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Vérifiez s'il existe deux éléments dans un tableau dont la somme est égale à la somme du reste du tableau

Nous avons un tableau d'entiers et nous devons trouver deux de ces éléments dans le tableau tels que la somme de ces deux éléments soit égale à la somme du reste des éléments du tableau. 

Exemples :  

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 

UN solution simple consiste à considérer chaque paire une par une, à trouver sa somme et à comparer la somme avec la somme du reste des éléments. Si nous trouvons une paire dont la somme est égale au reste des éléments, nous imprimons la paire et renvoyons vrai. La complexité temporelle de cette solution est O(n3)



Un solution efficace est de trouver la somme de tous les éléments du tableau. Que cette somme soit « somme ». Maintenant, la tâche se réduit à trouver une paire dont la somme est égale à somme/2. 

Une autre optimisation est qu'une paire ne peut exister que si la somme du tableau entier est paire, car nous la divisons essentiellement en deux parties de somme égale.

  1. Trouvez la somme du tableau entier. Que cette somme soit « somme » 
  2. Si la somme est impaire, renvoie faux. 
  3. Trouvez une paire dont la somme est égale à « somme/2 » en utilisant la méthode basée sur le hachage discutée ici comme méthode 2. Si une paire est trouvée, imprimez-la et renvoyez vrai. 
  4. Si aucune paire n’existe, retournez false.

Vous trouverez ci-dessous la mise en œuvre des étapes ci-dessus.

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>

Sortir
Pair elements are 4 and 11

Complexité temporelle : Sur) . ensemble_non ordonné est implémenté en utilisant le hachage. La recherche et l'insertion de hachage de complexité temporelle sont ici supposées être O(1).
Espace auxiliaire : Sur)

Une autre approche efficace (optimisation de l'espace) : Nous allons d'abord trier le tableau pour Recherche binaire . Ensuite, nous allons parcourir tout le tableau et vérifier s'il existe un index dans le tableau qui s'associe à i tel que arr[index] + a[i] == Somme restante du tableau. Nous pouvons utiliser la recherche binaire pour trouver un index dans le tableau en modifiant le programme de recherche binaire. Si une paire existe, imprimez cette paire. sinon, imprimez, aucune paire n'existe.

Vous trouverez ci-dessous la mise en œuvre de l'approche ci-dessus :

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) 

Sortir
Pair elements are 11 and 4

Complexité temporelle : O(n * connexion) 
Espace auxiliaire : O(1)

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