Choose an integer K such that maximum of the xor values of K with all Array elements is minimized Last Updated : 12 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given an array A consisting of N non-negative integers, the task is to choose an integer K such that the maximum of the xor values of K with all array elements is minimized. In other words find the minimum possible value of Z, where Z = max(A[i] xor K), 0 <= i <= n-1, for some value of K. Examples: Input: A = [1, 2, 3] Output: 2 Explanation: On choosing K = 3, max(A[i] xor 3) = 2, and this is the minimum possible value.Input: A = [3, 2, 5, 6] Output: 5 Approach: To solve the problem mentioned above we will use recursion. We will start from the most significant bit in the recursive function. In the recursive step, split the element into two sections - one having the current bit on and the other with current bit off. If any of the sections doesn't have a single element, then this particular bit for K can be chosen such that the final xor value has 0 at this bit position (since our aim is to minimise this value) and then proceed to the next bit in the next recursive step. If both the sections have some elements, then explore both the possibilities by placing 0 and 1 at this bit position and calculating the answer using the corresponding section in next recursive call. Let answer_on be the value if 1 is placed and answer_off be the value if 0 is placed at this position (pos). Since both sections are non empty whichever bit we choose for K, 2pos will be added to the final value. For each recursive step: answer = min(answer_on, answer_off) + 2pos Below is the implementation of the above approach: C++ // C++ implementation to find Minimum // possible value of the maximum xor // in an array by choosing some integer #include <bits/stdc++.h> using namespace std; // Function to calculate Minimum possible // value of the Maximum XOR in an array int calculate(vector<int>& section, int pos) { // base case if (pos < 0) return 0; // Divide elements into two sections vector<int> on_section, off_section; // Traverse all elements of current // section and divide in two groups for (auto el : section) { if (((el >> pos) & 1) == 0) off_section.push_back(el); else on_section.push_back(el); } // Check if one of the sections is empty if (off_section.size() == 0) return calculate(on_section, pos - 1); if (on_section.size() == 0) return calculate(off_section, pos - 1); // explore both the possibilities using recursion return min(calculate(off_section, pos - 1), calculate(on_section, pos - 1)) + (1 << pos); } // Function to calculate minimum XOR value int minXorValue(int a[], int n) { vector<int> section; for (int i = 0; i < n; i++) section.push_back(a[i]); // Start recursion from the // most significant pos position return calculate(section, 30); } // Driver code int main() { int N = 4; int A[N] = { 3, 2, 5, 6 }; cout << minXorValue(A, N); return 0; } Java // Java implementation to find Minimum // possible value of the maximum xor // in an array by choosing some integer import java.util.*; class GFG{ // Function to calculate Minimum possible // value of the Maximum XOR in an array static int calculate(Vector<Integer> section, int pos) { // Base case if (pos < 0) return 0; // Divide elements into two sections Vector<Integer> on_section = new Vector<Integer>(), off_section = new Vector<Integer>(); // Traverse all elements of current // section and divide in two groups for(int el : section) { if (((el >> pos) & 1) == 0) off_section.add(el); else on_section.add(el); } // Check if one of the sections is empty if (off_section.size() == 0) return calculate(on_section, pos - 1); if (on_section.size() == 0) return calculate(off_section, pos - 1); // Explore both the possibilities using recursion return Math.min(calculate(off_section, pos - 1), calculate(on_section, pos - 1)) + (1 << pos); } // Function to calculate minimum XOR value static int minXorValue(int a[], int n) { Vector<Integer> section = new Vector<Integer>(); for(int i = 0; i < n; i++) section.add(a[i]); // Start recursion from the // most significant pos position return calculate(section, 30); } // Driver code public static void main(String[] args) { int N = 4; int A[] = { 3, 2, 5, 6 }; System.out.print(minXorValue(A, N)); } } // This code is contributed by Princi Singh Python3 # Python3 implementation to find Minimum # possible value of the maximum xor # in an array by choosing some integer # Function to calculate Minimum possible # value of the Maximum XOR in an array def calculate(section, pos): # base case if (pos < 0): return 0 # Divide elements into two sections on_section = [] off_section = [] # Traverse all elements of current # section and divide in two groups for el in section: if (((el >> pos) & 1) == 0): off_section.append(el) else: on_section.append(el) # Check if one of the sections is empty if (len(off_section) == 0): return calculate(on_section, pos - 1) if (len(on_section) == 0): return calculate(off_section, pos - 1) # explore both the possibilities using recursion return min(calculate(off_section, pos - 1), calculate(on_section, pos - 1))+ (1 << pos) # Function to calculate minimum XOR value def minXorValue(a, n): section = [] for i in range( n): section.append(a[i]); # Start recursion from the # most significant pos position return calculate(section, 30) # Driver code if __name__ == "__main__": N = 4 A = [ 3, 2, 5, 6 ] print(minXorValue(A, N)) # This code is contributed by chitranayal C# // C# implementation to find minimum // possible value of the maximum xor // in an array by choosing some integer using System; using System.Collections.Generic; class GFG{ // Function to calculate minimum possible // value of the maximum XOR in an array static int calculate(List<int> section, int pos) { // Base case if (pos < 0) return 0; // Divide elements into two sections List<int> on_section = new List<int>(), off_section = new List<int>(); // Traverse all elements of current // section and divide in two groups foreach(int el in section) { if (((el >> pos) & 1) == 0) off_section.Add(el); else on_section.Add(el); } // Check if one of the sections is empty if (off_section.Count == 0) return calculate(on_section, pos - 1); if (on_section.Count == 0) return calculate(off_section, pos - 1); // Explore both the possibilities using recursion return Math.Min(calculate(off_section, pos - 1), calculate(on_section, pos - 1)) + (1 << pos); } // Function to calculate minimum XOR value static int minXorValue(int []a, int n) { List<int> section = new List<int>(); for(int i = 0; i < n; i++) section.Add(a[i]); // Start recursion from the // most significant pos position return calculate(section, 30); } // Driver code public static void Main(String[] args) { int N = 4; int []A = { 3, 2, 5, 6 }; Console.Write(minXorValue(A, N)); } } // This code is contributed by Princi Singh JavaScript <script> // Javascript implementation to find Minimum // possible value of the maximum xor // in an array by choosing some integer // Function to calculate Minimum possible // value of the Maximum XOR in an array function calculate(section, pos) { // base case if (pos < 0) return 0; // Divide elements into two sections let on_section = [], off_section = []; // Traverse all elements of current // section and divide in two groups for (let el = 0; el < section.length; el++) { if (((section[el] >> pos) & 1) == 0) off_section.push(section[el]); else on_section.push(section[el]); } // Check if one of the sections is empty if (off_section.length == 0) return calculate(on_section, pos - 1); if (on_section.length == 0) return calculate(off_section, pos - 1); // explore both the possibilities using recursion return Math.min(calculate(off_section, pos - 1), calculate(on_section, pos - 1)) + (1 << pos); } // Function to calculate minimum XOR value function minXorValue(a, n) { let section = []; for (let i = 0; i < n; i++) section.push(a[i]); // Start recursion from the // most significant pos position return calculate(section, 30); } // Driver code let N = 4; let A = [ 3, 2, 5, 6 ]; document.write(minXorValue(A, N)); </script> Output: 5 Time Complexity: O(N * log(max(Ai))Space complexity: O(NlogN) Comment More infoAdvertise with us Next Article Analysis of Algorithms K king_tsar Follow Improve Article Tags : DSA Bitwise-XOR Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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