Givet en array som representerar element av geometrisk progression i ordning. Ett element saknas i progressionen, hitta det saknade numret. Det kan antas att en term alltid saknas och att den saknade termen inte är den första eller sista i serien.
Exempel:
Input : arr[] = {1 3 27 81} Output : 9 Input : arr[] = {4 16 64 1024}; Output : 256 A Enkel lösning är att linjärt korsa matrisen och hitta det saknade talet. Tidskomplexiteten för denna lösning är O(n).
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En effektiv lösning för att lösa detta problem på O(Log n) tid med hjälp av binär sökning. Tanken är att gå till mittelementet. Kontrollera om förhållandet mellan mitten och intill mitten är lika med det vanliga förhållandet eller inte, om inte så ligger det saknade elementet mellan mitten och mitten+1. Om mittelementet är lika med n/2:e termen i Geometric Series (låt n vara antalet element i inmatningsmatrisen) så ligger det saknade elementet i höger halva. Annat element ligger i vänster halva.
Genomförande:
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>
Produktion
16
Tidskomplexitet: O(logga)
Hjälputrymme: O(logga)
Obs: Nackdelar med den här lösningen är: För större värden eller för större array kan det orsaka översvämning och/eller kan ta längre tid att starta datorn.
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