Gitt en kvadratisk matrise av størrelse N*N hvor hver celle er knyttet til en spesifikk kostnad. En bane er definert som en spesifikk sekvens av celler som starter fra den øverste venstre cellen beveger seg bare til høyre eller ned og ender i cellen nederst til høyre. Vi ønsker å finne en vei med maksimalt gjennomsnitt over alle eksisterende stier. Gjennomsnitt beregnes som totalkostnad delt på antall besøkte celler i banen. 

Eksempler:  

 Input : Matrix = [1 2 3   
  4 5 6   
  7 8 9]   
 Output : 5.8   
 Path with maximum average is 1 -> 4 -> 7 -> 8 -> 9   
 Sum of the path is 29 and average is 29/5 = 5.8   
 
 En interessant observasjon er at de eneste tillatte trekkene er nede, og til høyre trenger vi N-1 nedtrekk og N-1 høyretrekk for å nå målet (nederst til høyre). Så enhver vei fra øverste venstre hjørne til nederste høyre hjørne krever 2N - 1 celler. I    gjennomsnittlig    verdi nevneren er fast og vi trenger bare å maksimere telleren. Derfor må vi i utgangspunktet finne den maksimale sumveien. Å beregne maksimal sum av banen er et klassisk dynamisk programmeringsproblem hvis dp[i][j] representerer maksimal sum til celle (i j) fra (0 0), så oppdaterer vi ved hver celle (i j) dp[i][j] som nedenfor  
  
 for all i 1  <= i  <= N   
  dp[i][0] = dp[i-1][0] + cost[i][0];    
 for all j 1  <= j  <= N   
  dp[0][j] = dp[0][j-1] + cost[0][j];    
 otherwise    
  dp[i][j] = max(dp[i-1][j] dp[i][j-1]) + cost[i][j];    
 
 Når vi får maksimal sum av alle stier, deler vi denne summen med (2N - 1), og vi vil få vårt maksimale gjennomsnitt.   
  
