تحقق مما إذا كان من الممكن تشكيل التقدم الحسابي من المصفوفة المحددة

تحقق مما إذا كان من الممكن تشكيل التقدم الحسابي من المصفوفة المحددة
جربه على ممارسة GfG

نظرا لمجموعة من ن الأعداد الصحيحة. وتتمثل المهمة في التحقق مما إذا كان من الممكن تشكيل تقدم حسابي باستخدام جميع العناصر المحددة. إذا أمكن، اطبع "نعم" وإلا اطبع "لا".

أمثلة:  

مدخل : آر[] = {0 12 4 8}
الإخراج : نعم
أعد ترتيب المصفوفة المحددة كـ {0 4 8 12} والتي تشكل تقدمًا حسابيًا.

مدخل : آر[] = {12 40 11 20}
الإخراج : لا

استخدام الفرز - وقت O(n Log n).

والفكرة هي فرز مجموعة معينة. بعد الفرز، تحقق مما إذا كانت الاختلافات بين العناصر المتتالية متماثلة أم لا. إذا كانت جميع الاختلافات هي نفس التقدم الحسابي فمن الممكن. يرجى الرجوع - برنامج للتحقق من التقدم الحسابي لتنفيذ هذا النهج.

استخدام فرز العد - O(n) Time وO(n) Space

يمكننا تقليل المساحة المطلوبة في الطريقة الثالثة إذا كان من الممكن تعديل المصفوفة المحددة. 

  1. العثور على أصغر العناصر وثاني أصغر العناصر.
  2. أوجد d = Second_smallest - الأصغر
  3. اطرح أصغر عنصر من جميع العناصر.
  4. الآن، إذا كانت المصفوفة المعطاة تمثل AP، فيجب أن تكون جميع العناصر من النموذج i*d حيث يتراوح i من 0 إلى n-1.
  5. قم بتقسيم جميع العناصر المختزلة واحدًا تلو الآخر باستخدام d. إذا كان أي عنصر غير قابل للقسمة على d، فسيتم إرجاع خطأ.
  6. الآن، إذا كانت المصفوفة تمثل AP، فيجب أن تكون تبديلًا للأرقام من 0 إلى n-1. يمكننا التحقق من ذلك بسهولة باستخدام الفرز العدي.

وفيما يلي تنفيذ هذه الطريقة:

