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Überprüfen Sie, ob in einem Array zwei Elemente vorhanden sind, deren Summe gleich der Summe des Rests des Arrays ist

Wir haben ein Array von ganzen Zahlen und müssen zwei solcher Elemente im Array finden, sodass die Summe dieser beiden Elemente gleich der Summe der übrigen Elemente im Array ist. 

Beispiele:  

Input : arr[] = {2 11 5 1 4 7} Output : Elements are 4 and 11 Note that 4 + 11 = 2 + 5 + 1 + 7 Input : arr[] = {2 4 2 1 11 15} Output : Elements do not exist 

A einfache Lösung besteht darin, jedes Paar einzeln zu betrachten, seine Summe zu ermitteln und die Summe mit der Summe der übrigen Elemente zu vergleichen. Wenn wir ein Paar finden, dessen Summe gleich dem Rest der Elemente ist, geben wir das Paar aus und geben „true“ zurück. Die zeitliche Komplexität dieser Lösung beträgt O(n3)



Ein effiziente Lösung besteht darin, die Summe aller Array-Elemente zu ermitteln. Diese Summe sei „Summe“. Jetzt reduziert sich die Aufgabe darauf, ein Paar zu finden, dessen Summe gleich sum/2 ist. 

Eine weitere Optimierung besteht darin, dass ein Paar nur dann existieren kann, wenn die Summe des gesamten Arrays gerade ist, da wir es grundsätzlich in zwei Teile mit gleicher Summe teilen.

  1. Finden Sie die Summe des gesamten Arrays. Diese Summe sei „Summe“ 
  2. Wenn die Summe ungerade ist, wird „false“ zurückgegeben. 
  3. Finden Sie mithilfe der besprochenen Hashing-basierten Methode ein Paar, dessen Summe gleich „sum/2“ ist Hier wie Methode 2. Wenn ein Paar gefunden wird, drucken Sie es aus und geben Sie true zurück. 
  4. Wenn kein Paar vorhanden ist, wird „false“ zurückgegeben.

Nachfolgend finden Sie die Umsetzung der oben genannten Schritte.

