Parell de valors XOR mínim
Prova-ho a GfG Practice 
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#practiceLinkDiv { mostrar: cap !important; } Donada una matriu de nombres enters. Trobeu la parella en una matriu que tingui un valor XOR mínim.
Exemples:
Input : arr[] = {9 5 3} Output : 6 All pair with xor value (9 ^ 5) => 12 (5 ^ 3) => 6 (9 ^ 3) => 10. Minimum XOR value is 6 Input : arr[] = {1 2 3 4 5} Output : 1 Recommended Practice Parell de valors XOR mínim Prova-ho! A Solució senzilla és generar tots els parells de la matriu donada i calcular XOR els seus valors. Finalment retorna el valor XOR mínim. Aquesta solució pren O(n 2 ) temps.
Implementació:
C++ // C++ program to find minimum XOR value in an array. #include using namespace std ; // Returns minimum xor value of pair in arr[0..n-1] int minXOR ( int arr [] int n ) { int min_xor = INT_MAX ; // Initialize result // Generate all pair of given array for ( int i = 0 ; i < n ; i ++ ) for ( int j = i + 1 ; j < n ; j ++ ) // update minimum xor value if required min_xor = min ( min_xor arr [ i ] ^ arr [ j ]); return min_xor ; } // Driver program int main () { int arr [] = { 9 5 3 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < minXOR ( arr n ) < < endl ; return 0 ; }
Java // Java program to find minimum XOR value in an array. class GFG { // Returns minimum xor value of pair in arr[0..n-1] static int minXOR ( int arr [] int n ) { int min_xor = Integer . MAX_VALUE ; // Initialize result // Generate all pair of given array for ( int i = 0 ; i < n ; i ++ ) for ( int j = i + 1 ; j < n ; j ++ ) // update minimum xor value if required min_xor = Math . min ( min_xor arr [ i ] ^ arr [ j ] ); return min_xor ; } // Driver program public static void main ( String args [] ) { int arr [] = { 9 5 3 }; int n = arr . length ; System . out . println ( minXOR ( arr n )); } } // This code is contributed by Sumit Ghosh
Python3 # Python program to find minimum # XOR value in an array. # Function to find minimum XOR pair def minXOR ( arr n ): # Sort given array arr . sort (); min_xor = 999999 val = 0 # calculate min xor of # consecutive pairs for i in range ( 0 n - 1 ): for j in range ( i + 1 n - 1 ): # update minimum xor value # if required val = arr [ i ] ^ arr [ j ] min_xor = min ( min_xor val ) return min_xor # Driver program arr = [ 9 5 3 ] n = len ( arr ) print ( minXOR ( arr n )) # This code is contributed by Sam007.
C# // C# program to find minimum // XOR value in an array. using System ; class GFG { // Returns minimum xor value of // pair in arr[0..n-1] static int minXOR ( int [] arr int n ) { // Initialize result int min_xor = int . MaxValue ; // Generate all pair of given array for ( int i = 0 ; i < n ; i ++ ) for ( int j = i + 1 ; j < n ; j ++ ) // update minimum xor value if required min_xor = Math . Min ( min_xor arr [ i ] ^ arr [ j ]); return min_xor ; } // Driver program public static void Main () { int [] arr = { 9 5 3 }; int n = arr . Length ; Console . WriteLine ( minXOR ( arr n )); } } // This code is contributed by Sam007
PHP // PHP program to find minimum // XOR value in an array. // Returns minimum xor value // of pair in arr[0..n-1] function minXOR ( $arr $n ) { // Initialize result $min_xor = PHP_INT_MAX ; // Generate all pair of given array for ( $i = 0 ; $i < $n ; $i ++ ) for ( $j = $i + 1 ; $j < $n ; $j ++ ) // update minimum xor // value if required $min_xor = min ( $min_xor $arr [ $i ] ^ $arr [ $j ]); return $min_xor ; } // Driver Code $arr = array ( 9 5 3 ); $n = count ( $arr ); echo minXOR ( $arr $n ); // This code is contributed by anuj_67. ?>
JavaScript < script > // Javascript program to find // minimum XOR value in an array. // Returns minimum xor value of pair in arr[0..n-1] function minXOR ( arr n ) { // Initialize result let min_xor = Number . MAX_VALUE ; // Generate all pair of given array for ( let i = 0 ; i < n ; i ++ ) for ( let j = i + 1 ; j < n ; j ++ ) // update minimum xor value if required min_xor = Math . min ( min_xor arr [ i ] ^ arr [ j ]); return min_xor ; } // Driver program let arr = [ 9 5 3 ]; let n = arr . length ; document . write ( minXOR ( arr n )); < /script>
Sortida
6
Complexitat espacial: O(1)
An Solució eficient pot resoldre aquest problema en temps O(nlogn).
