Controlla se esistono due elementi in un array la cui somma è uguale alla somma del resto dell'array
Abbiamo un array di numeri interi e dobbiamo trovare due di questi elementi nell'array in modo tale che la somma di questi due elementi sia uguale alla somma del resto degli elementi nell'array.
Esempi:
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 soluzione semplice consiste nel considerare ogni coppia una per una, trovare la sua somma e confrontare la somma con la somma del resto degli elementi. Se troviamo una coppia la cui somma è uguale al resto degli elementi, stampiamo la coppia e restituiamo true. La complessità temporale di questa soluzione è O(n 3 )
UN soluzione efficiente è trovare la somma di tutti gli elementi dell'array. Lascia che questa somma sia 'somma'. Ora il compito si riduce a trovare una coppia con somma uguale a somma/2.
Un'altra ottimizzazione è che una coppia può esistere solo se la somma dell'intero array è pari perché sostanzialmente lo stiamo dividendo in due parti con la stessa somma.
- Trova la somma dell'intero array. Lascia che questa somma sia 'somma'
- Se la somma è dispari, restituisce falso.
- Trova una coppia con somma uguale a "somma/2" utilizzando il metodo basato sull'hashing discusso Qui come metodo 2. Se viene trovata una coppia, stampala e restituisce true.
- Se non esiste alcuna coppia, restituisce false.
Di seguito è riportata l'implementazione dei passaggi precedenti.
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 %d n ' 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 ( arr n ) { // 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>
Produzione
Pair elements are 4 and 11
Complessità temporale: SU) . insieme_non ordinato viene implementato utilizzando l'hashing. In questo caso la ricerca e l'inserimento dell'hash di complessità temporale vengono presupposti come O(1).
Spazio ausiliario: SU)
Un altro approccio efficiente (ottimizzazione dello spazio): Per prima cosa ordineremo l'array per Ricerca binaria . Quindi ripeteremo l'intero array e controlleremo se esiste un indice nell'array che si accoppia con i tale che arr[index] + a[i] == Somma restante dell'array. Possiamo utilizzare la ricerca binaria per trovare un indice nell'array modificando il programma di ricerca binaria. Se esiste una coppia, stampa quella coppia. altrimenti print non esiste alcuna coppia.
Di seguito è riportata l'implementazione dell'approccio di cui sopra:
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 i Totalsum ) ; 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 i Totalsum ) ; 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)
Produzione
Pair elements are 11 and 4
Complessità temporale: O(n * log)
Spazio ausiliario: O(1)
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