Paire de valeurs XOR minimales
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#practiceLinkDiv { display : aucun !important; } Étant donné un tableau d’entiers. Recherchez la paire dans un tableau qui a une valeur XOR minimale.
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 Paire de valeurs XOR minimales Essayez-le ! UN Solution simple consiste à générer toutes les paires du tableau donné et à calculer XOR leurs valeurs. Enfin, renvoyez la valeur XOR minimale. Cette solution prend O(n 2 ) temps.
Mise en œuvre:
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>
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Complexité spatiale : O(1)
Un Solution efficace peut résoudre ce problème en un temps O(nlogn).
Algorithme:
- Trier le tableau donné
- Parcourez et vérifiez XOR pour chaque paire consécutive
Vous trouverez ci-dessous la mise en œuvre de l’approche ci-dessus :
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>
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Complexité temporelle : O(N*logN)
Complexité spatiale : O(1)
Un autre plus Solution efficace peut résoudre le problème ci-dessus en un temps O(n) en supposant que les entiers prennent un nombre fixe de bits à stocker. L'idée est d'utiliser Trie Data Structure.
Algorithme:
- Créez un essai vide. Chaque nœud de trie contient deux enfants pour 0 et 1 bits.
- Initialiser min_xor = INT_MAX insérer arr[0] dans trie
- Parcourez tous les éléments du tableau un par un à partir de la seconde.
- Trouvez d'abord la valeur de différence de mise minimale dans trie
- faire xor de l'élément actuel avec la différence minimale de setbit avec cette valeur
- mettre à jour la valeur min_xor si nécessaire
- insérer l'élément de tableau actuel dans le trie
- retourner min_xor
Vous trouverez ci-dessous la mise en œuvre de l'algorithme ci-dessus.
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__
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Complexité temporelle O(n)
Complexité spatiale : O(n*INT_SIZE)
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