Algorisme:
- Ordena la matriu donada
- Travessa i comproveu XOR per a cada parell consecutiu
A continuació es mostra la implementació de l'enfocament anterior:
C++ #include using namespace std ; // Function to find minimum XOR pair int minXOR ( int arr [] int n ) { // Sort given array sort ( arr arr + n ); int minXor = INT_MAX ; int val = 0 ; // calculate min xor of consecutive pairs for ( int i = 0 ; i < n - 1 ; i ++ ) { val = arr [ i ] ^ arr [ i + 1 ]; minXor = min ( minXor val ); } return minXor ; } // Driver program int main () { int arr [] = { 9 5 3 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < minXOR ( arr n ) < < endl ; return 0 ; }
Java import java.util.Arrays ; class GFG { // Function to find minimum XOR pair static int minXOR ( int arr [] int n ) { // Sort given array Arrays . parallelSort ( arr ); int minXor = Integer . MAX_VALUE ; int val = 0 ; // calculate min xor of consecutive pairs for ( int i = 0 ; i < n - 1 ; i ++ ) { val = arr [ i ] ^ arr [ i + 1 ] ; minXor = Math . min ( minXor val ); } return minXor ; } // Driver program public static void main ( String args [] ) { int arr [] = { 9 5 3 }; int n = arr . length ; System . out . println ( minXOR ( arr n )); } } // This code is contributed by Sumit Ghosh
Python3 import sys # Function to find minimum XOR pair def minXOR ( arr n ): # Sort given array arr . sort () minXor = int ( sys . float_info . max ) val = 0 # calculate min xor of consecutive pairs for i in range ( 0 n - 1 ): val = arr [ i ] ^ arr [ i + 1 ]; minXor = min ( minXor val ); return minXor # Driver program arr = [ 9 5 3 ] n = len ( arr ) print ( minXOR ( arr n )) # This code is contributed by Sam007.
C# // C# program to find minimum // XOR value in an array. using System ; class GFG { // Function to find minimum XOR pair static int minXOR ( int [] arr int n ) { // Sort given array Array . Sort ( arr ); int minXor = int . MaxValue ; int val = 0 ; // calculate min xor of consecutive pairs for ( int i = 0 ; i < n - 1 ; i ++ ) { val = arr [ i ] ^ arr [ i + 1 ]; minXor = Math . Min ( minXor val ); } return minXor ; } // Driver program public static void Main () { int [] arr = { 9 5 3 }; int n = arr . Length ; Console . WriteLine ( minXOR ( arr n )); } } // This code is contributed by Sam007
PHP // Function to find minimum XOR pair function minXOR ( $arr $n ) { // Sort given array sort ( $arr ); $minXor = PHP_INT_MAX ; $val = 0 ; // calculate min xor // of consecutive pairs for ( $i = 0 ; $i < $n - 1 ; $i ++ ) { $val = $arr [ $i ] ^ $arr [ $i + 1 ]; $minXor = min ( $minXor $val ); } return $minXor ; } // Driver Code $arr = array ( 9 5 3 ); $n = count ( $arr ); echo minXOR ( $arr $n ); // This code is contributed by Smitha. ?>
JavaScript < script > // Function to find minimum XOR pair function minXOR ( arr n ) { // Sort given array arr . sort (); let minXor = Number . MAX_VALUE ; let val = 0 ; // calculate min xor of consecutive pairs for ( let i = 0 ; i < n - 1 ; i ++ ) { val = arr [ i ] ^ arr [ i + 1 ]; minXor = Math . min ( minXor val ); } return minXor ; } // Driver program let arr = [ 9 5 3 ]; let n = arr . length ; document . write ( minXOR ( arr n )); < /script>
Sortida
6
Complexitat temporal: O(N*logN)
Complexitat espacial: O(1)
Un més més Solució eficient pot resoldre el problema anterior en temps O(n) en el supòsit que els enters necessiten un nombre fix de bits per emmagatzemar. La idea és utilitzar Trie Data Structure.
