НАМЕРЕТЕ ДВЕ ЛИПСВАЩИ ЧИСЛА | КОМПЛЕКТ 1 (ИНТЕРЕСНО РЕШЕНИЕ ЗА ЛИНЕЙНО ВРЕМЕ)

Даден е масив от n уникални цели числа, където всеки елемент в масива е в диапазона [1 n]. Масивът има всички различни елементи и размерът на масива е (n-2). Следователно две числа от диапазона липсват в този масив. Намерете двете липсващи числа.

Примери:

Input : arr[] = {1 3 5 6} Output : 2 4 Input : arr[] = {1 2 4} Output : 3 5 Input : arr[] = {1 2} Output : 3 4

Метод 1 - O(n) времева сложност и O(n) допълнително пространство

Стъпка 1: Вземете булев масив марка който следи всички елементи, присъстващи в масива.
Стъпка 2: Повторете от 1 до n проверете за всеки елемент дали е маркиран като true в булевия масив, ако не е, тогава просто покажете този елемент.

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>

Изход

Two Missing Numbers are 2 4

Метод 2 - O(n) времева сложност и O(1) допълнително пространство

Идеята се основава на това популярно решение за намиране на едно липсващо число. Разширяваме решението, така че да се отпечатат два липсващи елемента.
Нека намерим сумата на 2 липсващи числа:

  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;

Едно от числата ще бъде по-малко или равно на ср докато другият ще бъде строго по-голям от ср . Две числа никога не могат да бъдат равни, тъй като всички дадени числа са различни.
Можем да намерим първото липсващо число като сбор от естествени числа от 1 до ср т.е. ср.*(ср.+1)/2 минус сумата от елементи на масива, по-малка от ср
Можем да намерим второто липсващо число, като извадим първото липсващо число от сбора на липсващите числа

Помислете за пример за по-добро изясняване

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

По-долу е изпълнението на горната идея.

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>

Изход

Two Missing Numbers are 2 4

Забележка: Може да има проблеми с препълването в горното решение.

В набор 2 по-долу се обсъжда друго решение, което е O(n) време O(1) пространство и не причинява проблеми с препълването.
Намерете две липсващи числа | Комплект 2 (решение, базирано на XOR)

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Намерете две липсващи числа | Комплект 1 (Интересно решение за линейно време)