C++
   // C++ program to check if a given array   // can form arithmetic progression   #include          using     namespace     std  ;   // Checking if array is permutation    // of 0 to n-1 using counting sort   bool     countingsort  (  int     arr  []     int     n  )   {      int     count  [  n  ]     =     {     0     };          // Counting the frequency      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      count  [  arr  [  i  ]]  ++  ;      }          // Check if each frequency is 1 only      for     (  int     i     =     0  ;     i      <=     n  -1  ;     i  ++  )     {      if     (  count  [  i  ]     !=     1  )      return     false  ;      }          return     true  ;   }   // Returns true if a permutation of arr[0..n-1]   // can form arithmetic progression   bool     checkIsAP  (  int     arr  []     int     n  )   {      int     smallest     =     INT_MAX       second_smallest     =     INT_MAX  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {          // Find the smallest and       // update second smallest      if     (  arr  [  i  ]      <     smallest  )     {      second_smallest     =     smallest  ;      smallest     =     arr  [  i  ];      }          // Find second smallest      else     if     (  arr  [  i  ]     !=     smallest      &&     arr  [  i  ]      <     second_smallest  )      second_smallest     =     arr  [  i  ];      }      // Find the difference between smallest and second      // smallest      int     diff     =     second_smallest     -     smallest  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      arr  [  i  ]  =  arr  [  i  ]  -  smallest  ;      }          for  (  int     i  =  0  ;  i   <  n  ;  i  ++  )      {      if  (  arr  [  i  ]  %  diff  !=  0  )      {      return     false  ;      }      else      {      arr  [  i  ]  =  arr  [  i  ]  /  diff  ;      }      }          // If array represents AP it must be a       // permutation of numbers from 0 to n-1.      // Check this using counting sort.      if  (  countingsort  (  arr    n  ))      return     true  ;      else      return     false  ;   }   // Driven Program   int     main  ()   {      int     arr  []     =     {     20       15       5       0       10     };      int     n     =     sizeof  (  arr  )     /     sizeof  (  arr  [  0  ]);      (  checkIsAP  (  arr       n  ))     ?     (  cout      < <     'Yes'      < <     endl  )      :     (  cout      < <     'No'      < <     endl  );      return     0  ;      // This code is contributed by Pushpesh Raj   }   
Java
   // Java program to check if a given array   // can form arithmetic progression   import     java.io.*  ;   class   GFG     {      // Checking if array is permutation      // of 0 to n-1 using counting sort      static     boolean     countingsort  (  int     arr  []       int     n  )      {      int  []     count     =     new     int  [  n  ]  ;      for  (  int     i     =     0  ;     i      <     n  ;     i  ++  )      count  [  i  ]     =     0  ;      // Counting the frequency      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      count  [  arr  [  i  ]]++  ;      }      // Check if each frequency is 1 only      for     (  int     i     =     0  ;     i      <=     n  -  1  ;     i  ++  )     {      if     (  count  [  i  ]     !=     1  )      return     false  ;      }      return     true  ;      }      // Returns true if a permutation of arr[0..n-1]      // can form arithmetic progression      static     boolean     checkIsAP  (  int     arr  []       int     n  )      {      int     smallest     =     Integer  .  MAX_VALUE       second_smallest     =     Integer  .  MAX_VALUE     ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      // Find the smallest and      // update second smallest      if     (  arr  [  i  ]      <     smallest  )     {      second_smallest     =     smallest  ;      smallest     =     arr  [  i  ]  ;      }      // Find second smallest      else     if     (  arr  [  i  ]     !