Algorisme:
- Crea un assaig buit. Cada node de trie conté dos fills per a 0 i 1 bits.
- Inicialitzar min_xor = INT_MAX inserir arr[0] a trie
- Travessa tots els elements de la matriu un per un a partir del segon.
- Primer trobeu el valor mínim de diferència setbet a trie
- fer xor de l'element actual amb un valor mínim de setbit diferencial
- actualitzeu el valor min_xor si cal
- inseriu l'element de matriu actual a trie
- torna min_xor
A continuació es mostra la implementació de l'algorisme anterior.
C++ // C++ program to find minimum XOR value in an array. #include using namespace std ; #define INT_SIZE 32 // A Trie Node struct TrieNode { int value ; // used in leaf node TrieNode * Child [ 2 ]; }; // Utility function to create a new Trie node TrieNode * getNode () { TrieNode * newNode = new TrieNode ; newNode -> value = 0 ; newNode -> Child [ 0 ] = newNode -> Child [ 1 ] = NULL ; return newNode ; } // utility function insert new key in trie void insert ( TrieNode * root int key ) { TrieNode * temp = root ; // start from the most significant bit insert all // bit of key one-by-one into trie for ( int i = INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix bool current_bit = ( key & ( 1 < < i )); // Add a new Node into trie if ( temp -> Child [ current_bit ] == NULL ) temp -> Child [ current_bit ] = getNode (); temp = temp -> Child [ current_bit ]; } // store value at leafNode temp -> value = key ; } // Returns minimum XOR value of an integer inserted // in Trie and given key. int minXORUtil ( TrieNode * root int key ) { TrieNode * temp = root ; for ( int i = INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix bool current_bit = ( key & ( 1 < < i )); // Traversal Trie look for prefix that has // same bit if ( temp -> Child [ current_bit ] != NULL ) temp = temp -> Child [ current_bit ]; // if there is no same bit.then looking for // opposite bit else if ( temp -> Child [ 1 - current_bit ] != NULL ) temp = temp -> Child [ 1 - current_bit ]; } // return xor value of minimum bit difference value // so we get minimum xor value return key ^ temp -> value ; } // Returns minimum xor value of pair in arr[0..n-1] int minXOR ( int arr [] int n ) { int min_xor = INT_MAX ; // Initialize result // create a True and insert first element in it TrieNode * root = getNode (); insert ( root arr [ 0 ]); // Traverse all array element and find minimum xor // for every element for ( int i = 1 ; i < n ; i ++ ) { // Find minimum XOR value of current element with // previous elements inserted in Trie min_xor = min ( min_xor minXORUtil ( root arr [ i ])); // insert current array value into Trie insert ( root arr [ i ]); } return min_xor ; } // Driver code int main () { int arr [] = { 9 5 3 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); cout < < minXOR ( arr n ) < < endl ; return 0 ; }
Java // Java program to find minimum XOR value in an array. class GFG { static final int INT_SIZE = 32 ; // A Trie Node static class TrieNode { int value ; // used in leaf node TrieNode [] Child = new TrieNode [ 2 ] ; public TrieNode () { value = 0 ; Child [ 0 ] = null ; Child [ 1 ] = null ; } } static TrieNode root ; // utility function insert new key in trie static void insert ( int key ) { TrieNode temp = root ; // start from the most significant bit insert all // bit of key one-by-one into trie for ( int i = INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix int current_bit = ( key & ( 1 < < i )) >= 1 ? 