k日後のアクティブセルと非アクティブセル

サイズ n のバイナリ配列が与えられるとします。 n > 3 。配列内の true (または 1) 値はアクティブを意味し、false (または 0) は非アクティブを意味します。数値 k が与えられると、タスクは k 日後のアクティブなセルと非アクティブなセルの数を見つけることです。毎日の後、左右のセルが同じでない場合は i 番目のセルのステータスがアクティブになり、左右のセルが同じ (両方 0 または両方 1) の場合は非アクティブになります。 

左端のセルの前および右端のセルの後にセルがないため、左端のセルの前および右端のセルの後の値セルは常に 0 (または非アクティブ) とみなされます。

例:  

Input : cells[] = {1 0 1 1} k = 2 Output : Active cells = 3 Inactive cells = 1 After 1 day cells[] = {0 0 1 1} After 2 days cells[] = {0 1 1 1} Input : cells[] = {0 1 0 1 0 1 0 1} k = 3 Output: Active Cells = 2  Inactive Cells = 6 Explanation : After 1 day cells[] = {1 0 0 0 0 0 0 0} After 2 days cells[] = {0 1 0 0 0 0 0 0} After 3 days cells[] = {1 0 1 0 0 0 0 0} Input : cells[] = {0 1 1 1 0 1 1 0} k = 4 Output: Active Cells = 3  Inactive Cells = 5 

唯一重要なことは、翌日に更新するために以前の値が必要であるため、指定された配列のコピーを確実に保持することです。以下に詳細な手順を示します。 

  1. まず、cells[] 配列を temp[] 配列にコピーし、指定された条件に従って temp[] 配列を変更します。
  2. この条件では、i 番目のセルのすぐ左と右のセルが非アクティブまたはアクティブである場合、翌日 i は非アクティブになると仮定します。 (セル[i-1] == 0 およびセル[i+1] == 0) または (セル[i-1] == 1 およびセル[i+1] == 1) の場合、セル[i] = 0 これらの条件は、セル[i-1] とセル[i+1] の XOR を使用して適用できます。
  3. 0 番目のインデックス セルの場合は temp[0] = 0^cells[1]、(n-1) 番目のインデックス セルの場合は temp[n-1] = 0^cells[n-2] です。
  4. ここで、インデックス 1 から n-2 に対して次の操作を実行します。 temp[i] = cell[i-1] ^ cell[i+1]
  5. k 日が完了するまでこのプロセスを繰り返します。