C++
// C++ program to find whether two elements exist // whose sum is equal to sum of rest of the elements. #include    using namespace std; // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. bool checkPair(int arr[] int n) {  // Find sum of whole array  int sum = 0;  for (int i = 0; i < n; i++)  sum += arr[i];  // If sum of array is not even then we can not  // divide it into two part  if (sum % 2 != 0)  return false;  sum = sum / 2;  // For each element arr[i] see if there is  // another element with value sum - arr[i]  unordered_set<int> s;  for (int i = 0; i < n; i++) {  int val = sum - arr[i];  // If element exist than return the pair  if (s.find(val) != s.end()) {  printf('Pair elements are %d and %dn' arr[i]  val);  return true;  }  s.insert(arr[i]);  }  return false; } // Driver program. int main() {  int arr[] = { 2 11 5 1 4 7 };  int n = sizeof(arr) / sizeof(arr[0]);  if (checkPair(arr n) == false)  printf('No pair found');  return 0; } 
Java
// Java program to find whether two elements exist // whose sum is equal to sum of rest of the elements. import java.util.*; class GFG {  // Function to check whether two elements exist  // whose sum is equal to sum of rest of the elements.  static boolean checkPair(int arr[] int n)  {  // Find sum of whole array  int sum = 0;  for (int i = 0; i < n; i++) {  sum += arr[i];  }  // If sum of array is not even then we can not  // divide it into two part  if (sum % 2 != 0) {  return false;  }  sum = sum / 2;  // For each element arr[i] see if there is  // another element with value sum - arr[i]  HashSet<Integer> s = new HashSet<Integer>();  for (int i = 0; i < n; i++) {  int val = sum - arr[i];  // If element exist than return the pair  if (s.contains(val)  && val == (int)s.toArray()[s.size() - 1]) {  System.out.printf(  'Pair elements are %d and %dn' arr[i]  val);  return true;  }  s.add(arr[i]);  }  return false;  }  // Driver program.  public static void main(String[] args)  {  int arr[] = { 2 11 5 1 4 7 };  int n = arr.length;  if (checkPair(arr n) == false) {  System.out.printf('No pair found');  }  } } /* This code contributed by PrinciRaj1992 */ 
Python3
# Python3 program to find whether  # two elements exist whose sum is # equal to sum of rest of the elements.  # Function to check whether two  # elements exist whose sum is equal  # to sum of rest of the elements.  def checkPair(arr n): s = set() sum = 0 # Find sum of whole array  for i in range(n): sum += arr[i] # / If sum of array is not  # even then we can not  # divide it into two part  if sum % 2 != 0: return False sum = sum / 2 # For each element arr[i] see if  # there is another element with  # value sum - arr[i]  for i in range(n): val = sum - arr[i] if arr[i] not in s: s.add(arr[i]) # If element exist than  # return the pair  if val in s: print('Pair elements are' arr[i] 'and' int(val)) # Driver Code  arr = [2 11 5 1 4 7] n = len(arr) if checkPair(arr n) == False: print('No pair found') # This code is contributed  # by Shrikant13 
C#
// C# program to find whether two elements exist // whose sum is equal to sum of rest of the elements. using System;  using System.Collections.Generic;  class GFG  {    // Function to check whether two elements exist  // whose sum is equal to sum of rest of the elements.  static bool checkPair(int []arr int n)  {  // Find sum of whole array  int sum = 0;  for (int i = 0; i < n; i++)  {  sum += arr[i];  }  // If sum of array is not even then we can not  // divide it into two part  if (sum % 2 != 0)   {  return false;  }  sum = sum / 2;  // For each element arr[i] see if there is  // another element with value sum - arr[i]  HashSet<int> s = new HashSet<int>();  for (int i = 0; i < n; i++)  {  int val = sum - arr[i];  // If element exist than return the pair  if (s.Contains(val))  {  Console.Write('Pair elements are {0} and {1}n'  arr[i] val);  return true;  }  s.Add(arr[i]);  }  return false;  }  // Driver code  public static void Main(String[] args)  {  int []arr = {2 11 5 1 4 7};  int n = arr.Length;  if (checkPair(arr n) == false)   {  Console.Write('No pair found');  }  } } // This code contributed by Rajput-Ji 
PHP
 // PHP program to find whether two elements exist // whose sum is equal to sum of rest of the elements. // Function to check whether two elements exist // whose sum is equal to sum of rest of the elements. function checkPair(&$arr $n) { // Find sum of whole array $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $arr[$i]; // If sum of array is not even then we  // can not divide it into two part if ($sum % 2 != 0) return false; $sum = $sum / 2; // For each element arr[i] see if there is // another element with value sum - arr[i] $s = array(); for ($i = 0; $i < $n; $i++) { $val = $sum - $arr[$i]; // If element exist than return the pair if (array_search($val $s)) { echo 'Pair elements are ' . $arr[$i] . ' and ' . $val . 'n'; return true; } array_push($s $arr[$i]); } return false; } // Driver Code $arr = array(2 11 5 1 4 7); $n = sizeof($arr); if (checkPair($arr $n) == false) echo 'No pair found'; // This code is contributed by ita_c ?> 
JavaScript
<script> // Javascript program to find  // whether two elements exist // whose sum is equal to sum of rest // of the elements.     // Function to check whether   // two elements exist  // whose sum is equal to sum of   // rest of the elements.  function checkPair(arrn)  {  // Find sum of whole array  let sum = 0;  for (let i = 0; i < n; i++)  {  sum += arr[i];  }    // If sum of array is not even then we can not  // divide it into two part  if (sum % 2 != 0)   {  return false;  }    sum = Math.floor(sum / 2);    // For each element arr[i] see if there is  // another element with value sum - arr[i]  let s = new Set();  for (let i = 0; i < n; i++)  {  let val = sum - arr[i];    // If element exist than return the pair    if(!s.has(arr[i]))  {  s.add(arr[i])  }    if (s.has(val) )   {  document.write('Pair elements are '+  arr[i]+' and '+ val+'  
'
); return true; } s.add(arr[i]); } return false; } // Driver program. let arr=[2 11 5 1 4 7]; let n = arr.length; if (checkPair(arr n) == false) { document.write('No pair found'); } // This code is contributed by rag2127 </script>

