Având în vedere o matrice de numere întregi pozitive distincte și un număr x, găsiți numărul de perechi de numere întregi din tabloul al cărui XOR este egal cu x.
Exemple:
Input : arr[] = {5 4 10 15 7 6} x = 5 Output : 1 Explanation : (10 ^ 15) = 5 Input : arr[] = {3 6 8 10 15 50} x = 5 Output : 2 Explanation : (3 ^ 6) = 5 and (10 ^ 15) = 5 O Soluție simplă este de a parcurge fiecare element și de a verifica dacă există un alt număr al cărui XOR cu el este egal cu x. Această soluție ia O(n2) timp. Un solutie eficienta pentru această problemă este nevoie de timp O(n). Ideea se bazează pe faptul că arr[i] ^ arr[j] este egal cu x dacă și numai dacă arr[i] ^ x este egal cu arr[j].
1) Initialize result as 0. 2) Create an empty hash set 's'. 3) Do following for each element arr[i] in arr[] (a) If x ^ arr[i] is in 's' then increment result by 1. (b) Insert arr[i] into the hash set 's'. 3) return result.
Implementare:
C++// C++ program to Count all pair with given XOR // value x #include
// Java program to Count all pair with // given XOR value x import java.util.*; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int arr[] int n int x) { int result = 0; // Initialize result // create empty set that stores the visiting // element of array. // Refer below post for details of unordered_set // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/ HashSet<Integer> s = new HashSet<Integer>(); for (int i = 0; i < n; i++) { // If there exist an element in set s // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (s.contains(x ^ arr[i]) && (x ^ arr[i]) == (int) s.toArray()[s.size() - 1]) { result++; } // Make element visited s.add(arr[i]); } // return total count of // pairs with XOR equal to x return result; } // Driver code public static void main(String[] args) { int arr[] = {5 4 10 15 7 6}; int n = arr.length; int x = 5; System.out.print('Count of pairs with given XOR = ' + xorPairCount(arr n x)); } } // This code contributed by Rajput-Ji
# Python3 program to count all the pair # with given xor # Returns count of pairs in arr[0..n-1] # with XOR value equals to x. def xorPairCount(arr n x): result = 0 # Initialize result # create empty set that stores the # visiting element of array. s = set() for i in range(0 n): # If there exist an element in set s # with XOR equals to x^arr[i] that # means there exist an element such # that the XOR of element with arr[i] # is equal to x then increment count. if(x ^ arr[i] in s): result = result + 1 # Make element visited s.add(arr[i]) return result # Driver Code if __name__ == '__main__': arr = [5 4 10 15 7 6] n = len(arr) x = 5 print('Count of pair with given XOR = ' + str(xorPairCount(arr n x))) # This code is contributed by Anubhav Natani
// C# program to Count all pair with // given XOR value x using System; using System.Collections.Generic; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int []arr int n int x) { int result = 0; // Initialize result // create empty set that stores the visiting // element of array. // Refer below post for details of unordered_set // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/ HashSet<int> s = new HashSet<int>(); for (int i = 0; i < n; i++) { // If there exist an element in set s // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (s.Contains(x ^ arr[i])) { result++; } // Make element visited s.Add(arr[i]); } // return total count of // pairs with XOR equal to x return result; } // Driver code public static void Main() { int []arr = {5 4 10 15 7 6}; int n = arr.Length; int x = 5; Console.WriteLine('Count of pairs with given XOR = ' + xorPairCount(arr n x)); } } /* This code contributed by PrinciRaj1992 */
<script> // Javascript program to Count all pair with // given XOR value x // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. function xorPairCount(arrnx) { let result = 0; // Initialize result // create empty set that stores the visiting // element of array. // Refer below post for details of unordered_set // https://www.geeksforgeeks.org/cpp/unordered_set-in-cpp-stl/ let s = new Set(); for (let i = 0; i < n; i++) { // If there exist an element in set s // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (s.has(x ^ arr[i]) ) { result++; } // Make element visited s.add(arr[i]); } // return total count of // pairs with XOR equal to x return result; } // Driver code let arr=[5 4 10 15 7 6]; let n = arr.length; let x = 5; document.write('Count of pairs with given XOR = ' + xorPairCount(arr n x)); // This code is contributed by unknown2108 </script>
Ieșire
Count of pairs with given XOR = 1
Complexitatea timpului: Pe)
Spațiu auxiliar: O(n)
Cum să gestionezi duplicatele?
