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Etsi puuttuva luku geometriasta etenemisestä

Annettu taulukko, joka edustaa geometrisen etenemisen elementtejä järjestyksessä. Etenemisestä puuttuu yksi elementti, etsi puuttuva numero. Voidaan olettaa, että yksi termi puuttuu aina ja puuttuva termi ei ole sarjan ensimmäinen tai viimeinen.

Esimerkkejä:  

Input : arr[] = {1 3  27 81} Output : 9 Input : arr[] = {4 16 64 1024}; Output : 256

A Yksinkertainen Ratkaisu on kulkea lineaarisesti taulukon läpi ja löytää puuttuva luku. Tämän ratkaisun aikamonimutkaisuus on O(n).



istunto on vanhentunut

An tehokas ratkaisu ratkaista tämä ongelma O(Log n) -ajassa käyttämällä binaarihakua. Ajatuksena on siirtyä keskielementtiin. Tarkista, onko keski- ja vieressä-keskisuhde yhtä suuri kuin yhteinen suhde vai ei, jos ei, puuttuva elementti on mid- ja mid+1 välissä. Jos keskimmäinen elementti on yhtä suuri kuin n/2. termi Geometrisessa sarjassa (olkoon n syöttötaulukon elementtien lukumäärä), puuttuva elementti on oikealla puoliskolla. Muu elementti sijaitsee vasemmalla puoliskolla.

Toteutus:

