S obzirom na matricu veličine m*n, zadatak je prebrojati sve retke u matrici koji su poredani ili u strogo rastućem ili u strogo opadajućem redoslijedu?

Primjeri:  

 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 je jednostavna i uključuje dva obilaska matrice.   
 
 
 Pomaknite se s lijeve strane matrice da biste izbrojali sve redove koji su unutra     strogo rastući poredak       
 
 Krećite se s desne strane matrice da biste izbrojali sve redove koji su unutra     strogo opadajući poredak    
 
 
 Ispod je implementacija gornje ideje.   
 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.   ?>   
    
  Izlaz   
3 
 
    Vremenska složenost: O(m*n)      
    Pomoćni prostor:          O(1)    
 
 Ako imate neki drugi optimizirani pristup rješavanju ovog problema, podijelite ga u komentarima.  
 
    
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