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기하학적 진행에서 누락된 숫자 찾기

기하학적 수열 요소를 순서대로 나타내는 배열이 제공됩니다. 진행 중 하나의 요소가 누락되어 누락된 숫자를 찾습니다. 하나의 항이 항상 결측되어 있고 결측된 항이 계열의 첫 번째 또는 마지막 항이 아니라고 가정할 수 있습니다.

예:  

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

에이 간단한 솔루션 배열을 선형적으로 탐색하여 누락된 숫자를 찾는 것입니다. 이 솔루션의 시간 복잡도는 O(n)입니다.



.다음 자바

효율적인 솔루션 이진 검색을 사용하여 O(Log n) 시간 안에 이 문제를 해결합니다. 아이디어는 중간 요소로 이동하는 것입니다. 중간과 중앙 옆의 비율이 공통 비율과 같은지 확인하고 그렇지 않은 경우 누락된 요소는 중간과 중간+1 사이에 있습니다. 중간 요소가 기하 급수에서 n/2번째 항과 같은 경우(n을 입력 배열의 요소 수로 가정) 누락된 요소는 오른쪽 절반에 놓입니다. 다른 요소는 왼쪽 절반에 있습니다.

구현:

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>  

산출
16

시간 복잡도: 오(로그인)

보조 공간: 오(로그인)

메모 : 이 솔루션의 단점은 다음과 같습니다. 값이 크거나 배열이 크면 오버플로가 발생하거나 컴퓨터 전원 공급에 더 많은 시간이 걸릴 수 있습니다.

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