=     smallest      &&     arr  [  i  ]      <     second_smallest  )      second_smallest     =     arr  [  i  ]  ;      }      // Find the difference between smallest and second      // smallest      int     diff     =     second_smallest     -     smallest  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      arr  [  i  ]     =     arr  [  i  ]     -     smallest  ;      }      for  (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      if  (  arr  [  i  ]     %     diff     !=     0  )      {      return     false  ;      }      else      {      arr  [  i  ]     =     arr  [  i  ]/  diff  ;      }      }      // If array represents AP it must be a      // permutation of numbers from 0 to n-1.      // Check this using counting sort.      if  (  countingsort  (  arr    n  ))      return     true  ;      else      return     false  ;      }      // Driven Program      public     static     void     main     (  String  []     args  )      {      int     arr  []     =     {     20       15       5       0       10     };      int     n     =     arr  .  length  ;      if  (  checkIsAP  (  arr       n  ))         System  .  out  .  println  (  'Yes'  );      else     System  .  out  .  println  (  'No'  );      }   }   // This code is contributed by Utkarsh   
Python
   # Python program to check if a given array   # can form arithmetic progression   import   sys   # Checking if array is permutation    # of 0 to n-1 using counting sort   def   countingsort  (   arr     n  ):   count   =   [  0  ]  *  n  ;   # Counting the frequency   for   i   in   range  (  0     n  ):   count  [  arr  [  i  ]]   +=   1  ;   # Check if each frequency is 1 only   for   i   in   range  (  0     n   -   1  ):   if   (  count  [  i  ]   !=   1  ):   return   False  ;   return   True  ;   # Returns true if a permutation of arr[0..n-1]   # can form arithmetic progression   def   checkIsAP  (   arr     n  ):   smallest   =   sys  .  maxsize  ;   second_smallest   =   sys  .  maxsize  ;   for   i   in   range  (  0    n  ):   # Find the smallest and    # update second smallest   if   (  arr  [  i  ]    <   smallest  )   :   second_smallest   =   smallest  ;   smallest   =   arr  [  i  ];   # Find second smallest   elif   (  arr  [  i  ]   !=   smallest   and   arr  [  i  ]    <   second_smallest  ):   second_smallest   =   arr  [  i  ];   # Find the difference between smallest and second   # smallest   diff   =   second_smallest   -   smallest  ;   for   i   in   range  (  0    n  ):   arr  [  i  ]  =  arr  [  i  ]  -  smallest  ;   for   i   in   range  (  0    n  ):   if  (  arr  [  i  ]  %  diff  !=  0  ):   return   False  ;   else  :   arr  [  i  ]  =  (  int  )(  arr  [  i  ]  /  diff  );   # If array represents AP it must be a    # permutation of numbers from 0 to n-1.   # Check this using counting sort.   if  (  countingsort  (  arr    n  )):   return   True  ;   else  :   return   False  ;   # Driven Program   arr   =   [   20     15     5     0     10   ];   n   =   len  (  arr  );   if  (  checkIsAP  (  arr     n  )):   print  (  'Yes'  );   else  :   print  (  'NO'  );   # This code is contributed by ratiagrawal.   
C#
   using     System  ;      class     GFG      {      // Checking if array is permutation      // of 0 to n-1 using counting sort      static     bool     CountingSort  (  int  []     arr       int     n  )      {      // Counting the frequency      int  []     count     =     new     int  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      count  [  arr  [  i  ]]  ++  ;      }      // Check if each frequency is 1 only      for     (  int     i     =     0  ;     i      <=     n     -     1  ;     i  ++  )      {      if     (  count  [  i  ]     !=     1  )      {      return     false  ;      }      }      return     true  ;      }  // Returns true if a permutation of arr[0..n-1]      // can form arithmetic progression      static     bool     CheckIsAP  (  int  []     arr       int     n  )      {  // Find the smallest and      // update second smallest      int     smallest     =     int  .  