Count all sorted rows in a matrix

Given a matrix of m*n size, the task is to count all the rows in a matrix that are sorted either in strictly increasing order or in strictly decreasing order?

Examples :

Input : m = 4,  n = 5
        mat[m][n] = 1 2 3 4 5
                    4 3 1 2 6
                    8 7 6 5 4
                    5 7 8 9 10
Output: 3 



The idea is simple and involves two traversals of matrix.
1) Traverse from left side of the matrix to count all the row which are in strictly increasing order
2) Traverse from right side of the matrix to count all the row which are in strictly decreasing order

Below is the implementation of above idea.

C++

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// C++ program to find number of sorted rows
#include <bits/stdc++.h>
#define MAX 100
using namespace std;
  
// Function to count all sorted rows in a matrix
int sortedCount(int mat[][MAX], int r, int c)
{
    int result = 0; // Initialize result
  
    // Start from left side of matrix to
    // count increasing order rows
    for (int i=0; i<r; i++)
    {
        // Check if there is any pair ofs element
        // that are  not in increasing order.
        int j;
        for (j=0; j<c-1; j++)
            if (mat[i][j+1] <= mat[i][j])
                break;
  
        // If the loop didn't break (All elements
        // of current row were in increasing order)
        if (j == c-1)
            result++;
    }
  
    // Start from right side of matrix to
    // count increasing order rows ( reference
    // to left these are in decreasing order )
    for (int i=0; i<r; i++)
    {
        // Check if there is any pair ofs element
        // that are  not in decreasing order.
        int j;
        for (j=c-1; j>0; j--)
            if (mat[i][j-1] <= mat[i][j])
                break;
  
        // Note c > 1 condition is required to make
        // sure that a single column row is not counted
        // twice (Note that a single column row is sorted
        // both in increasing and decreasing order) 
        if (c > 1 && j == 0)
            result++;
    }
    return result;
}
  
// Driver program to run the case
int main()
{
    int m = 4, n = 5;
    int mat[][MAX] = {{1, 2, 3, 4, 5},
                      {4, 3, 1, 2, 6},
                      {8, 7, 6, 5, 4},
                      {5, 7, 8, 9, 10}};
    cout << sortedCount(mat, m, n);
    return 0;
}

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Java

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// Java program to find number of sorted rows
  
class GFG {
      
    static int MAX = 100;
  
    // Function to count all sorted rows in a matrix
    static int sortedCount(int mat[][], int r, int c)
    {
        int result = 0; // Initialize result
  
        // Start from left side of matrix to
        // count increasing order rows
        for (int i = 0; i < r; i++) {
              
            // Check if there is any pair ofs element
            // that are not in increasing order.
            int j;
            for (j = 0; j < c - 1; j++)
                if (mat[i][j + 1] <= mat[i][j])
                    break;
  
            // If the loop didn't break (All elements
            // of current row were in increasing order)
            if (j == c - 1)
                result++;
        }
  
        // Start from right side of matrix to
        // count increasing order rows ( reference
        // to left these are in decreasing order )
        for (int i = 0; i < r; i++) {
              
            // Check if there is any pair ofs element
            // that are not in decreasing order.
            int j;
            for (j = c - 1; j > 0; j--)
                if (mat[i][j - 1] <= mat[i][j])
                    break;
  
            // Note c > 1 condition is required to make
            // sure that a single column row is not counted
            // twice (Note that a single column row is sorted
            // both in increasing and decreasing order)
            if (c > 1 && j == 0)
                result++;
        }
        return result;
    }
      
    // Driver code
    public static void main(String arg[])
    {
        int m = 4, n = 5;
        int mat[][] = { { 1, 2, 3, 4, 5 },
                        { 4, 3, 1, 2, 6 },
                        { 8, 7, 6, 5, 4 },
                        { 5, 7, 8, 9, 10 } };
        System.out.print(sortedCount(mat, m, n));
    }
}
  
// This code is contributed by Anant Agarwal.

