Finden Sie zwei fehlende Zahlen | Satz 1 (Eine interessante lineare Zeitlösung)
Gegeben sei ein Array mit n eindeutigen Ganzzahlen, wobei jedes Element im Array im Bereich [1 n] liegt. Das Array enthält alle unterschiedlichen Elemente und die Größe des Arrays beträgt (n-2). Daher fehlen in diesem Array zwei Zahlen aus dem Bereich. Finden Sie die beiden fehlenden Zahlen.
Beispiele:
Input : arr[] = {1 3 5 6} Output : 2 4 Input : arr[] = {1 2 4} Output : 3 5 Input : arr[] = {1 2} Output : 3 4 Methode 1 – O(n) Zeitkomplexität und O(n) Extra Space
Schritt 1: Nehmen Sie ein boolesches Array markieren Das verfolgt alle im Array vorhandenen Elemente.
Schritt 2: Iterieren Sie von 1 bis n und prüfen Sie für jedes Element, ob es im booleschen Array als wahr markiert ist. Wenn nicht, zeigen Sie dieses Element einfach an.
// C++ Program to find two Missing Numbers using O(n) // extra space #include using namespace std ; // Function to find two missing numbers in range // [1 n]. This function assumes that size of array // is n-2 and all array elements are distinct void findTwoMissingNumbers ( int arr [] int n ) { // Create a boolean vector of size n+1 and // mark all present elements of arr[] in it. vector < bool > mark ( n + 1 false ); for ( int i = 0 ; i < n -2 ; i ++ ) mark [ arr [ i ]] = true ; // Print two unmarked elements cout < < 'Two Missing Numbers are n ' ; for ( int i = 1 ; i <= n ; i ++ ) if ( ! mark [ i ]) cout < < i < < ' ' ; cout < < endl ; } // Driver program to test above function int main () { int arr [] = { 1 3 5 6 }; // Range of numbers is 2 plus size of array int n = 2 + sizeof ( arr ) / sizeof ( arr [ 0 ]); findTwoMissingNumbers ( arr n ); return 0 ; }
Java // Java Program to find two Missing Numbers using O(n) // extra space import java.util.* ; class GFG { // Function to find two missing numbers in range // [1 n]. This function assumes that size of array // is n-2 and all array elements are distinct static void findTwoMissingNumbers ( int arr [] int n ) { // Create a boolean vector of size n+1 and // mark all present elements of arr[] in it. boolean [] mark = new boolean [ n + 1 ] ; for ( int i = 0 ; i < n - 2 ; i ++ ) mark [ arr [ i ]] = true ; // Print two unmarked elements System . out . println ( 'Two Missing Numbers are' ); for ( int i = 1 ; i <= n ; i ++ ) if ( ! mark [ i ] ) System . out . print ( i + ' ' ); System . out . println (); } // Driver code public static void main ( String [] args ) { int arr [] = { 1 3 5 6 }; // Range of numbers is 2 plus size of array int n = 2 + arr . length ; findTwoMissingNumbers ( arr n ); } } // This code is contributed by 29AjayKumar
Python3 # Python3 program to find two Missing Numbers using O(n) # extra space # Function to find two missing numbers in range # [1 n]. This function assumes that size of array # is n-2 and all array elements are distinct def findTwoMissingNumbers ( arr n ): # Create a boolean vector of size n+1 and # mark all present elements of arr[] in it. mark = [ False for i in range ( n + 1 )] for i in range ( 0 n - 2 1 ): mark [ arr [ i ]] = True # Print two unmarked elements print ( 'Two Missing Numbers are' ) for i in range ( 1 n + 1 1 ): if ( mark [ i ] == False ): print ( i end = ' ' ) print ( ' n ' ) # Driver program to test above function if __name__ == '__main__' : arr = [ 1 3 5 6 ] # Range of numbers is 2 plus size of array n = 2 + len ( arr ) findTwoMissingNumbers ( arr n ); # This code is contributed by # Surendra_Gangwar
C# // C# Program to find two Missing Numbers // using O(n) extra space using System ; using System.Collections.Generic ; class GFG { // Function to find two missing numbers in range // [1 n]. This function assumes that size of array // is n-2 and all array elements are distinct static void findTwoMissingNumbers ( int [] arr int n ) { // Create a boolean vector of size n+1 and // mark all present elements of arr[] in it. Boolean [] mark = new Boolean [ n + 1 ]; for ( int i = 0 ; i < n - 2 ; i ++ ) mark [ arr [ i ]] = true ; // Print two unmarked elements Console . WriteLine ( 'Two Missing Numbers are' ); for ( int i = 1 ; i <= n ; i ++ ) if ( ! mark [ i ]) Console . Write ( i + ' ' ); Console . WriteLine (); } // Driver code public static void Main ( String [] args ) { int [] arr = { 1 3 5 6 }; // Range of numbers is 2 plus size of array int n = 2 + arr . Length ; findTwoMissingNumbers ( arr n ); } } // This code contributed by Rajput-Ji