MaxValue  ;      int     secondSmallest     =     int  .  MaxValue  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      if     (  arr  [  i  ]      <     smallest  )      {      secondSmallest     =     smallest  ;      smallest     =     arr  [  i  ];      }      else     if     (  arr  [  i  ]     !=     smallest     &&     arr  [  i  ]      <     secondSmallest  )      {      secondSmallest     =     arr  [  i  ];      }      }      int     diff     =     secondSmallest     -     smallest  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      arr  [  i  ]     =     arr  [  i  ]     -     smallest  ;      }      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      if     (  arr  [  i  ]     %     diff     !=     0  )      {      return     false  ;      }      else      {      arr  [  i  ]     =     arr  [  i  ]     /     diff  ;      }      }   // If array represents AP it must be a      // permutation of numbers from 0 to n-1.      // Check this using counting sort.      if     (  CountingSort  (  arr       n  ))      {      return     true  ;      }      else      {      return     false  ;      }      }   // Driven Program      static     void     Main  (  string  []     args  )      {      int  []     arr     =     new     int  []     {     20       15       5       0       10     };      int     n     =     arr  .  Length  ;      Console  .  WriteLine  (  CheckIsAP  (  arr       n  )     ?     'Yes'     :     'No'  );      }      }   
JavaScript
   // Javascript program to check if a given array   // can form arithmetic progression   // Checking if array is permutation    // of 0 to n-1 using counting sort   function     countingsort  (     arr       n  )   {      let     count  =  new     Array  (  n  ).  fill  (  0  );          // Counting the frequency      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      count  [  arr  [  i  ]]  ++  ;      }          // Check if each frequency is 1 only      for     (  let     i     =     0  ;     i      <=     n  -  1  ;     i  ++  )     {      if     (  count  [  i  ]     !=     1  )      return     false  ;      }          return     true  ;   }   // Returns true if a permutation of arr[0..n-1]   // can form arithmetic progression   function     checkIsAP  (     arr       n  )   {      let     smallest     =     Number  .  MAX_SAFE_INTEGER       second_smallest     =     Number  .  MAX_SAFE_INTEGER  ;      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {          // Find the smallest and       // update second smallest      if     (  arr  [  i  ]      <     smallest  )     {      second_smallest     =     smallest  ;      smallest     =     arr  [  i  ];      }          // Find second smallest      else     if     (  arr  [  i  ]     !=     smallest      &&     arr  [  i  ]      <     second_smallest  )      second_smallest     =     arr  [  i  ];      }      // Find the difference between smallest and second      // smallest      let     diff     =     second_smallest     -     smallest  ;      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      arr  [  i  ]  =  arr  [  i  ]  -  smallest  ;      }          for  (  let     i  =  0  ;  i   <  n  ;  i  ++  )      {      if  (  arr  [  i  ]  %  diff  !=  0  )      {      return     false  ;      }      else      {      arr  [  i  ]  =  arr  [  i  ]  /  diff  ;      }      }          // If array represents AP it must be a       // permutation of numbers from 0 to n-1.      // Check this using counting sort.      if  (  countingsort  (  arr    n  ))      return     true  ;      else      return     false  ;   }   // Driven Program   let     arr     =     [  20       15       5       0       10     ];   let     n     =     arr  .  length  ;   (  checkIsAP  (  arr       n  ))     ?     (  console  .  log  (  'Yesn'  ))      :     (  console  .  log  (  'Non'  ));          // // This code was contributed by poojaagrawal2.   