Soluția eficientă de mai sus nu funcționează dacă există duplicate în matricea de intrare. De exemplu, soluția de mai sus produce rezultate diferite pentru {2 2 5} și {5 2 2}. Pentru a gestiona duplicatele, stocăm numărul de apariții ale tuturor elementelor. Folosim unordered_map în loc de unordered_set.
Implementare:
C++// C++ program to Count all pair with given XOR // value x #include
// Java program to Count all pair with given XOR // value x import java.util.*; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int arr[] int n int x) { int result = 0; // Initialize result // create empty map that stores counts of // individual elements of array. Map<IntegerInteger> m = new HashMap<>(); for (int i = 0; i < n ; i++) { int curr_xor = x^arr[i]; // If there exist an element in map m // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (m.containsKey(curr_xor)) result += m.get(curr_xor); // Increment count of current element if(m.containsKey(arr[i])) { m.put(arr[i] m.get(arr[i]) + 1); } else{ m.put(arr[i] 1); } } // return total count of pairs with XOR equal to x return result; } // Driver code public static void main(String[] args) { int arr[] = {2 5 2}; int n = arr.length; int x = 0; System.out.println('Count of pairs with given XOR = ' + xorPairCount(arr n x)); } } // This code has been contributed by 29AjayKumar
# Python3 program to Count all pair with # given XOR value x # Returns count of pairs in arr[0..n-1] # with XOR value equals to x. def xorPairCount(arr n x): result = 0 # Initialize result # create empty map that stores counts # of individual elements of array. m = dict() for i in range(n): curr_xor = x ^ arr[i] # If there exist an element in map m # with XOR equals to x^arr[i] that # means there exist an element such that # the XOR of element with arr[i] is equal # to x then increment count. if (curr_xor in m.keys()): result += m[curr_xor] # Increment count of current element if arr[i] in m.keys(): m[arr[i]] += 1 else: m[arr[i]] = 1 # return total count of pairs # with XOR equal to x return result # Driver Code arr = [2 5 2] n = len(arr) x = 0 print('Count of pairs with given XOR = ' xorPairCount(arr n x)) # This code is contributed by Mohit Kumar
// C# program to Count all pair with given XOR // value x using System; using System.Collections.Generic; class GFG { // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. static int xorPairCount(int []arr int n int x) { int result = 0; // Initialize result // create empty map that stores counts of // individual elements of array. Dictionary<intint> m = new Dictionary<intint>(); for (int i = 0; i < n ; i++) { int curr_xor = x^arr[i]; // If there exist an element in map m // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (m.ContainsKey(curr_xor)) result += m[curr_xor]; // Increment count of current element if(m.ContainsKey(arr[i])) { var val = m[arr[i]]; m.Remove(arr[i]); m.Add(arr[i] val + 1); } else { m.Add(arr[i] 1); } } // return total count of pairs with XOR equal to x return result; } // Driver code public static void Main(String[] args) { int []arr = {2 5 2}; int n = arr.Length; int x = 0; Console.WriteLine('Count of pairs with given XOR = ' + xorPairCount(arr n x)); } } // This code has been contributed by 29AjayKumar
<script> // Javascript program to Count all pair with given XOR // value x // Returns count of pairs in arr[0..n-1] with XOR // value equals to x. function xorPairCount(arr n x) { let result = 0; // Initialize result // create empty map that stores counts of // individual elements of array. let m = new Map(); for (let i = 0; i < n ; i++) { let curr_xor = x^arr[i]; // If there exist an element in map m // with XOR equals to x^arr[i] that means // there exist an element such that the // XOR of element with arr[i] is equal to // x then increment count. if (m.has(curr_xor)) result += m.get(curr_xor); // Increment count of current element if(m.has(arr[i])) { m.set(arr[i] m.get(arr[i]) + 1); } else{ m.set(arr[i] 1); } } // return total count of pairs with XOR equal to x return result; } // Driver program let arr = [2 5 2]; let n = arr.length; let x = 0; document.write('Count of pairs with given XOR = ' + xorPairCount(arr n x)); </script>
Ieșire
Count of pairs with given XOR = 1
Complexitatea timpului: O(n)
Spațiu auxiliar: O(n)