;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
if(mat[i][j] == x)
{
cout<<"Element found at ("<<
i << ", " << j << ")\n";
return 1;
}
}
cout << "n Element not found";
return 0;
}
int main()
{
int mat[4][4] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
return 0;
}
|
Java
class GFG {
static int search(int[][] mat, int n, int x)
{
if (n == 0)
return -1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
if (mat[i][j] == x) {
System.out.print("Element found at ("
+ i + ", " + j
+ ")\n");
return 1;
}
}
System.out.print(" Element not found");
return 0;
}
public static void main(String[] args)
{
int mat[][] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
}
}
|
Python3
def search(mat, n, x):
if(n == 0):
return -1
for i in range(n):
for j in range(n):
if(mat[i][j] == x):
print("Element found at (", i, ",", j, ")")
return 1
print(" Element not found")
return 0
mat = [[10, 20, 30, 40], [15, 25, 35, 45],[27, 29, 37, 48],[32, 33, 39, 50]]
search(mat, 4, 29)
|
C#
using System;
class GFG{
static int search(int[,] mat, int n, int x)
{
if (n == 0)
return -1;
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
if (mat[i,j] == x)
{
Console.Write("Element found at (" + i +
", " + j + ")\n");
return 1;
}
}
Console.Write(" Element not found");
return 0;
}
static public void Main()
{
int[,] mat = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
}
}
|
Javascript
<script>
function search(mat,n,x) {
if (n == 0)
return -1;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++)
if (mat[i][j] == x) {
document.write("Element found at ("
+ i + ", " + j
+ ")<br>");
return 1;
}
}
document.write(" Element not found");
return 0;
}
let mat = [[ 10, 20, 30, 40 ],
[15, 25, 35, 45] ,
[ 27, 29, 37, 48 ],
[ 32, 33, 39, 50 ]];
search(mat, 4, 29);
</script>
|
OutputElement found at (2, 1)
A better solution is use Divide and Conquer to find the element which has a time complexity of O(n1.58). Please refer here for details.
Efficient Solution
- Approach: The simple 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 three possible cases.
- The given number is greater than the current number: This will ensure that all the elements in the current row are smaller than the given number as the pointer is already at the right-most elements and the row is sorted. Thus, the entire row gets eliminated and continues the search for the next row. Here, elimination means that a row needs not be searched.
- The given number is smaller than the current number: This will ensure that all the elements in the current column are greater than the given number. Thus, the entire column gets eliminated and continues the search for 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.
- Algorithm:
- Let the given element be x, create two variable i = 0, j = n-1 as index of row and column
- Run a loop until i = n
- Check if the current element is greater than x then decrease the count of j. Exclude the current column.
- Check if the current element is less than x then increase the count of i. Exclude the current row.
- If the element is equal, then print the position and end.
- Thanks to devendraiiit for suggesting the approach below.
- Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int search(int mat[4][4], int n, int x)
{
if (n == 0)
return -1;
int smallest = mat[0][0], largest = mat[n - 1][n - 1];
if (x < smallest || x > largest)
return -1;
int i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (mat[i][j] == x)
{
cout << "n Found at "
<< i << ", " << j;
return 1;
}
if (mat[i][j] > x)
j--;
else
i++;
}
cout << "n Element not found";
return 0;
}
int main()
{
int mat[4][4] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
return 0;
}
|
C
#include <stdio.h>
int search(int mat[4][4], int n, int x)
{
if (n == 0)
return -1;
int smallest = mat[0][0], largest = mat[n - 1][n - 1];
if (x < smallest || x > largest)
return -1;
int i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (mat[i][j] == x)
{
printf("\n Found at %d, %d", i, j);
return 1;
}
if (mat[i][j] > x)
j--;
else
i++;
}
printf("n Element not found");
return 0;
}
int main()
{
int mat[4][4] = {
{ 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 },
};
search(mat, 4, 29);
return 0;
}
|
Java
class GFG {
private static void search(int[][] mat,
int n, int x)
{
int i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (mat[i][j] == x)
{
System.out.print("n Found at " +
i + " " + j);
return;
}
if (mat[i][j] > x)
j--;
else
i++;
}
System.out.print("n Element not found");
return;
}
public static void main(String[] args)
{
int mat[][] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
}
}
|
Python3
def search(mat, n, x):
i = 0
j = n - 1
while ( i < n and j >= 0 ):
if (mat[i][j] == x ):
print("n Found at ", i, ", ", j)
return 1
if (mat[i][j] > x ):
j -= 1
else:
i += 1
print("Element not found")
return 0
mat = [ [10, 20, 30, 40],
[15, 25, 35, 45],
[27, 29, 37, 48],
[32, 33, 39, 50] ]
search(mat, 4, 29)
|
C#
using System;
class GFG
{
private static void search(int[, ] mat,
int n, int x)
{
int i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (mat[i, j] == x)
{
Console.Write("n Found at "
+ i + ", " + j);
return;
}
if (mat[i, j] > x)
j--;
else
i++;
}
Console.Write("n Element not found");
return;
}
public static void Main()
{
int[, ] mat = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
search(mat, 4, 29);
}
}
|
PHP
<?php
function search(&$mat, $n, $x)
{
$i = 0;
$j = $n - 1;
while ($i < $n && $j >= 0)
{
if ($mat[$i][$j] == $x)
{
echo "n found at " . $i.
", " . $j;
return 1;
}
if ($mat[$i][$j] > $x)
$j--;
else
$i++;
}
echo "n Element not found";
return 0;
}
$mat = array(array(10, 20, 30, 40),
array(15, 25, 35, 45),
array(27, 29, 37, 48),
array(32, 33, 39, 50));
search($mat, 4, 29);
?>
|
Javascript
<script>
function search(mat,n,x){
let i = 0, j = n - 1;
while (i < n && j >= 0)
{
if (mat[i][j] == x)
{
document.write("n Found at " +
i + " " + j);
return;
}
if (mat[i][j] > x)
j--;
else
i++;
}
document.write("n Element not found");
return;
}
let mat = [[10, 20, 30, 40 ],
[ 15, 25, 35, 45 ],
[ 27, 29, 37, 48 ],
[ 32, 33, 39, 50 ]];
search(mat, 4, 29);
</script>
|
Time Complexity: O(n).
Only one traversal is needed, i.e, i from 0 to n and j from n-1 to 0 with at most 2*n steps.
The above approach will also work for m x n matrix (not only for n x n). Complexity would be O(m + n).
Space Complexity: O(1).
No extra space is required.
Related Article :
Search element in a sorted matrix
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