الإخراج
Yes 

تعقيد الوقت - O(n) 
المساحة المساعدة - O(n)

التجزئة مع مرور واحد - O(n) الوقت وO(n) الفضاء

الفكرة الأساسية هي إيجاد الفرق المشترك بين AP من خلال معرفة الحد الأقصى والحد الأدنى لعنصر المصفوفة. بعد ذلك ابدأ من القيمة القصوى واستمر في تقليل القيمة بالفرق المشترك مع التأكد من وجود هذه القيمة الجديدة في الهاشماب أم لا. إذا لم تكن القيمة موجودة في مجموعة التجزئة في أي وقت، فقم بكسر الحلقة. الوضع المثالي بعد كسر الحلقة هو أن جميع العناصر n قد تمت تغطيتها، وإذا كانت الإجابة بنعم، فارجع صحيحًا وإلا قم بإرجاع خطأ. 

C++
   // C++ program for above approach   #include          using     namespace     std  ;   bool     checkIsAP  (  int     arr  []     int     n  )   {      unordered_set   <  int  >     st  ;      int     maxi     =     INT_MIN  ;      int     mini     =     INT_MAX  ;      for     (  int     i  =  0  ;  i   <  n  ;  i  ++  )     {      maxi     =     max  (  arr  [  i  ]     maxi  );      mini     =     min  (  arr  [  i  ]     mini  );      st  .  insert  (  arr  [  i  ]);      }          // FINDING THE COMMON DIFFERENCE      int     diff     =     (  maxi     -     mini  )     /     (  n     -     1  );      int     count     =     0  ;      // CHECK TERMS OF AP PRESENT IN THE HASHSET      while     (  st  .  find  (  maxi  )  !=  st  .  end  ())     {      count  ++  ;      maxi     =     maxi     -     diff  ;      }          if     (  count     ==     n  )      return     true  ;      return     false  ;   }   // Driver Code   int     main  ()   {      int     arr  []     =     {     0       12       4       8     };      int     n     =     4  ;      cout      < <     boolalpha      < <     checkIsAP  (  arr       n  );      return     0  ;   }   // This code is contributed by Rohit Pradhan   
Java
   /*package whatever //do not write package name here */   import     java.io.*  ;   import     java.util.*  ;   class   GFG     {      public     static     void     main  (  String  []     args  )      {      int  []     arr     =     {     0       12       4       8     };      int     n     =     arr  .  length  ;      System  .  out  .  println  (  checkIsAP  (  arr       n  ));      }      static     boolean     checkIsAP  (  int     arr  []       int     n  )      {      HashSet   <  Integer  >     set     =     new     HashSet   <  Integer  >  ();      int     max     =     Integer  .  MIN_VALUE  ;      int     min     =     Integer  .  MAX_VALUE  ;      for     (  int     i     :     arr  )     {      max     =     Math  .  max  (  i       max  );      min     =     Math  .  min  (  i       min  );      set  .  add  (  i  );      }          // FINDING THE COMMON DIFFERENCE      int     diff     =     (  max     -     min  )     /     (  n     -     1  );      int     count     =     0  ;      // CHECK IF TERMS OF AP PRESENT IN THE HASHSET       while     (  set  .  contains  (  max  ))     {      count  ++  ;      max     =     max     -     diff  ;      }      if     (  count     ==     arr  .  length  )      return     true  ;      return     false  ;      }   }   
Python
   import   sys   def   checkIsAP  (  arr     n  ):   Set   =   set  ()   Max   =   -  sys  .  maxsize   -   1   Min   =   sys  .  maxsize   for   i   in   arr  :   Max   =   max  (  i     Max  )   Min   =   min  (  i     Min  )   Set  .  add  (  i  )   # FINDING THE COMMON DIFFERENCE   diff   =   (  Max   -   Min  )   //   (  n   -   1  )   count   =   0   # CHECK IF TERMS OF AP PRESENT IN THE HASHSET    while   (  Max   in   Set  ):   count   +=   1   Max   =   Max   -   diff   if   (  count   ==   len  (  arr  )):   return   True   return   False   # driver code   arr   =   [   0     12     4     8   ]   n   =   len  (  arr  )   print  (  checkIsAP  (  arr     n  ))   # This code is contributed by shinjanpatra   
C#
   using     System  ;   using     System.Collections.Generic  ;   public     class     GFG      {      // C# program for above approach      static     bool     checkIsAP  (  int  []     arr       int     n  )      {      HashSet   <  int  >     st     =     new     HashSet   <  int  >  ();      int     maxi     =     int  .  MinValue  ;      int     mini     =     int  .  MaxValue  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )     {      maxi     =     Math  .  Max  (  arr  [  i  ]     maxi  );      mini     =     Math  .  Min  (  arr  [  i  ]     mini  );      st  .  Add  (  arr  [  i  ]);      }          // FINDING THE COMMON DIFFERENCE      int     diff     =     (  maxi     -     mini  )     /     (  n     -     1  );      int     count     =     0  ;      // CHECK IF TERMS OF AP PRESENT IN THE HASHSET       while     (  st  .  Contains  (  maxi  ))     {      count  ++  ;      maxi     =     maxi     -     diff  ;      }      if     (  count     ==     n  )     {      return     true  ;      }      return     false  ;      }      // Driver Code      internal     static     void     Main  ()      {      int  []     arr     =     {     0       12       4       8     };      int     n     =     4  ;      Console  .  Write  (  checkIsAP  (  arr       n  ));      }      // This code is contributed by Aarti_Rathi   }   
JavaScript
   function     checkIsAP  (  arr       n  ){      set     =     new     Set  ()      let     Max     =     Number  .  MIN_VALUE      let     Min     =     Number  .  MAX_VALUE      for  (  let     i     of     arr  ){      Max     =     Math  .  max  (  i       Max  )      Min     =     Math  .  min  (  i       Min  )      set  .  add  (  i  )      }          // FINDING THE COMMON DIFFERENCE      let     diff     =     Math  .  floor  ((  Max     -     Min  )     /     (  n     -     1  ))      let     count     =     0      // CHECK IF TERMS OF AP PRESENT IN THE HASHSET       while     (  set  .  has  (  Max  )){      count     +=     1      Max     =     Max     -     diff      }      if     (  count     ==     arr  .  length  )      return     true      return     false   }   // driver code   let     arr     =     [     0       12       4       8     ]   let     n     =     arr  .  length   console  .  log  (  checkIsAP  (  arr       n  ))   

الإخراج
true 
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