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Controleer of er twee elementen in een array bestaan ​​waarvan de som gelijk is aan de som van de rest van de array

We hebben een array van gehele getallen en we moeten twee van dergelijke elementen in de array vinden, zodat de som van deze twee elementen gelijk is aan de som van de rest van de elementen in de array. 

Voorbeelden:  

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 eenvoudige oplossing is om van elk paar één voor één de som te vinden en de som te vergelijken met de som van de rest van de elementen. Als we een paar vinden waarvan de som gelijk is aan de rest van de elementen, printen we het paar en retourneren we waar. De tijdscomplexiteit van deze oplossing is O(n3)



Een efficiënte oplossing is om de som van alle array-elementen te vinden. Laat deze som 'som' zijn. Nu wordt de taak beperkt tot het vinden van een paar waarvan de som gelijk is aan som/2. 

Een andere optimalisatie is dat een paar alleen kan bestaan ​​als de som van de hele array gelijk is, omdat we deze in feite in twee delen verdelen met een gelijke som.

  1. Vind de som van de hele array. Laat deze som 'som' zijn 
  2. Als de som oneven is, retourneert u false. 
  3. Vind een paar waarvan de som gelijk is aan 'sum/2' met behulp van de besproken hashing-gebaseerde methode hier als methode 2. Als er een paar wordt gevonden, druk het dan af en retourneer waar. 
  4. Als er geen paar bestaat, retourneert u false.

Hieronder vindt u de implementatie van bovenstaande stappen.

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>

Uitvoer
Pair elements are 4 and 11

Tijdcomplexiteit: Op) . ongeordende_set wordt geïmplementeerd met behulp van hashing. Het zoeken en invoegen van hash met tijdcomplexiteit wordt hier aangenomen als O(1).
Hulpruimte: Op)

Een andere efficiënte aanpak (ruimteoptimalisatie): Eerst zullen we de array sorteren Binaire zoekopdracht . Vervolgens zullen we de hele array herhalen en controleren of er een index in de array bestaat die met i paart, zodat arr[index] + a[i] == Restsom van de array. We kunnen binair zoeken gebruiken om een ​​index in de array te vinden door het binaire zoekprogramma aan te passen. Als er een paar bestaat, druk dan dat paar af. anders print er bestaat geen paar.

Hieronder vindt u de implementatie van de bovenstaande aanpak:

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) 

Uitvoer
Pair elements are 11 and 4

Tijdcomplexiteit: O(n * ingelogd) 
Hulpruimte: O(1)

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