    Implementering:    
 C++   
   //C/C++ program to find maximum average cost path   #include          using     namespace     std  ;   // Maximum number of rows and/or columns   const     int     M     =     100  ;   // method returns maximum average of all path of   // cost matrix   double     maxAverageOfPath  (  int     cost  [  M  ][  M  ]     int     N  )   {      int     dp  [  N  +  1  ][  N  +  1  ];      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ];      /* Initialize first column of total cost(dp) array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -1  ][  0  ]     +     cost  [  i  ][  0  ];      /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -1  ]     +     cost  [  0  ][  j  ];      /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <=     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     max  (  dp  [  i  -1  ][  j  ]      dp  [  i  ][  j  -1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N  -1  ][  N  -1  ]     /     (  2  *  N  -1  );   }   /* Driver program to test above functions */   int     main  ()   {      int     cost  [  M  ][  M  ]     =     {     {  1       2       3  }      {  6       5       4  }      {  7       3       9  }      };      printf  (  '%f'       maxAverageOfPath  (  cost       3  ));      return     0  ;   }   
  Java   
   // JAVA Code for Path with maximum average   // value   import     java.io.*  ;   class   GFG     {          // method returns maximum average of all      // path of cost matrix      public     static     double     maxAverageOfPath  (  int     cost  [][]        int     N  )      {      int     dp  [][]     =     new     int  [  N  +  1  ][  N  +  1  ]  ;      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ]  ;          /* Initialize first column of total cost(dp)    array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -  1  ][  0  ]     +     cost  [  i  ][  0  ]  ;          /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -  1  ]     +     cost  [  0  ][  j  ]  ;          /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     Math  .  max  (  dp  [  i  -  1  ][  j  ]        dp  [  i  ][  j  -  1  ]  )     +     cost  [  i  ][  j  ]  ;          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N  -  1  ][  N  -  1  ]     /     (  2     *     N     -     1  );      }          /* Driver program to test above function */      public     static     void     main  (  String  []     args  )         {      int     cost  [][]     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }};          System  .  out  .  println  (  maxAverageOfPath  (  cost       3  ));      }   }   // This code is contributed by Arnav Kr. Mandal.   
  C#   
   // C# Code for Path with maximum average   // value   using     System  ;   class     GFG     {          // method returns maximum average of all      // path of cost matrix      public     static     double     maxAverageOfPath  (  int     []  cost        int     N  )      {      int     []  dp     =     new     int  [  N  +  1    N  +  1  ];      dp  [  0    0  ]     =     cost  [  0    0  ];          /* Initialize first column of total cost(dp)    array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i       0  ]     =     dp  [  i     -     1    0  ]     +     cost  [  i       0  ];          /* Initialize first row of dp array */      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0       j  ]     =     dp  [  0    j     -     1  ]     +     cost  [  0       j  ];          /* Construct rest of the dp array */      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i       j  ]     =     Math  .  Max  (  dp  [  i     -     1       j  ]      dp  [  i    j     -     1  ])     +     cost  [  i       j  ];          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  dp  [  N     -     1       N     -     1  ]     /     (  2     *     N     -     1  );      }          // Driver Code      public     static     void     Main  ()         {      int     []  cost     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }};          Console  .  Write  (  maxAverageOfPath  (  cost       3  ));      }   }   // This code is contributed by nitin mittal.   
  JavaScript   
    <  script  >      // JavaScript Code for Path with maximum average value          // method returns maximum average of all      // path of cost matrix      function     maxAverageOfPath  (  cost       N  )      {      let     dp     =     new     Array  (  N  +  1  );      for     (  let     i     =     0  ;     i      <     N     +     1  ;     i  ++  )      {      dp  [  i  ]     =     new     Array  (  N     +     1  );      for     (  let     j     =     0  ;     j      <     N     +     1  ;     j  ++  )      {      dp  [  i  ][  j  ]     =     0  ;      }      }      dp  [  0  ][  0  ]     =     cost  [  0  ][  0  ];          /* Initialize first column of total cost(dp)    array */      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      dp  [  i  ][  0  ]     =     dp  [  i  -  1  ][  0  ]     +     cost  [  i  ][  0  ];          /* Initialize first row of dp array */      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  0  ][  j  ]     =     dp  [  0  ][  j  -  1  ]     +     cost  [  0  ][  j  ];          /* Construct rest of the dp array */      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      dp  [  i  ][  j  ]     =     Math  .  max  (  dp  [  i  -  1  ][  j  ]      dp  [  i  ][  j  -  1  ])     +     cost  [  i  ][  j  ];          // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     dp  [  N  -  1  ][  N  -  1  ]     /     (  2     *     N     -     1  );      }          let     cost     =     [[  1       2       3  ]      [  6       5       4  ]      [  7       3       9  ]];          document  .  write  (  maxAverageOfPath  (  cost       3  ));    <  /script>   
  PHP   
      // Php program to find maximum average cost path    // method returns maximum average of all path of    // cost matrix    function   maxAverageOfPath  (  $cost     $N  )   {   $dp   =   array  (  array  ())   ;   $dp  [  0  ][  0  ]   =   $cost  [  0  ][  0  ];   /* Initialize first column of total cost(dp) array */   for   (  $i   =   1  ;   $i    <   $N  ;   $i  ++  )   $dp  [  $i  ][  0  ]   =   $dp  [  $i  -  1  ][  0  ]   +   $cost  [  $i  ][  0  ];   /* Initialize first row of dp array */   for   (  $j   =   1  ;   $j    <   $N  ;   $j  ++  )   $dp  [  0  ][  $j  ]   =   $dp  [  0  ][  $j  -  1  ]   +   $cost  [  0  ][  $j  ];   /* Construct rest of the dp array */   for   (  $i   =   1  ;   $i    <   $N  ;   $i  ++  )   {   for   (  $j   =   1  ;   $j    <=   $N  ;   $j  ++  )   $dp  [  $i  ][  $j  ]   =   max  (  $dp  [  $i  -  1  ][  $j  ]  $dp  [  $i  ][  $j  -  1  ])   +   $cost  [  $i  ][  $j  ];   }   // divide maximum sum by constant path    // length : (2N - 1) for getting average    return   $dp  [  $N  -  1  ][  $N  -  1  ]   /   (  2  *  $N  -  1  );   }   // Driver code   $cost   =   array  (  array  (  1     2     3  )   array  (   6     5     4  )   array  (  7     3     9  )   )   ;   echo   maxAverageOfPath  (  $cost     3  )   ;   // This code is contributed by Ryuga   ?>   
  Python3   
   # Python program to find    # maximum average cost path   # Maximum number of rows    # and/or columns   M   =   100   # method returns maximum average of    # all path of cost matrix   def   maxAverageOfPath  (  cost     N  ):   dp   =   [[  0   for   i   in   range  (  N   +   1  )]   for   j   in   range  (  N   +   1  )]   dp  [  0  ][  0  ]   =   cost  [  0  ][  0  ]   # Initialize first column of total cost(dp) array   for   i   in   range  (  1     N  ):   dp  [  i  ][  0  ]   =   dp  [  i   -   1  ][  0  ]   +   cost  [  i  ][  0  ]   # Initialize first row of dp array   for   j   in   range  (  1     N  ):   dp  [  0  ][  j  ]   =   dp  [  0  ][  j   -   1  ]   +   cost  [  0  ][  j  ]   # Construct rest of the dp array   for   i   in   range  (  1     N  ):   for   j   in   range  (  1     N  ):   dp  [  i  ][  j  ]   =   max  (  dp  [  i   -   1  ][  j  ]   dp  [  i  ][  j   -   1  ])   +   cost  [  i  ][  j  ]   # divide maximum sum by constant path   # length : (2N - 1) for getting average   return   dp  [  N   -   1  ][  N   -   1  ]   /   (  2   *   N   -   1  )   # Driver program to test above function   cost   =   [[  1     2     3  ]   [  6     5     4  ]   [  7     3     9  ]]   print  (  maxAverageOfPath  (  cost     3  ))   # This code is contributed by Soumen Ghosh.   
    