以下は、上記の手順の実装です。 

C++
   // C++ program to count active and inactive cells after k   // days   #include       using     namespace     std  ;   // cells[] - store current status of cells   // n - Number of cells   // temp[] - to perform intermediate operations   // k - number of days   // active - count of active cells after k days   // inactive - count of active cells after k days   void     activeAndInactive  (  bool     cells  []     int     n       int     k  )   {      // copy cells[] array into temp [] array      bool     temp  [  n  ];      for     (  int     i  =  0  ;     i   <  n     ;     i  ++  )      temp  [  i  ]     =     cells  [  i  ];      // Iterate for k days      while     (  k  --  )      {      // Finding next values for corner cells      temp  [  0  ]     =     0  ^  cells  [  1  ];      temp  [  n  -1  ]     =     0  ^  cells  [  n  -2  ];      // Compute values of intermediate cells      // If both cells active or inactive then temp[i]=0      // else temp[i] = 1.      for     (  int     i  =  1  ;     i   <=  n  -2  ;     i  ++  )      temp  [  i  ]     =     cells  [  i  -1  ]     ^     cells  [  i  +  1  ];      // Copy temp[] to cells[] for next iteration      for     (  int     i  =  0  ;     i   <  n  ;     i  ++  )      cells  [  i  ]     =     temp  [  i  ];      }      // count active and inactive cells      int     active     =     0       inactive     =     0  ;      for     (  int     i  =  0  ;     i   <  n  ;     i  ++  )      (  cells  [  i  ]     ==     1  )  ?     active  ++     :     inactive  ++  ;      printf  (  'Active Cells = %d Inactive Cells = %d'        active       inactive  );   }   // Driver program to check the test case   int     main  ()   {      bool     cells  []     =     {  0       1       0       1       0       1       0       1  };      int     k     =     3  ;      int     n     =     sizeof  (  cells  )  /  sizeof  (  cells  [  0  ]);      activeAndInactive  (  cells       n       k  );      return     0  ;   }   
Java
   // Java program to count active and    // inactive cells after k days   class   GFG     {       // cells[] - store current status    // of cells n - Number of cells   // temp[] - to perform intermediate operations   // k - number of days   // active - count of active cells after k days   // inactive - count of active cells after k days   static     void     activeAndInactive  (  boolean     cells  []           int     n       int     k  )   {      // copy cells[] array into temp [] array      boolean     temp  []     =     new     boolean  [  n  ]  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      temp  [  i  ]     =     cells  [  i  ]  ;      // Iterate for k days      while     (  k  --     >     0  )     {          // Finding next values for corner cells      temp  [  0  ]     =     false     ^     cells  [  1  ]  ;      temp  [  n     -     1  ]     =     false     ^     cells  [  n     -     2  ]  ;      // Compute values of intermediate cells      // If both cells active or inactive then       // temp[i]=0 else temp[i] = 1.      for     (  int     i     =     1  ;     i      <=     n     -     2  ;     i  ++  )      temp  [  i  ]     =     cells  [  i     -     1  ]     ^     cells  [  i     +     1  ]  ;      // Copy temp[] to cells[] for next iteration      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      cells  [  i  ]     =     temp  [  i  ]  ;      }      // count active and inactive cells      int     active     =     0       inactive     =     0  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      if     (  cells  [  i  ]     ==     true  )      active  ++  ;      else      inactive  ++  ;      System  .  out  .  print  (  'Active Cells = '     +     active     +     ' '     +         'Inactive Cells = '     +     inactive  );   }   // Driver code   public     static     void     main  (  String  []     args  )      {      boolean     cells  []     =     {  false       true       false       true        false       true       false       true  };      int     k     =     3  ;      int     n     =     cells  .  length  ;      activeAndInactive  (  cells       n       k  );   }   }   // This code is contributed by Anant Agarwal.   
Python3
   # Python program to count   # active and inactive cells after k   # days   # cells[] - store current   # status of cells   # n - Number of cells   # temp[] - to perform   # intermediate operations   # k - number of days   # active - count of active   # cells after k days   # inactive - count of active   # cells after k days   def   activeAndInactive  (  cells    n    k  ):   # copy cells[] array into temp [] array   temp  =  []   for   i   in   range  (  n  +  1  ):   temp  .  append  (  False  )   for   i   in   range  (  n  ):   temp  [  i  ]   =   cells  [  i  ]   # Iterate for k days   while   (  k   >  0  ):   # Finding next values for corner cells   temp  [  0  ]   =   False  ^  cells  [  1  ]   temp  [  n  -  1  ]   =   False  ^  cells  [  n  -  2  ]   # Compute values of intermediate cells   # If both cells active or   # inactive then temp[i]=0   # else temp[i] = 1.   for   i   in   range  (  1    n  -  2  +  1  ):   temp  [  i  ]   =   cells  [  i  -  1  ]   ^   cells  [  i  +  1  ]   # Copy temp[] to cells[]   # for next iteration   for   i   in   range  (  n  ):   cells  [  i  ]   =   temp  [  i  ]   k  -=  1   # count active and inactive cells   active   =   0   inactive   =   0  ;   for   i   in   range  (  n  ):   if  (  cells  [  i  ]   ==   True  ):   active  +=  1   else  :   inactive  +=  1   print  (  'Active Cells ='    active    '  '      'Inactive Cells ='     inactive  )   # Driver code   cells   =   [  False     True     False     True     False     True     False     True  ]   k   =   3   n   =  len  (  cells  )   activeAndInactive  (  cells     n     k  )   # This code is contributed   # by Anant Agarwal.   
C#