Ausgabe
Pair elements are 4 and 11

Zeitkomplexität: An) . unordered_set wird mittels Hashing implementiert. Die zeitliche Komplexität der Hash-Suche und -Einfügung wird hier als O(1) angenommen.
Hilfsraum: An)

Ein weiterer effizienter Ansatz (Raumoptimierung): Zuerst sortieren wir das Array nach Binäre Suche . Dann iterieren wir das gesamte Array und prüfen, ob im Array ein Index vorhanden ist, der mit i gepaart ist, sodass arr[index] + a[i] == Restsumme des Arrays ist. Wir können die binäre Suche verwenden, um einen Index im Array zu finden, indem wir das Programm für die binäre Suche modifizieren. Wenn ein Paar vorhanden ist, drucken Sie dieses Paar aus. Sonst gibt es kein Paar.

Nachfolgend finden Sie die Umsetzung des oben genannten Ansatzes:

C++
// C++ program for the above approach #include   using namespace std; // Function to Find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array int binarysearch(int arr[] int n int i int Totalsum) {   int l = 0 r = n - 1  index = -1;//initialize as -1  while (l <= r)   { int mid = (l + r) / 2;    int Pairsum = arr[mid] + arr[i];//pair sum  int Restsum = Totalsum - Pairsum;//Rest sum    if ( Pairsum == Restsum )  {  if( index != i )// checking a pair has same position or not  { index = mid; }//Then update index -1 to mid    // Checking for adjacent element  else if(index == i && mid>0 && arr[mid-1]==arr[i])  { index = mid-1; }//Then update index -1 to mid-1    else if(index == i && mid<n-1 && arr[mid+1]==arr[i])   { index = mid+1; } //Then update index-1 to mid+1   break;   }  else if (Pairsum > Restsum)   { // If pair sum is greater than rest sum  our index will  // be in the Range [mid+1R]  l = mid + 1;  }  else {  // If pair sum is smaller than rest sum  our index will  // be in the Range [Lmid-1]  r = mid - 1;  }  }  // return index=-1 if a pair not exist with arr[i]  // else return index such that arr[i]+arr[index] == sum of rest of arr  return index; } // Function to check if a pair exist such their sum  // equal to rest of the array or not  bool checkPair(int arr[]int n) { int Totalsum=0;  sort(arr  arr + n);//sort arr for Binary search    for(int i=0;i<n;i++)  { Totalsum+=arr[i]; } //Finding total sum of the arr    for(int i=0;i<n;i++)  { // If index is -1  Means arr[i] can't pair with any element   // else arr[i]+a[index] == sum of rest of the arr  int index = binarysearch(arr n iTotalsum) ;    if(index != -1) {   cout<<'Pair elements are '<< arr[i]<<' and '<< arr[index];  return true;  }  }  return false;//Return false if a pair not exist } // Driver Code int main() {  int arr[] = {2 11 5 1 4 7};  int n = sizeof(arr)/sizeof(arr[0]);    //Function call  if (checkPair(arr n) == false)  { cout<<'No pair found'; }  return 0; } // This Approach is contributed by nikhilsainiofficial546  
Java
// Java program for the above approach import java.util.*; class GFG {  // Function to Find if a index exist in array such that  // arr[index] + a[i] == Rest sum of the array  static int binarysearch(int arr[] int n int i  int Totalsum)  {  int l = 0 r = n - 1 index = -1; // initialize as -1  while (l <= r) {  int mid = (l + r) / 2;  int Pairsum = arr[mid] + arr[i]; // pair sum  int Restsum = Totalsum - Pairsum; // Rest sum  if (Pairsum == Restsum) {  if (index != i) // checking a pair has same  // position or not  {  index = mid;  } // Then update index -1 to mid  // Checking for adjacent element  else if (index == i && mid > 0  && arr[mid - 1] == arr[i]) {  index = mid - 1;  } // Then update index -1 to mid-1  else if (index == i && mid < n - 1  && arr[mid + 1] == arr[i]) {  index = mid + 1;  } // Then update index-1 to mid+1  break;  }  else if (Pairsum > Restsum) {  // If pair sum is greater than rest sum   // our index will be in the Range [mid+1R]  l = mid + 1;  }  else {  // If pair sum is smaller than rest sum   // our index will be in the Range [Lmid-1]  r = mid - 1;  }  }  // return index=-1 if a pair not exist with arr[i]  // else return index such that arr[i]+arr[index] ==  // sum of rest of arr  return index;  }  // Function to check if a pair exist such their sum  // equal to rest of the array or not  static boolean checkPair(int arr[] int n)  {  int Totalsum = 0;  Arrays.sort(arr); // sort arr for Binary search  for (int i = 0; i < n; i++) {  Totalsum += arr[i];  } // Finding total sum of the arr  for (int i = 0; i < n; i++) {  // If index is -1  Means arr[i] can't pair with  // any element else arr[i]+a[index] == sum of  // rest of the arr  int index = binarysearch(arr n i Totalsum);  if (index != -1) {  System.out.println('Pair elements are '  + arr[i] + ' and '  + arr[index]);  return true;  }  }  return false; // Return false if a pair not exist  }  // Driver Code  public static void main(String[] args)  {  int arr[] = { 2 11 5 1 4 7 };  int n = arr.length;  // Function call  if (checkPair(arr n) == false) {  System.out.println('No pair found');  }  } } 
Python3
# Python program for the above approach # Function to find if a index exist in array such that # arr[index] + a[i] == Rest sum of the array def binarysearch(arr n i Totalsum): l = 0 r = n - 1 index = -1 # Initialize as -1 while l <= r: mid = (l + r) // 2 Pairsum = arr[mid] + arr[i] # Pair sum Restsum = Totalsum - Pairsum # Rest sum if Pairsum == Restsum: if index != i: # Checking if a pair has the same position or not index = mid # Then update index -1 to mid # Checking for adjacent element elif index == i and mid > 0 and arr[mid - 1] == arr[i]: index = mid - 1 # Then update index -1 to mid-1 elif index == i and mid < n - 1 and arr[mid + 1] == arr[i]: index = mid + 1 # Then update index-1 to mid+1 break elif Pairsum > Restsum: # If pair sum is greater than rest sum our index will # be in the Range [mid+1R] l = mid + 1 else: # If pair sum is smaller than rest sum our index will # be in the Range [Lmid-1] r = mid - 1 # Return index=-1 if a pair not exist with arr[i] # else return index such that arr[i]+arr[index] == sum of rest of arr return index # Function to check if a pair exists such that their sum # equals to rest of the array or not def checkPair(arr n): Totalsum = 0 arr = sorted(arr) # Sort arr for Binary search for i in range(n): Totalsum += arr[i] # Finding total sum of the arr for i in range(n): # If index is -1 means arr[i] can't pair with any element # else arr[i]+a[index] == sum of rest of the arr index = binarysearch(arr n i Totalsum) if index != -1: print('Pair elements are' arr[i] 'and' arr[index]) return True return False # Return false if a pair not exist # Driver Code arr = [2 11 5 1 4 7] n = len(arr) # Function call if checkPair(arr n) == False: print('No pair found') 
C#
using System; class GFG {  // Function to Find if a index exist in array such that  // arr[index] + a[i] == Rest sum of the array  static int BinarySearch(int[] arr int n int i int totalSum)  {  int l = 0 r = n - 1 index = -1; // initialize as -1  while (l <= r)  {  int mid = (l + r) / 2;  int pairSum = arr[mid] + arr[i]; // pair sum  int restSum = totalSum - pairSum; // rest sum  if (pairSum == restSum)  {  if (index != i) // checking a pair has same  // position or not  {  index = mid;  } // Then update index -1 to mid  // Checking for adjacent