Tarkista, onko taulukossa kaksi elementtiä, joiden summa on yhtä suuri kuin taulukon loppuosan summa
Meillä on joukko kokonaislukuja ja meidän on löydettävä taulukosta kaksi sellaista elementtiä, että näiden kahden elementin summa on yhtä suuri kuin taulukon muiden elementtien summa.
Esimerkkejä:
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 yksinkertainen ratkaisu on tarkastella jokaista paria yksitellen löytää sen summa ja verrata summaa muiden elementtien summaan. Jos löydämme parin, jonka summa on yhtä suuri kuin loput alkiot, tulostetaan pari ja palautetaan tosi. Tämän ratkaisun aikamonimutkaisuus on O(n 3 )
An tehokas ratkaisu on löytää taulukon kaikkien elementtien summa. Olkoon tämä summa "summa". Nyt tehtävä pelkistyy niin, että löydetään pari, jonka summa on yhtä suuri kuin summa/2.
Toinen optimointi on, että pari voi olla olemassa vain, jos koko taulukon summa on parillinen, koska jaamme sen periaatteessa kahteen osaan yhtä suurella summalla.
- Etsi koko taulukon summa. Olkoon tämä summa "summa"
- Jos summa on pariton, palauta epätosi.
- Etsi pari, jonka summa on yhtä suuri kuin 'summa/2' käyttämällä käsiteltyä hajautusmenetelmää tässä kuten menetelmä 2. Jos pari löytyy, tulosta se ja palauta tosi.
- Jos paria ei ole, palauta false.
Alla on yllä olevien vaiheiden toteutus.
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>
Lähtö
Pair elements are 4 and 11
Aika monimutkaisuus: O(n) . unordered_set toteutetaan hajautustekniikalla. Aikamonimutkaisuuden hash-haku ja lisäys oletetaan tässä muodossa O(1).
Aputila: O(n)
Toinen tehokas lähestymistapa (tilan optimointi): Ensin lajittelemme taulukon Binäärihaku . Sitten iteroidaan koko taulukko ja tarkistetaan, onko taulukossa indeksi, joka muodostaa parin i:n kanssa siten, että arr[index] + a[i] == Taulukon loppusumma . Voimme käyttää binaarihakua löytääksemme indeksin taulukosta muokkaamalla binaarihakuohjelmaa . Jos pari on olemassa, tulosta se. muuten tulosta paria ei ole olemassa.
Alla on yllä olevan lähestymistavan toteutus:
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)
Lähtö
Pair elements are 11 and 4
Aika monimutkaisuus: O(n * kirjautuminen)
Aputila: O(1)
Luo tietokilpailu
