Dots masīvs, kas attēlo ģeometriskās progresijas elementus secībā. Progresijā trūkst viena elementa, atrodiet trūkstošo skaitli. Var pieņemt, ka vienmēr trūkst viena termina un trūkstošais termins nav pirmais vai pēdējais no sērijas.
Piemēri:
Input : arr[] = {1 3 27 81} Output : 9 Input : arr[] = {4 16 64 1024}; Output : 256 A Vienkāršs risinājums ir lineāri šķērsot masīvu un atrast trūkstošo skaitli. Šī risinājuma laika sarežģītība ir O(n).
unix augšējā komanda
An efektīvs risinājums lai atrisinātu šo problēmu O(Log n) laikā, izmantojot bināro meklēšanu. Ideja ir pāriet uz vidējo elementu. Pārbaudiet, vai vidējā un blakus vidējā attiecība ir vienāda ar kopējo attiecību vai nē, ja nē, tad trūkstošais elements atrodas starp vidu un vidu +1. Ja vidējais elements ir vienāds ar n/2. ģeometriskās sērijas vienību (lai n ir elementu skaits ievades masīvā), tad trūkstošais elements atrodas labajā pusē. Cits elements atrodas kreisajā pusē.
Īstenošana:
C++// C++ program to find missing number in // geometric progression #include
// 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 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# 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 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 ?> <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>
Izvade
16
Laika sarežģītība: O(pieteikties)
Palīgtelpa: O(pieteikties)
Piezīme: Šī risinājuma trūkumi ir šādi: lielākām vērtībām vai lielākam masīvam tas var izraisīt pārplūdi un/vai var aizņemt vairāk laika datora jaudai.
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