Maximum XOR using K numbers from 1 to n Last Updated : 15 Feb, 2023 Comments Improve Suggest changes Like Article Like Report Given a positive integer n and k. Find maximum xor of 1 to n using at most k numbers. Xor sum of 1 to n is defined as 1 ^ 2 ^ 3 ^ ... ^ n.Examples : Input : n = 4, k = 3 Output : 7 Explanation Maximum possible xor sum is 1 ^ 2 ^ 4 = 7. Input : n = 11, k = 1 Output : 11 Explanation Maximum Possible xor sum is 11. If we have k = 1 then the maximum possible xor sum is 'n' itself. Now for k > 1, we can always have a number with its all bits set till the most significant set bit in 'n'. C++ // CPP program to find max xor sum // of 1 to n using atmost k numbers #include <bits/stdc++.h> using namespace std; typedef long long int ll; // To return max xor sum of 1 to n // using at most k numbers ll maxXorSum(ll n, ll k) { // If k is 1 then maximum // possible sum is n if (k == 1) return n; // Finding number greater than // or equal to n with most significant // bit same as n. For example, if n is // 4, result is 7. If n is 5 or 6, result // is 7 ll res = 1; while (res <= n) res <<= 1; // Return res - 1 which denotes // a number with all bits set to 1 return res - 1; } // Driver program int main() { ll n = 4, k = 3; cout << maxXorSum(n, k); return 0; } Java // Java program to find max xor sum // of 1 to n using atmost k numbers public class Main { // To return max xor sum of 1 to n // using at most k numbers static int maxXorSum(int n, int k) { // If k is 1 then maximum // possible sum is n if (k == 1) return n; // Finding number greater than // or equal to n with most significant // bit same as n. For example, if n is // 4, result is 7. If n is 5 or 6, result // is 7 int res = 1; while (res <= n) res <<= 1; // Return res - 1 which denotes // a number with all bits set to 1 return res - 1; } // Driver program to test maxXorSum() public static void main(String[] args) { int n = 4, k = 3; System.out.print(maxXorSum(n, k)); } } Python # Python3 code to find max xor sum # of 1 to n using atmost k numbers # To return max xor sum of 1 to n # using at most k numbers def maxXorSum( n , k ): # If k is 1 then maximum # possible sum is n if k == 1: return n # Finding number greater than # or equal to n with most significant # bit same as n. For example, if n is # 4, result is 7. If n is 5 or 6, result # is 7 res = 1 while res <= n: res <<= 1 # Return res - 1 which denotes # a number with all bits set to 1 return res - 1 # Driver code n = 4 k = 3 print( maxXorSum(n, k) ) # This code is contributed by Abhishek Sharma44. C# // C# program to find max xor sum // of 1 to n using atmost k numbers using System; public class main { // To return max xor sum of 1 to n // using at most k numbers static int maxXorSum(int n, int k) { // If k is 1 then maximum // possible sum is n if (k == 1) return n; // Finding number greater than // or equal to n with most significant // bit same as n. For example, if n is // 4, result is 7. If n is 5 or 6, result // is 7 int res = 1; while (res <= n) res <<= 1; // Return res - 1 which denotes // a number with all bits set to 1 return res - 1; } // Driver program public static void Main() { int n = 4, k = 3; Console.WriteLine(maxXorSum(n, k)); } } // This code is contributed by vt_m. PHP <?php // PHP program to find max xor sum // of 1 to n using atmost k numbers // To return max xor sum of 1 to n // using at most k numbers function maxXorSum($n, $k) { // If k is 1 then maximum // possible sum is n if ($k == 1) return $n; // Finding number greater than // or equal to n with most // significant bit same as n. // For example, if n is 4, result // is 7. If n is 5 or 6, result is 7 $res = 1; while ($res <= $n) $res <<= 1; // Return res - 1 which denotes // a number with all bits set to 1 return $res - 1; } // Driver code $n = 4; $k = 3; echo maxXorSum($n, $k); // This code is contributed by Mithun Kumar ?> JavaScript <script> // JavaScript program to find max xor sum // of 1 to n using atmost k numbers // To return max xor sum of 1 to n // using at most k numbers function maxXorSum(n, k) { // If k is 1 then maximum // possible sum is n if (k == 1) return n; // Finding number greater than // or equal to n with most significant // bit same as n. For example, if n is // 4, result is 7. If n is 5 or 6, result // is 7 let res = 1; while (res <= n) res <<= 1; // Return res - 1 which denotes // a number with all bits set to 1 return res - 1; } // Driver code let n = 4, k = 3; document.write(maxXorSum(n, k)); </script> Output7 Time Complexity : O(logn)Auxiliary Space : O(1) Maximum XOR using K numbers from 1 to n Comment More infoAdvertise with us Next Article Analysis of Algorithms N nuclode Follow Improve Article Tags : Bit Magic DSA Bitwise-XOR Practice Tags : Bit Magic 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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