JavaScript < script > // Javascript Program to find two // Missing Numbers using O(n) extra space // Function to find two missing numbers in range // [1 n]. This function assumes that size of array // is n-2 and all array elements are distinct function findTwoMissingNumbers ( arr n ) { // Create a boolean vector of size n+1 and // mark all present elements of arr[] in it. let mark = new Array ( n + 1 ); for ( let i = 0 ; i < n - 2 ; i ++ ) mark [ arr [ i ]] = true ; // Print two unmarked elements document . write ( 'Two Missing Numbers are' + ' ' ); for ( let i = 1 ; i <= n ; i ++ ) if ( ! mark [ i ]) document . write ( i + ' ' ); document . write ( ' ' ); } let arr = [ 1 3 5 6 ]; // Range of numbers is 2 plus size of array let n = 2 + arr . length ; findTwoMissingNumbers ( arr n ); < /script>
Ausgabe
Two Missing Numbers are 2 4
Methode 2 – O(n) Zeitkomplexität und O(1) Extra Space
Die Idee basiert auf Das beliebte Lösung zum Auffinden einer fehlenden Nummer. Wir erweitern die Lösung so, dass zwei fehlende Elemente gedruckt werden.
Lassen Sie uns die Summe zweier fehlender Zahlen ermitteln:
arrSum => Sum of all elements in the array sum (Sum of 2 missing numbers) = (Sum of integers from 1 to n) - arrSum = ((n)*(n+1))/2 – arrSum avg (Average of 2 missing numbers) = sum / 2;
- Eine der Zahlen wird kleiner oder gleich sein Durchschn während der andere streng größer sein wird als Durchschn . Zwei Zahlen können niemals gleich sein, da alle gegebenen Zahlen verschieden sind.
- Wir können die erste fehlende Zahl als Summe natürlicher Zahlen von 1 bis finden Durchschn d.h. avg*(avg+1)/2 Minus die Summe der Array-Elemente kleiner als Durchschn
- Wir können die zweite fehlende Zahl finden, indem wir die erste fehlende Zahl von der Summe der fehlenden Zahlen subtrahieren
Betrachten Sie zur besseren Verdeutlichung ein Beispiel
Input : 1 3 5 6 n = 6 Sum of missing integers = n*(n+1)/2 - (1+3+5+6) = 6. Average of missing integers = 6/2 = 3. Sum of array elements less than or equal to average = 1 + 3 = 4 Sum of natural numbers from 1 to avg = avg*(avg + 1)/2 = 3*4/2 = 6 First missing number = 6 - 4 = 2 Second missing number = Sum of missing integers-First missing number Second missing number = 6-2= 4
Nachfolgend finden Sie die Umsetzung der obigen Idee.
C++ // C++ Program to find 2 Missing Numbers using O(1) // extra space #include using namespace std ; // Returns the sum of the array int getSum ( int arr [] int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += arr [ i ]; return sum ; } // Function to find two missing numbers in range // [1 n]. This function assumes that size of array // is n-2 and all array elements are distinct void findTwoMissingNumbers ( int arr [] int n ) { // Sum of 2 Missing Numbers int sum = ( n * ( n + 1 )) / 2 - getSum ( arr n -2 ); // Find average of two elements int avg = ( sum / 2 ); // Find sum of elements smaller than average (avg) // and sum of elements greater than average (avg) int sumSmallerHalf = 0 sumGreaterHalf = 0 ; for ( int i = 0 ; i < n -2 ; i ++ ) { if ( arr [ i ] <= avg ) sumSmallerHalf += arr [ i ]; else sumGreaterHalf += arr [ i ]; } cout < < 'Two Missing Numbers are n ' ; // The first (smaller) element = (sum of natural // numbers upto avg) - (sum of array elements // smaller than or equal to avg) int totalSmallerHalf = ( avg * ( avg + 1 )) / 2 ; int smallerElement = totalSmallerHalf - sumSmallerHalf ; cout < < smallerElement < < ' ' ; // The second (larger) element = (sum of both // the elements) - smaller element cout < < sum - smallerElement ; } // Driver program to test above function int main () { int arr [] = { 1 3 5 6 }; // Range of numbers is 2 plus size of array int n = 2 + sizeof ( arr ) / sizeof ( arr [ 0 ]); findTwoMissingNumbers ( arr n ); return 0 ; }