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Python

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# Python3 program to find number 
# of sorted rows
def sortedCount(mat, r, c):
      
    result = 0
      
    # Start from left side of matrix to 
    # count increasing order rows 
    for i in range(r):
          
        # Check if there is any pair ofs element 
        # that are not in increasing order.
        j = 0
        for j in range(c - 1):
            if mat[i][j + 1] <= mat[i][j]:
                break
      
        # If the loop didn't break (All elements 
        # of current row were in increasing order)
        if j == c - 2:
            result += 1
  
    # Start from right side of matrix to 
    # count increasing order rows ( reference 
    # to left these are in decreasing order )
    for i in range(0, r):
  
        # Check if there is any pair ofs element 
        # that are not in decreasing order.
        j = 0
        for j in range(c - 1, 0, -1):
            if mat[i][j - 1] <= mat[i][j]:
                break
          
        # Note c > 1 condition is required to 
        # make sure that a single column row 
        # is not counted twice (Note that a 
        # single column row is sorted both 
        # in increasing and decreasing order)
        if c > 1 and j == 1:
            result += 1
      
    return result     
  
# Driver code
m, n = 4, 5
  
mat = [[1, 2, 3, 4, 5],
       [4, 3, 1, 2, 6],
       [8, 7, 6, 5, 4],
       [5, 7, 8, 9, 10]]
  
print(sortedCount(mat, m, n))
  
# This code is contributed by
# Mohit kumar 29 (IIIT gwalior)

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C#

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// C# program to find number of sorted rows
using System;
  
class GFG {
      
// static int MAX = 100;
  
    // Function to count all sorted rows in 
    // a matrix
    static int sortedCount(int [,]mat, int r,
                                       int c)
    {
        int result = 0; // Initialize result
  
        // Start from left side of matrix to
        // count increasing order rows
        for (int i = 0; i < r; i++) {
              
            // Check if there is any pair of
            // element that are not in
            // increasing order.
            int j;
            for (j = 0; j < c - 1; j++)
                if (mat[i,j + 1] <= mat[i,j])
                    break;
  
            // If the loop didn't break (All
            // elements of current row were 
            // in increasing order)
            if (j == c - 1)
                result++;
        }
  
        // Start from right side of matrix 
        // to count increasing order rows 
        // ( reference to left these are in 
        // decreasing order )
        for (int i = 0; i < r; i++) {
              
            // Check if there is any pair 
            // ofs element that are not in
            // decreasing order.
            int j;
            for (j = c - 1; j > 0; j--)
                if (mat[i,j - 1] <= mat[i,j])
                    break;
  
            // Note c > 1 condition is 
            // required to make sure that a 
            // single column row is not 
            // counted twice (Note that a 
            // single column row is sorted
            // both in increasing and
            // decreasing order)
            if (c > 1 && j == 0)
                result++;
        }
        return result;
    }
      
    // Driver code
    public static void Main()
    {
        int m = 4, n = 5;
        int [,]mat = { { 1, 2, 3, 4, 5 },
                       { 4, 3, 1, 2, 6 },
                       { 8, 7, 6, 5, 4 },
                       { 5, 7, 8, 9, 10 } };
                         
        Console.WriteLine(
                   sortedCount(mat, m, n));
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to find 
// number of sorted rows
  
$MAX = 100;
  
// Function to count all 
// sorted rows in a matrix
function sortedCount($mat
                     $r, $c)
{
    // Initialize result
    $result = 0; 
  
    // Start from left side of
    // matrix to count increasing 
    // order rows
    for ( $i = 0; $i < $r; $i++)
    {
        // Check if there is any 
        // pair ofs element that 
        // are not in increasing order.
        $j;
        for ($j = 0; $j < $c - 1; $j++)
            if ($mat[$i][$j + 1] <= $mat[$i][$j])
                break;
  
        // If the loop didn't break 
        // (All elements of current 
        // row were in increasing order)
        if ($j == $c - 1)
            $result++;
    }
  
    // Start from right side of 
    // matrix to count increasing 
    // order rows ( reference to left
    // these are in decreasing order )
    for ($i = 0; $i < $r; $i++)
    {
        // Check if there is any pair 
        // ofs element that are not
        // in decreasing order.
        $j;
        for ($j = $c - 1; $j > 0; $j--)
            if ($mat[$i][$j - 1] <= $mat[$i][$j])
                break;
  
        // Note c > 1 condition is 
        // required to make sure that 
        // a single column row is not 
        // counted twice (Note that a 
        // single column row is sorted
        // both in increasing and 
        // decreasing order) 
        if ($c > 1 && $j == 0)
            $result++;
    }
    return $result;
}
  
// Driver Code
$m = 4; $n = 5;
$mat = array(array(1, 2, 3, 4, 5),
             array(4, 3, 1, 2, 6),
             array(8, 7, 6, 5, 4),
             array(5, 7, 8, 9, 10));
echo sortedCount($mat, $m, $n);
  
// This code is contributed by anuj_67.
?>

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Output :

3

Time Complexity : O(m*n)
Auxiliary space : O(1)

If you have another optimized approach to solve this problem then please share in comments.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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