   // C# program to count active and    // inactive cells after k days   using     System  ;   class     GFG     {       // cells[] - store current status    // of cells n - Number of cells   // temp[] - to perform intermediate    // operations k - number of days   // active - count of active cells    // after k days inactive - count   // of active cells after k days   static     void     activeAndInactive  (  bool     []  cells           int     n       int     k  )   {          // copy cells[] array into      // temp [] array      bool     []  temp     =     new     bool  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      temp  [  i  ]     =     cells  [  i  ];      // Iterate for k days      while     (  k  --     >     0  )     {          // Finding next values       // for corner cells      temp  [  0  ]     =     false     ^     cells  [  1  ];      temp  [  n     -     1  ]     =     false     ^     cells  [  n     -     2  ];      // Compute values of intermediate cells      // If both cells active or inactive then       // temp[i]=0 else temp[i] = 1.      for     (  int     i     =     1  ;     i      <=     n     -     2  ;     i  ++  )      temp  [  i  ]     =     cells  [  i     -     1  ]     ^     cells  [  i     +     1  ];      // Copy temp[] to cells[]       // for next iteration      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      cells  [  i  ]     =     temp  [  i  ];      }      // count active and inactive cells      int     active     =     0       inactive     =     0  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      if     (  cells  [  i  ]     ==     true  )      active  ++  ;      else      inactive  ++  ;      Console  .  Write  (  'Active Cells = '     +     active     +     ' '     +         'Inactive Cells = '     +     inactive  );   }   // Driver code   public     static     void     Main  ()      {      bool     []  cells     =     {  false       true       false       true        false       true       false       true  };      int     k     =     3  ;      int     n     =     cells  .  Length  ;      activeAndInactive  (  cells       n       k  );   }   }   // This code is contributed by Nitin Mittal.   
PHP
      // PHP program to count active    // and inactive cells after k   // days   // cells[] - store current status    // of cells n - Number of cells   // temp[] - to perform intermediate    // operations k - number of days   // active - count of active cells    // after k days inactive - count of   // active cells after k days   function   activeAndInactive  (  $cells     $n     $k  )   {   // copy cells[] array into   // temp [] array   $temp   =   array  ();   for   (  $i   =   0  ;   $i    <   $n   ;   $i  ++  )   $temp  [  $i  ]   =   $cells  [  $i  ];   // Iterate for k days   while   (  $k  --  )   {   // Finding next values    // for corner cells   $temp  [  0  ]   =   0   ^   $cells  [  1  ];   $temp  [  $n   -   1  ]   =   0   ^   $cells  [  $n   -   2  ];   // Compute values of    // intermediate cells   // If both cells active    // or inactive then temp[i]=0   // else temp[i] = 1.   for   (  $i   =   1  ;   $i    <=   $n   -   2  ;   $i  ++  )   $temp  [  $i  ]   =   $cells  [  $i   -   1  ]   ^   $cells  [  $i   +   1  ];   // Copy temp[] to cells[]    // for next iteration   for   (  $i   =   0  ;   $i    <   $n  ;   $i  ++  )   $cells  [  $i  ]   =   $temp  [  $i  ];   }   // count active and    // inactive cells   $active   =   0  ;  $inactive   =   0  ;   for   (  $i   =   0  ;   $i    <   $n  ;   $i  ++  )   (  $cells  [  $i  ]   ==   1  )  ?   $active  ++   :   $inactive  ++  ;   echo   'Active Cells = '     $active     ' Inactive Cells = '     $inactive  ;   }   // Driver Code   $cells  =   array  (  0     1     0     1     0     1     0     1  );   $k   =   3  ;   $n   =   count  (  $cells  );   activeAndInactive  (  $cells     $n     $k  );   // This code is contributed by anuj_67.   ?>   
JavaScript
    <  script  >   // javascript program to count active and    // inactive cells after k days      // cells - store current status      // of cells n - Number of cells      // temp - to perform intermediate operations      // k - number of days      // active - count of active cells after k days      // inactive - count of active cells after k days      function     activeAndInactive  (  cells          n          k  )         {          // copy cells array into temp array      var     temp     =     Array  (  n  ).  fill  (  false  );      for     (  i     =     0  ;     i      <     n  ;     i  ++  )      temp  [  i  ]     =     cells  [  i  ];      // Iterate for k days      while     (  k  --     >     0  )      {      // Finding next values for corner cells      temp  [  0  ]     =     false     ^     cells  [  1  ];      temp  [  n     -     1  ]     =     false     ^     cells  [  n     -     2  ];      // Compute values of intermediate cells      // If both cells active or inactive then      // temp[i]=0 else temp[i] = 1.      for     (  i     =     1  ;     i      <=     n     -     2  ;     i  ++  )      temp  [  i  ]     =     cells  [  i     -     1  ]     ^     cells  [  i     +     1  ];      // Copy temp to cells for next iteration      for     (  i     =     0  ;     i      <     n  ;     i  ++  )      cells  [  i  ]     =     temp  [  i  ];      }      // count active and inactive cells      var     active     =     0       inactive     =     0  ;      for     (  i     =     0  ;     i      <     n  ;     i  ++  )      if     (  cells  [  i  ]     ==     true  )      active  ++  ;      else      inactive  ++  ;      document  .  write  (  'Active Cells = '     +     active     +     ' '     +     'Inactive Cells = '     +     inactive  );      }      // Driver code      var     cells     =     [     false       true       false       true       false       true       false       true     ];      var     k     =     3  ;      var     n     =     cells  .  length  ;      activeAndInactive  (  cells       n       k  );   // This code is contributed by Rajput-Ji    <  /script>   

出力
Active Cells = 2 Inactive Cells = 6 

時間計算量: O(N*K) ここで、N は配列のサイズ、K は日数です。
補助スペース:O(N)

この記事は geeksforgeeks チームによってレビューされています。この問題に対するより良いアプローチがある場合は、共有してください。