Donada una matriu que representa elements de progressió geomètrica en ordre. Falta un element a la progressió trobar el número que falta. Es pot suposar que sempre falta un terme i el terme que falta no és el primer ni l'últim de la sèrie.
Exemples:
Input : arr[] = {1 3 27 81} Output : 9 Input : arr[] = {4 16 64 1024}; Output : 256 A Solució senzilla és recórrer linealment la matriu i trobar el nombre que falta. La complexitat temporal d'aquesta solució és O(n).
desavantatges de la banca en línia
An solució eficient per resoldre aquest problema en temps O(Log n) mitjançant la cerca binària. La idea és anar a l'element mitjà. Comproveu si la proporció del mig i al costat del mig és igual a la relació comuna o no, si no, l'element que falta es troba entre mig i mig+1. Si l'element central és igual a n/2è terme de la sèrie geomètrica (sigui n el nombre d'elements de la matriu d'entrada), l'element que falta es troba a la meitat dreta. Un altre element es troba a la meitat esquerra.
Implementació:
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>
Sortida
16
Complexitat temporal: O (inici de sessió)
Espai auxiliar: O (inici de sessió)
Nota: Els inconvenients d'aquesta solució són: Per a valors més grans o per a una matriu més gran, pot provocar un desbordament i/o pot trigar més temps a les potències de l'ordinador.
primer cerca de profunditat de l'algorisme
Crea un qüestionari