element  else if (index == i && mid > 0  && arr[mid - 1] == arr[i])  {  index = mid - 1;  } // Then update index -1 to mid-1  else if (index == i && mid < n - 1  && arr[mid + 1] == arr[i])  {  index = mid + 1;  } // Then update index-1 to mid+1  break;  }  else if (pairSum > restSum)  {  // If pair sum is greater than rest sum   // our index will be in the Range [mid+1R]  l = mid + 1;  }  else  {  // If pair sum is smaller than rest sum   // our index will be in the Range [Lmid-1]  r = mid - 1;  }  }  // return index=-1 if a pair not exist with arr[i]  // else return index such that arr[i]+arr[index] ==  // sum of rest of arr  return index;  }  // Function to check if a pair exist such their sum  // equal to rest of the array or not  static bool CheckPair(int[] arr int n)  {  int totalSum = 0;  Array.Sort(arr); // sort arr for Binary search  for (int i = 0; i < n; i++)  {  totalSum += arr[i];  } // Finding total sum of the arr  for (int i = 0; i < n; i++)  {  // If index is -1  Means arr[i] can't pair with  // any element else arr[i]+a[index] == sum of  // rest of the arr  int index = BinarySearch(arr n i totalSum);  if (index != -1)  {  Console.WriteLine('Pair elements are ' + arr[i] + ' and ' + arr[index]);  return true;  }  }  return false; // Return false if a pair not exist  }  // Driver Code  static void Main(string[] args)  {  int[] arr = { 2 11 5 1 4 7 };  int n = arr.Length;  // Function call  if (!CheckPair(arr n))  {  Console.WriteLine('No pair found');  }  } } 
JavaScript
// JavaScript program for the above approach // function to find if a index exist in array such that // arr[index] + a[i] == Rest sum of the array function binarysearch(arr n i TotalSum){  let l = 0;  let r = n-1;  let index = -1;    while(l <= r){  let mid = parseInt((l+r)/2);  let Pairsum = arr[mid] + arr[i];  let Restsum = TotalSum - Pairsum;    if ( Pairsum == Restsum )  {  if( index != i )// checking a pair has same position or not  { index = mid; }//Then update index -1 to mid    // Checking for adjacent element  else if(index == i && mid>0 && arr[mid-1]==arr[i])  { index = mid-1; }//Then update index -1 to mid-1    else if(index == i && mid<n-1 && arr[mid+1]==arr[i])  { index = mid+1; } //Then update index-1 to mid+1   break;  }  else if (Pairsum > Restsum)  { // If pair sum is greater than rest sum  our index will  // be in the Range [mid+1R]  l = mid + 1;  }  else {  // If pair sum is smaller than rest sum  our index will  // be in the Range [Lmid-1]  r = mid - 1;  }  }  // return index=-1 if a pair not exist with arr[i]  // else return index such that arr[i]+arr[index] == sum of rest of arr  return index; } // Function to check if a pair exist such their sum // equal to rest of the array or not function checkPair(arr n){  let Totalsum = 0;  arr.sort(function(a b){return a - b});    for(let i=0;i<n;i++)  { Totalsum+=arr[i]; } //Finding total sum of the arr    for(let i=0;i<n;i++)  { // If index is -1  Means arr[i] can't pair with any element  // else arr[i]+a[index] == sum of rest of the arr  let index = binarysearch(arr n iTotalsum) ;    if(index != -1) {  console.log('Pair elements are ' + arr[i] + ' and ' + arr[index]);  return true;  }  }  return false;//Return false if a pair not exist } // driver code to test above function let arr = [2 11 5 1 4 7]; let n = arr.length; // function call if(checkPair(arr n) == false)   console.log('No Pair Found')    // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) 

Ausgabe
Pair elements are 11 and 4

Zeitkomplexität: O(n * logn) 
Hilfsraum: O(1)

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