Ņemot vērā m*n lieluma matricu, uzdevums ir saskaitīt visas matricas rindas, kas ir sakārtotas vai nu stingri augošā, vai stingri dilstošā secībā?

Piemēri:  

 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  
 Ideja ir vienkārša un ietver divus matricas šķērsojumus.   
 
 
 Pārejiet no matricas kreisās puses, lai saskaitītu visas tajā esošās rindas     stingri palielinot kārtību       
 
 Pārejiet no matricas labās puses, lai saskaitītu visas tajā esošās rindas     stingri dilstošā secībā    
 
 
 Zemāk ir iepriekš minētās idejas īstenošana.   
 C++   
   // 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   
   // 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.   
  Python   
   # 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#   
   // 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       // 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  ()      {      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.   
  JavaScript   
    <  script  >   // Javascript program to find number of sorted rows          let     MAX     =     100  ;          // Function to count all sorted rows in a matrix      function     sortedCount  (  mat    r    c  )      {      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   
      // 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.   ?>   
    
  Izvade   
3 
 
    Laika sarežģītība: O(m*n)      
    Palīgtelpa:          O(1)    
 
 Ja jums ir cita optimizēta pieeja šīs problēmas risināšanai, lūdzu, dalieties komentāros.  
 
    
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