  Produksjon   
5.200000  
 
    Tidskompleksitet    : O(N   2   ) for gitt inndata N   
    Hjelpeplass:    PÅ   2   ) for gitt inngang N.  
  
    Metode - 2: Uten å bruke Ekstra N*N mellomrom     
  
 
 Vi kan bruke input cost array som en dp for å lagre ans. så på denne måten trenger vi ikke en ekstra dp-array eller ikke den ekstra plassen.  
  
 
 En observasjon er at de eneste tillatte trekkene er nedover og til høyre trenger vi N-1 nedtrekk og N-1 høyretrekk for å nå målet (nederst til høyre). Så enhver bane fra øverste venstre hjørne til nederste høyre hjørne krever 2N - 1 celle. I    gjennomsnittlig    verdi nevneren er fast og vi trenger bare å maksimere telleren. Derfor må vi i utgangspunktet finne den maksimale sumveien. Å beregne maksimal sum av banen er et klassisk dynamisk programmeringsproblem. Vi trenger heller ikke noen prev cost[i][j] verdi etter å ha beregnet dp[i][j] slik at vi kan endre kostnads[i][j] verdien slik at vi ikke trenger ekstra plass for dp[i][j].  
  
 for all i 1  <= i  < N   
  cost[i][0] = cost[i-1][0] + cost[i][0];    
 for all j 1  <= j  < N   
  cost[0][j] = cost[0][j-1] + cost[0][j];    
 otherwise    
  cost[i][j] = max(cost[i-1][j] cost[i][j-1]) + cost[i][j];    
 
