// C++ program to find number of sorted rows #include    #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 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 of 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; } 

// 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 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 of 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. 

# 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 of 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 of 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) 

// 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[ij + 1] <= mat[ij])  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   // of element that are not in  // decreasing order.  int j;  for (j = c - 1; j > 0; j--)  if (mat[ij - 1] <= mat[ij])  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. 

<script> // Javascript program to find number of sorted rows    let MAX = 100;    // Function to count all sorted rows in a matrix  function sortedCount(matrc)  {  let result = 0; // Initialize result    // Start from left side of matrix to  // count increasing order rows  for (let i = 0; i < r; i++) {    // Check if there is any pair of element  // that are not in increasing order.  let 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 (let i = 0; i < r; i++) {    // Check if there is any pair of element  // that are not in decreasing order.  let 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   let m = 4 n = 5;    let mat = [[1 2 3 4 5]  [4 3 1 2 6]  [8 7 6 5 4]  [5 7 8 9 10]]  document.write(sortedCount(mat m n))    // This code is contributed by unknown2108 </script> 

 // 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 of 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  // of 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. ?>