Java // Java Program to find 2 Missing // Numbers using O(1) extra space import java.io.* ; class GFG { // Returns the sum of the array static int getSum ( int arr [] int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += arr [ i ] ; return sum ; } // Function to find two missing // numbers in range [1 n]. This // function assumes that size of // array is n-2 and all array // elements are distinct static void findTwoMissingNumbers ( int arr [] int n ) { // Sum of 2 Missing Numbers int sum = ( n * ( n + 1 )) / 2 - getSum ( arr n - 2 ); // Find average of two elements int avg = ( sum / 2 ); // Find sum of elements smaller // than average (avg) and sum of // elements greater than average (avg) int sumSmallerHalf = 0 sumGreaterHalf = 0 ; for ( int i = 0 ; i < n - 2 ; i ++ ) { if ( arr [ i ] <= avg ) sumSmallerHalf += arr [ i ] ; else sumGreaterHalf += arr [ i ] ; } System . out . println ( 'Two Missing ' + 'Numbers are' ); // The first (smaller) element = // (sum of natural numbers upto // avg) - (sum of array elements // smaller than or equal to avg) int totalSmallerHalf = ( avg * ( avg + 1 )) / 2 ; System . out . println ( totalSmallerHalf - sumSmallerHalf ); // The first (smaller) element = // (sum of natural numbers from // avg+1 to n) - (sum of array // elements greater than avg) System . out . println ((( n * ( n + 1 )) / 2 - totalSmallerHalf ) - sumGreaterHalf ); } // Driver Code public static void main ( String [] args ) { int arr [] = { 1 3 5 6 }; // Range of numbers is 2 // plus size of array int n = 2 + arr . length ; findTwoMissingNumbers ( arr n ); } } // This code is contributed by aj_36
Python3 # Python Program to find 2 Missing # Numbers using O(1) extra space # Returns the sum of the array def getSum ( arr n ): sum = 0 ; for i in range ( 0 n ): sum += arr [ i ] return sum # Function to find two missing # numbers in range [1 n]. This # function assumes that size of # array is n-2 and all array # elements are distinct def findTwoMissingNumbers ( arr n ): # Sum of 2 Missing Numbers sum = (( n * ( n + 1 )) / 2 - getSum ( arr n - 2 )); #Find average of two elements avg = ( sum / 2 ); # Find sum of elements smaller # than average (avg) and sum # of elements greater than # average (avg) sumSmallerHalf = 0 sumGreaterHalf = 0 ; for i in range ( 0 n - 2 ): if ( arr [ i ] <= avg ): sumSmallerHalf += arr [ i ] else : sumGreaterHalf += arr [ i ] print ( 'Two Missing Numbers are' ) # The first (smaller) element = (sum # of natural numbers upto avg) - (sum # of array elements smaller than or # equal to avg) totalSmallerHalf = ( avg * ( avg + 1 )) / 2 print ( str ( totalSmallerHalf - sumSmallerHalf ) + ' ' ) # The first (smaller) element = (sum # of natural numbers from avg+1 to n) - # (sum of array elements greater than avg) print ( str ((( n * ( n + 1 )) / 2 - totalSmallerHalf ) - sumGreaterHalf )) # Driver Code arr = [ 1 3 5 6 ] # Range of numbers is 2 # plus size of array n = 2 + len ( arr ) findTwoMissingNumbers ( arr n ) # This code is contributed # by Yatin Gupta
C# // C# Program to find 2 Missing // Numbers using O(1) extra space using System ; class GFG { // Returns the sum of the array static int getSum ( int [] arr int n ) { int sum = 0 ; for ( int i = 0 ; i < n ; i ++ ) sum += arr [ i ]; return sum ; } // Function to find two missing // numbers in range [1 n]. This // function assumes that size of // array is n-2 and all array // elements are distinct static void findTwoMissingNumbers ( int [] arr int n ) { // Sum of 2 Missing Numbers int sum = ( n * ( n + 1 )) / 2 - getSum ( arr n - 2 ); // Find average of two elements int avg = ( sum / 2 ); // Find sum of elements smaller // than average (avg) and sum of // elements greater than average (avg) int sumSmallerHalf = 0 sumGreaterHalf = 0 ; for ( int i = 0 ; i < n - 2 ; i ++ ) { if ( arr [ i ] <= avg ) sumSmallerHalf += arr [ i ]; else sumGreaterHalf += arr [ i ]; } Console . WriteLine ( 'Two Missing ' + 'Numbers are ' ); // The first (smaller) element = // (sum of natural numbers upto // avg) - (sum of array elements // smaller than or equal to avg) int totalSmallerHalf = ( avg * ( avg + 1 )) / 2 ; Console . WriteLine ( totalSmallerHalf - sumSmallerHalf ); // The first (smaller) element = // (sum of natural numbers from // avg+1 to n) - (sum of array // elements greater than avg) Console . WriteLine ((( n * ( n + 1 )) / 2 - totalSmallerHalf ) - sumGreaterHalf ); } // Driver Code static public void Main () { int [] arr = { 1 3 5 6 }; // Range of numbers is 2 // plus size of array int n = 2 + arr . Length ; findTwoMissingNumbers ( arr n ); } } // This code is contributed by ajit
PHP // PHP Program to find 2 Missing // Numbers using O(1) extra space // Returns the sum of the array function getSum ( $arr $n ) { $sum = 0 ; for ( $i = 0 ; $i < $n ; $i ++ ) $sum += $arr [ $i ]; return $sum ; } // Function to find two missing // numbers in range [1 n]. This // function assumes that size of // array is n-2 and all array // elements are distinct function findTwoMissingNumbers ( $arr $n ) { // Sum of 2 Missing Numbers $sum = ( $n * ( $n + 1 )) / 2 - getSum ( $arr $n - 2 ); // Find average of two elements $avg = ( $sum / 2 ); // Find sum of elements smaller // than average (avg) and sum of // elements greater than average (avg) $sumSmallerHalf = 0 ; $sumGreaterHalf = 0 ; for ( $i = 0 ; $i < $n - 2 ; $i ++ ) { if ( $arr [ $i ] <= $avg ) $sumSmallerHalf += $arr [ $i ]; else $sumGreaterHalf += $arr [ $i ]; } echo 'Two Missing Numbers are n ' ; // The first (smaller) element = // (sum of natural numbers upto avg) - // (sum of array elements smaller // than or equal to avg) $totalSmallerHalf = ( $avg * ( $avg + 1 )) / 2 ; echo ( $totalSmallerHalf - $sumSmallerHalf ) ' ' ; // The first (smaller) element = // (sum of natural numbers from avg + // 1 to n) - (sum of array elements // greater than avg) echo ((( $n * ( $n + 1 )) / 2 - $totalSmallerHalf ) - $sumGreaterHalf ); } // Driver Code $arr = array ( 1 3 5 6 ); // Range of numbers is // 2 plus size of array $n = 2 + sizeof ( $arr ); findTwoMissingNumbers ( $arr $n ); // This code is contributed by aj_36 ?>
JavaScript < script > // Javascript Program to find 2 Missing // Numbers using O(1) extra space // Returns the sum of the array function getSum ( arr n ) { let sum = 0 ; for ( let i = 0 ; i < n ; i ++ ) sum += arr [ i ]; return sum ; } // Function to find two missing // numbers in range [1 n]. This // function assumes that size of // array is n-2 and all array // elements are distinct function findTwoMissingNumbers ( arr n ) { // Sum of 2 Missing Numbers let sum = ( n * ( n + 1 )) / 2 - getSum ( arr n - 2 ); // Find average of two elements let avg = ( sum / 2 ); // Find sum of elements smaller // than average (avg) and sum of // elements greater than average (avg) let sumSmallerHalf = 0 sumGreaterHalf = 0 ; for ( let i = 0 ; i < n - 2 ; i ++ ) { if ( arr [ i ] <= avg ) sumSmallerHalf += arr [ i ]; else sumGreaterHalf += arr [ i ]; } document . write ( 'Two Missing ' + 'Numbers are ' + ' ' ); // The first (smaller) element = // (sum of natural numbers upto // avg) - (sum of array elements // smaller than or equal to avg) let totalSmallerHalf = ( avg * ( avg + 1 )) / 2 ; document . write ( ( totalSmallerHalf - sumSmallerHalf ) + ' ' ); // The first (smaller) element = // (sum of natural numbers from // avg+1 to n) - (sum of array // elements greater than avg) document . write ( (( n * ( n + 1 )) / 2 - totalSmallerHalf ) - sumGreaterHalf + ' ' ); } let arr = [ 1 3 5 6 ]; // Range of numbers is 2 // plus size of array let n = 2 + arr . length ; findTwoMissingNumbers ( arr n ); < /script>
Ausgabe
Two Missing Numbers are 2 4
Hinweis: Bei der oben genannten Lösung kann es zu Überlaufproblemen kommen.
Im folgenden Satz 2 wird eine weitere Lösung besprochen, die O(n) Zeit O(1) Raum hat und keine Überlaufprobleme verursacht.
Finden Sie zwei fehlende Zahlen | Satz 2 (XOR-basierte Lösung)