 Nedenfor er implementeringen av tilnærmingen ovenfor:  
 C++   
   // C++ program to find maximum average cost path   #include          using     namespace     std  ;   // Method returns maximum average of all path of cost matrix   double     maxAverageOfPath  (  vector   <  vector   <  int  >>  cost  )   {      int     N     =     cost  .  size  ();      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ];      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ];      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <=     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     max  (  cost  [  i     -     1  ][  j  ]     cost  [  i  ][  j     -     1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1  ][  N     -     1  ]     /     (  2     *     N     -     1  );   }   // Driver program   int     main  ()   {      vector   <  vector   <  int  >>     cost     =     {{  1       2       3  }      {  6       5       4  }      {  7       3       9  }      };      cout      < <     maxAverageOfPath  (  cost  );      return     0  ;   }   
  Java   
   // Java program to find maximum average cost path   import     java.io.*  ;   class   GFG     {      // Method returns maximum average of all path of cost      // matrix      static     double     maxAverageOfPath  (  int  [][]     cost  )      {      int     N     =     cost  .  length  ;      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ]  ;      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ]  ;      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     Math  .  max  (  cost  [  i     -     1  ][  j  ]        cost  [  i  ][  j     -     1  ]  )      +     cost  [  i  ][  j  ]  ;      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1  ][  N     -     1  ]     /     (  2     *     N     -     1  );      }      // Driver program      public     static     void     main  (  String  []     args  )      {      int  [][]     cost      =     {     {     1       2       3     }     {     6       5       4     }     {     7       3       9     }     };      System  .  out  .  println  (  maxAverageOfPath  (  cost  ));      }   }   // This code is contributed by karandeep1234   
  C#   
   // C# program to find maximum average cost path   using     System  ;   class     GFG     {      // Method returns maximum average of all path of cost      // matrix      static     double     maxAverageOfPath  (  int  [     ]     cost  )      {      int     N     =     cost  .  GetLength  (  0  );      // Initialize first column of total cost array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i       0  ]     =     cost  [  i       0  ]     +     cost  [  i     -     1       0  ];      // Initialize first row of array      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0       j  ]     =     cost  [  0       j     -     1  ]     +     cost  [  0       j  ];      // Construct rest of the array      for     (  int     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  int     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  i       j  ]     =     Math  .  Max  (  cost  [  i     -     1       j  ]      cost  [  i       j     -     1  ])      +     cost  [  i       j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  double  )  cost  [  N     -     1       N     -     1  ]     /     (  2     *     N     -     1  );      }      // Driver program      static     void     Main  (  string  []     args  )      {      int  [     ]     cost      =     {     {     1       2       3     }     {     6       5       4     }     {     7       3       9     }     };      Console  .  WriteLine  (  maxAverageOfPath  (  cost  ));      }   }   // This code is contributed by karandeep1234   
  JavaScript   
   // Method returns maximum average of all path of cost matrix   function     maxAverageOfPath  (  cost  )   {      let     N     =     cost  .  length  ;      // Initialize first column of total cost array      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      cost  [  i  ][  0  ]     =     cost  [  i  ][  0  ]     +     cost  [  i     -     1  ][  0  ];      // Initialize first row of array      for     (  let     j     =     1  ;     j      <     N  ;     j  ++  )      cost  [  0  ][  j  ]     =     cost  [  0  ][  j     -     1  ]     +     cost  [  0  ][  j  ];      // Construct rest of the array      for     (  let     i     =     1  ;     i      <     N  ;     i  ++  )      for     (  let     j     =     1  ;     j      <=     N  ;     j  ++  )      cost  [  i  ][  j  ]     =     Math  .  max  (  cost  [  i     -     1  ][  j  ]     cost  [  i  ][  j     -     1  ])     +     cost  [  i  ][  j  ];      // divide maximum sum by constant path      // length : (2N - 1) for getting average      return     (  cost  [  N     -     1  ][  N     -     1  ])     /     (  2.0     *     N     -     1  );   }   // Driver program   let     cost     =     [[  1       2       3  ]      [  6       5       4  ]      [  7       3       9  ]];   console  .  log  (  maxAverageOfPath  (  cost  ))   // This code is contributed by karandeep1234.   
  Python3   
   # Python program to find maximum average cost path   from   typing   import   List   def   maxAverageOfPath  (  cost  :   List  [  List  [  int  ]])   ->   float  :   N   =   len  (  cost  )   # Initialize first column of total cost array   for   i   in   range  (  1     N  ):   cost  [  i  ][  0  ]   =   cost  [  i  ][  0  ]   +   cost  [  i   -   1  ][  0  ]   # Initialize first row of array   for   j   in   range  (  1     N  ):   cost  [  0  ][  j  ]   =   cost  [  0  ][  j   -   1  ]   +   cost  [  0  ][  j  ]   # Construct rest of the array   for   i   in   range  (  1     N  ):   for   j   in   range  (  1     N  ):   cost  [  i  ][  j  ]   =   max  (  cost  [  i   -   1  ][  j  ]   cost  [  i  ][  j   -   1  ])   +   cost  [  i  ][  j  ]   # divide maximum sum by constant path   # length : (2N - 1) for getting average   return   cost  [  N   -   1  ][  N   -   1  ]   /   (  2   *   N   -   1  )   # Driver program   def   main  ():   cost   =   [[  1     2     3  ]   [  6     5     4  ]   [  7     3     9  ]]   print  (  maxAverageOfPath  (  cost  ))   if   __name__   ==   '__main__'  :   main  ()   
    
  Produksjon   
5.2  
 
    Tidskompleksitet:    O(N*N)   
    Hjelpeplass:    O(1)