1 : 0 ; // Add a new Node into trie if ( temp != null && temp . Child [ current_bit ] == null ) temp . Child [ current_bit ] = new TrieNode (); temp = temp . Child [ current_bit ] ; } // store value at leafNode temp . value = key ; } // Returns minimum XOR value of an integer inserted // in Trie and given key. static int minXORUtil ( int key ) { TrieNode temp = root ; for ( int i = INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix int current_bit = ( key & ( 1 < < i )) >= 1 ? 1 : 0 ; // Traversal Trie look for prefix that has // same bit if ( temp . Child [ current_bit ] != null ) temp = temp . Child [ current_bit ] ; // if there is no same bit.then looking for // opposite bit else if ( temp . Child [ 1 - current_bit ] != null ) temp = temp . Child [ 1 - current_bit ] ; } // return xor value of minimum bit difference value // so we get minimum xor value return key ^ temp . value ; } // Returns minimum xor value of pair in arr[0..n-1] static int minXOR ( int arr [] int n ) { int min_xor = Integer . MAX_VALUE ; // Initialize result // create a True and insert first element in it root = new TrieNode (); insert ( arr [ 0 ] ); // Traverse all array element and find minimum xor // for every element for ( int i = 1 ; i < n ; i ++ ) { // Find minimum XOR value of current element with // previous elements inserted in Trie min_xor = Math . min ( min_xor minXORUtil ( arr [ i ] )); // insert current array value into Trie insert ( arr [ i ] ); } return min_xor ; } // Driver code public static void main ( String args [] ) { int arr [] = { 9 5 3 }; int n = arr . length ; System . out . println ( minXOR ( arr n )); } } // This code is contributed by Sumit Ghosh
Python # class for the basic Trie Node class TrieNode : def __init__ ( self ): # Child array with 0 and 1 self . child = [ None ] * 2 # meant for the lead Node self . value = None class Trie : def __init__ ( self ): # initialise the root Node self . root = self . getNode () def getNode ( self ): # get a new Trie Node return TrieNode () # inserts a new element def insert ( self key ): temp = self . root # 32 bit valued binary digit for i in range ( 31 - 1 - 1 ): # finding the bit at ith position curr = ( key >> i ) & ( 1 ) # if the child is None create one if ( temp . child [ curr ] is None ): temp . child [ curr ] = self . getNode () temp = temp . child [ curr ] # add value to the leaf node temp . value = key # traverse the trie and xor with the most similar element def xorUtil ( self key ): temp = self . root # 32 bit valued binary digit for i in range ( 31 - 1 - 1 ): # finding the bit at ith position curr = ( key >> i ) & 1 # traverse for the same bit if ( temp . child [ curr ] is not None ): temp = temp . child [ curr ] # traverse if the same bit is not set in trie elif ( temp . child [ 1 - curr ] is not None ): temp = temp . child [ 1 - curr ] # return with the xor of the value return temp . value ^ key def minXor ( arr ): # set m to a large number m = 2 ** 30 # initialize Trie trie = Trie () # insert the first element trie . insert ( arr [ 0 ]) # for each element in the array for i in range ( 1 len ( arr )): # find the minimum xor value m = min ( m trie . xorUtil ( arr [ i ])) # insert the new element trie . insert ( arr [ i ]) return m # Driver Code if __name__ == '__main__' : sample = [ 9 5 3 ] print ( minXor ( sample )) #code contributed by Shushant Kumar
