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++

// 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 aren';  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(0n-21): mark[arr[i]] = True # Print two unmarked elements print('Two Missing Numbers are') for i in range(1n+11): if (mark[i] == False): print(iend = ' ') 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# 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 aren';  // 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(arrn): 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# 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 aren'; // 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)

TechCodeview

Finden Sie zwei fehlende Zahlen | Satz 1 (Eine interessante lineare Zeitlösung)