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:
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 found
Simple Solution
- Approach: The simple idea is to traverse the array and to search element 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
// C++ program to search an element in row-wise // and column-wise sorted matrix #include <bits/stdc++.h> using namespace std; /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */int search(int mat[4][4], int n, int x) { if (n == 0) return -1; //traverse through the matrix for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) //if the element is found if(mat[i][j] == x) { cout<<"Element found at ("<< i << ", " << j << ")\n"; return 1; } } cout << "n Element not found"; return 0; } // Driver code 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; } |
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Output
Element found at (2, 1)
A better solution is to 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 element and the row is sorted. Thus, the entire row gets eliminated and continue the search on the next row. Here elimination means that row needs not to 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 continue the search on the previous column i.e. the column at 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 = 0
- 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 below approach.
- Implementation:
C++
// C++ program to search an element in row-wise // and column-wise sorted matrix #include <bits/stdc++.h> using namespace std; /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */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; // set indexes for top right element 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--; // Check if mat[i][j] < x else i++; } cout << "n Element not found"; return 0; } // Driver code 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; } // This code is contributed // by Akanksha Rai(Abby_akku) |
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C
// C program to search an element in row-wise // and column-wise sorted matrix #include <stdio.h> /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */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; // set indexes for top right element 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 // if mat[i][j] < x i++; } printf("n Element not found"); return 0; // if ( i==n || j== -1 ) } // driver program to test above function 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; } |
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Java
// JAVA Code for Search in a row wise and // column wise sorted matrix class GFG { /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */ private static void search(int[][] mat, int n, int x) { // set indexes for top right int i = 0, j = n - 1; // element 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 // if mat[i][j] < x i++; } System.out.print("n Element not found"); return; // if ( i==n || j== -1 ) } // driver program to test above function 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); } } // This code is contributed by Arnav Kr. Mandal. |
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Python3
# Python3 program to search an element # in row-wise and column-wise sorted matrix # Searches the element x in mat[][]. If the # element is found, then prints its position # and returns true, otherwise prints "not found" # and returns false def search(mat, n, x): i = 0 # set indexes for top right element 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 # if mat[i][j] < x else: i += 1 print("Element not found") return 0 # if (i == n || j == -1 ) # Driver Code mat = [ [10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50] ] search(mat, 4, 29) # This code is contributed by Anant Agarwal. |
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C#
// C# Code for Search in a row wise and // column wise sorted matrix using System; class GFG { /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */ private static void search(int[, ] mat, int n, int x) { // set indexes for top right // element 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 // if mat[i][j] < x i++; } Console.Write("n Element not found"); return; // if ( i==n || j== -1 ) } // driver program to test above function 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); } } // This code is contributed by Sam007 |
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PHP
<?php // PHP program to search an // element in row-wise and // column-wise sorted matrix /* Searches the element $x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */function search(&$mat, $n, $x) { $i = 0; $j = $n - 1; // set indexes for // top right element while ($i < $n && $j >= 0) { if ($mat[$i][$j] == $x) { echo "n found at " . $i. ", " . $j; return 1; } if ($mat[$i][$j] > $x) $j--; else // if $mat[$i][$j] < $x $i++; } echo "n Element not found"; return 0; // if ( $i==$n || $j== -1 ) } // Driver Code $mat = array(array(10, 20, 30, 40), array(15, 25, 35, 45), array(27, 29, 37, 48), array(32, 33, 39, 50)); search($mat, 4, 29); // This code is contributed // by ChitraNayal ?> |
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Output
n Found at 2, 1
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.
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