C++
// C++ program to find missing number in // geometric progression #include    using namespace std; // It returns INT_MAX in case of error int findMissingRec(int arr[] int low  int high int ratio) {  if (low >= high)  return INT_MAX;  int mid = low + (high - low)/2;  // If element next to mid is missing  if (arr[mid+1]/arr[mid] != ratio)  return (arr[mid] * ratio);  // If element previous to mid is missing  if ((mid > 0) && (arr[mid]/arr[mid-1]) != ratio)  return (arr[mid-1] * ratio);  // If missing element is in right half  if (arr[mid] == arr[0] * (pow(ratio mid)) )  return findMissingRec(arr mid+1 high ratio);  return findMissingRec(arr low mid-1 ratio); } // Find ration and calls findMissingRec int findMissing(int arr[] int n) {  // Finding ration assuming that the missing term is  // not first or last term of series.  int ratio = (float) pow(arr[n-1]/arr[0] 1.0/n);  return findMissingRec(arr 0 n-1 ratio); } // Driver code int main(void) {  int arr[] = {2 4 8 32};  int n = sizeof(arr)/sizeof(arr[0]);  cout << findMissing(arr n);  return 0; } 
Java
// JAVA Code for Find the missing number // in Geometric Progression class GFG {    // It returns INT_MAX in case of error  public static int findMissingRec(int arr[] int low  int high int ratio)  {  if (low >= high)  return Integer.MAX_VALUE;  int mid = low + (high - low)/2;    // If element next to mid is missing  if (arr[mid+1]/arr[mid] != ratio)  return (arr[mid] * ratio);    // If element previous to mid is missing  if ((mid > 0) && (arr[mid]/arr[mid-1]) != ratio)  return (arr[mid-1] * ratio);    // If missing element is in right half  if (arr[mid] == arr[0] * (Math.pow(ratio mid)) )  return findMissingRec(arr mid+1 high ratio);    return findMissingRec(arr low mid-1 ratio);  }    // Find ration and calls findMissingRec  public static int findMissing(int arr[] int n)  {  // Finding ration assuming that the missing  // term is not first or last term of series.  int ratio =(int) Math.pow(arr[n-1]/arr[0] 1.0/n);    return findMissingRec(arr 0 n-1 ratio);  }     /* Driver program to test above function */  public static void main(String[] args)   {  int arr[] = {2 4 8 32};  int n = arr.length;    System.out.print(findMissing(arr n));  }  } // This code is contributed by Arnav Kr. Mandal. 
Python3
# Python3 program to find missing  # number in geometric progression # It returns INT_MAX in case of error def findMissingRec(arr low high ratio): if (low >= high): return 2147483647 mid = low + (high - low) // 2 # If element next to mid is missing if (arr[mid + 1] // arr[mid] != ratio): return (arr[mid] * ratio) # If element previous to mid is missing if ((mid > 0) and (arr[mid] / arr[mid-1]) != ratio): return (arr[mid - 1] * ratio) # If missing element is in right half if (arr[mid] == arr[0] * (pow(ratio mid)) ): return findMissingRec(arr mid+1 high ratio) return findMissingRec(arr low mid-1 ratio) # Find ration and calls findMissingRec def findMissing(arr n): # Finding ration assuming that  # the missing term is not first # or last term of series. ratio = int(pow(arr[n-1] / arr[0] 1.0 / n)) return findMissingRec(arr 0 n-1 ratio) # Driver code arr = [2 4 8 32] n = len(arr) print(findMissing(arr n)) # This code is contributed by Anant Agarwal. 
C#
// C# Code for Find the missing number // in Geometric Progression using System; class GFG {    // It returns INT_MAX in case of error  public static int findMissingRec(int []arr int low  int high int ratio)  {  if (low >= high)  return int.MaxValue;    int mid = low + (high - low)/2;    // If element next to mid is missing  if (arr[mid+1]/arr[mid] != ratio)  return (arr[mid] * ratio);    // If element previous to mid is missing  if ((mid > 0) && (arr[mid]/arr[mid-1]) != ratio)  return (arr[mid-1] * ratio);    // If missing element is in right half  if (arr[mid] == arr[0] * (Math.Pow(ratio mid)) )  return findMissingRec(arr mid+1 high ratio);    return findMissingRec(arr low mid-1 ratio);  }    // Find ration and calls findMissingRec  public static int findMissing(int []arr int n)  {    // Finding ration assuming that the missing  // term is not first or last term of series.  int ratio =(int) Math.Pow(arr[n-1]/arr[0] 1.0/n);    return findMissingRec(arr 0 n-1 ratio);  }     /* Driver program to test above function */  public static void Main()   {  int []arr = {2 4 8 32};  int n = arr.Length;    Console.Write(findMissing(arr n));  } } // This code is contributed by nitin mittal. 
PHP
 // PHP program to find missing number // in geometric progression // It returns INT_MAX in case of error function findMissingRec(&$arr $low $high $ratio) { if ($low >= $high) return PHP_INT_MAX; $mid = $low + intval(($high - $low) / 2); // If element next to mid is missing if ($arr[$mid+1]/$arr[$mid] != $ratio) return ($arr[$mid] * $ratio); // If element previous to mid is missing if (($mid > 0) && ($arr[$mid] / $arr[$mid - 1]) != $ratio) return ($arr[$mid - 1] * $ratio); // If missing element is in right half if ($arr[$mid] == $arr[0] * (pow($ratio $mid))) return findMissingRec($arr $mid + 1 $high $ratio); return findMissingRec($arr $low $mid - 1 $ratio); } // Find ration and calls findMissingRec function findMissing(&$arr $n) { // Finding ration assuming that the missing  // term is not first or last term of series. $ratio = (float) pow($arr[$n - 1] / $arr[0] 1.0 / $n); return findMissingRec($arr 0 $n - 1 $ratio); } // Driver code $arr = array(2 4 8 32); $n = sizeof($arr); echo findMissing($arr $n); // This code is contributed by ita_c ?> 
JavaScript
<script> // Javascript Code for Find the missing number // in Geometric Progression    // It returns INT_MAX in case of error  function findMissingRec(arrlowhighratio)  {  if (low >= high)  return Integer.MAX_VALUE;  let mid = Math.floor(low + (high - low)/2);    // If element next to mid is missing  if (arr[mid+1]/arr[mid] != ratio)  return (arr[mid] * ratio);    // If element previous to mid is missing  if ((mid > 0) && (arr[mid]/arr[mid-1]) != ratio)  return (arr[mid-1] * ratio);    // If missing element is in right half  if (arr[mid] == arr[0] * (Math.pow(ratio mid)) )  return findMissingRec(arr mid+1 high ratio);    return findMissingRec(arr low mid-1 ratio);  }    // Find ration and calls findMissingRec  function findMissing(arrn)  {  // Finding ration assuming that the missing  // term is not first or last term of series.  let ratio =Math.floor( Math.pow(arr[n-1]/arr[0] 1.0/n));    return findMissingRec(arr 0 n-1 ratio);  }    /* Driver program to test above function */  let arr=[2 4 8 32];  let n = arr.length;  document.write(findMissing(arr n));    // This code is contributed by rag2127   </script>  

Lähtö
16

Aika monimutkaisuus: O (kirjaudu sisään)

Aputila: O (kirjaudu sisään)

Huomautus: Tämän ratkaisun haittapuolia ovat: Suuremmille arvoille tai suuremmille matriisille se voi aiheuttaa ylivuotoa ja/tai voi viedä enemmän aikaa tietokoneen tehoon.

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