C# // Include namespace system using System ; // C# program to find minimum XOR value in an array. public class GFG { public const int INT_SIZE = 32 ; // A Trie Node public class TrieNode { public int value ; // used in leaf node public TrieNode [] Child = new TrieNode [ 2 ]; public TrieNode () { this . value = 0 ; this . Child [ 0 ] = null ; this . Child [ 1 ] = null ; } } public static TrieNode root ; // utility function insert new key in trie public static void insert ( int key ) { var temp = root ; // start from the most significant bit insert all // bit of key one-by-one into trie for ( int i = GFG . INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix var current_bit = ( key & ( 1 < < i )) >= 1 ? 1 : 0 ; // Add a new Node into trie if ( temp != null && temp . Child [ current_bit ] == null ) { temp . Child [ current_bit ] = new TrieNode (); } temp = temp . Child [ current_bit ]; } // store value at leafNode temp . value = key ; } // Returns minimum XOR value of an integer inserted // in Trie and given key. public static int minXORUtil ( int key ) { var temp = root ; for ( int i = GFG . INT_SIZE - 1 ; i >= 0 ; i -- ) { // Find current bit in given prefix var current_bit = ( key & ( 1 < < i )) >= 1 ? 1 : 0 ; // Traversal Trie look for prefix that has // same bit if ( temp . Child [ current_bit ] != null ) { temp = temp . Child [ current_bit ]; } else if ( temp . Child [ 1 - current_bit ] != null ) { temp = temp . Child [ 1 - current_bit ]; } } // return xor value of minimum bit difference value // so we get minimum xor value return key ^ temp . value ; } // Returns minimum xor value of pair in arr[0..n-1] public static int minXOR ( int [] arr int n ) { var min_xor = int . MaxValue ; // Initialize result // create a True and insert first element in it root = new TrieNode (); GFG . insert ( arr [ 0 ]); // Traverse all array element and find minimum xor // for every element for ( int i = 1 ; i < n ; i ++ ) { // Find minimum XOR value of current element with // previous elements inserted in Trie min_xor = Math . Min ( min_xor GFG . minXORUtil ( arr [ i ])); // insert current array value into Trie GFG . insert ( arr [ i ]); } return min_xor ; } // Driver code public static void Main ( String [] args ) { int [] arr = { 9 5 3 }; var n = arr . Length ; Console . WriteLine ( GFG . minXOR ( arr n )); } } // This code is contributed by aadityaburujwale.
JavaScript class TrieNode { constructor () { this . child = new Array ( 2 ); this . value = null ; } } class Trie { constructor () { this . root = this . getNode (); } getNode () { return new TrieNode (); } insert ( key ) { let temp = this . root ; for ( let i = 31 ; i >= 0 ; i -- ) { let curr = ( key >> i ) & 1 ; if ( ! temp . child [ curr ]) temp . child [ curr ] = this . getNode (); temp = temp . child [ curr ]; } temp . value = key ; } xorUtil ( key ) { let temp = this . root ; for ( let i = 31 ; i >= 0 ; i -- ) { let curr = ( key >> i ) & 1 ; if ( temp . child [ curr ]) temp = temp . child [ curr ]; else if ( temp . child [ 1 - curr ]) temp = temp . child [ 1 - curr ]; } return temp . value ^ key ; } } function minXor ( arr ) { let m = 2 ** 30 ; let trie = new Trie (); trie . insert ( arr [ 0 ]); for ( let i = 1 ; i < arr . length ; i ++ ) { m = Math . min ( m trie . xorUtil ( arr [ i ])); trie . insert ( arr [ i ]); } return m ; } if ( typeof module !== 'undefined' ) { module . exports = { minXor : minXor }; } console . log ( minXor ([ 9 5 3 ])); // This code is contributed by akashish__
Sortida
6
Complexitat temporal O(n)
Complexitat espacial: O(n*INT_SIZE)
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