# Maximize XOR by selecting 3 numbers in range [0, A], [0, B], and [0, C] respectively

Given 3 integers **A**, **B**, **C**, the task is to find the maximum XOR value of three number selected one from each range [0, A], [0, B], [0, C] respectively.

**Example: **

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Input:A = 1, B = 2, C = 4Output:7Explanation:Maximum XOR can be calculated by selecting 1 (from [0, 1]), 2 (from [0, 2]) & 4 (from [0, 4]) i.e.1âŠ• 2âŠ• 4 = 7

Input:A = 6, B = 2, C = 10Output:15Explanation:Maximum XOR can be calculated by selecting 5 (from [0, 6]), 1 (from [0, 2]) & 10 (from [0, 10]) i.e.5âŠ• 1âŠ• 10 = 15.

**Naive Approach:** Generate all possible triplets in the above ranges and calculate the maximum XOR possible.

**Time Complexity: O(A*B*C)**

**Auxiliary Space: O(1)**

**Efficient Approach: **

To find the largest value of an XOR operation, the value of XOR should have every bit to be a set bit i.e In a 32-bit number, the goal is to get the most 1 set starting left to right. This can be done by following the below steps:

- Create a variable
**ans**which will store the value of maximum xor possible and initialize it to zero. - To create a 32-bit value that would be maximum xor possible, run a loop for each of its bits from most significant bit to least significant bit.
- Now for each bit:
- Create a variable
**cur**that will store the minimum number in which that bit is set. - Check each range A, B, C and try to find a number which has a set bit on that position. This can be done by comparing A, B & C with cur, i.e. if any number from A, B & C is greater than or equal to cur then it will contain a set bit in that position. If a set bit is found in any of the number from A, B, C then maximum xor will also contain set bit at that position.
- Now, if a set bit is found, then add cur to ans to set that bit in it, and decrease the value of the number in which the set bit is found (from A, B, C) by cur to unset that bit as it cannot be used again.
- Do the above steps for each bit, to calculate the final value of answer.

- Create a variable
- After the above steps, print the required result.

**Implementation: **

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to calculate` `// maximum triplet XOR form 3 ranges` `int` `maximumTripletXOR(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` `// Initialize a variable` ` ` `// to store the answer` ` ` `int` `ans = 0;` ` ` `for` `(` `int` `i = 30; i >= 0; i--) {` ` ` `// create minimum number` ` ` `// that have a set bit` ` ` `// at ith position` ` ` `int` `cur = 1 << i;` ` ` `// Check for each number` ` ` `// and try to greedily choose` ` ` `// the bit if possible` ` ` `// If A >= cur then A also have a` ` ` `// set bit at ith position` ` ` `if` `(A >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from A` ` ` `A -= cur;` ` ` `}` ` ` `// Check for B` ` ` `else` `if` `(B >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from B` ` ` `B -= cur;` ` ` `}` ` ` `// Check for C` ` ` `else` `if` `(C >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from C` ` ` `C -= cur;` ` ` `}` ` ` `// If any of the above conditions` ` ` `// is not satisfied then` ` ` `// there is no way to turn that bit ON.` ` ` `}` ` ` `return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `A = 6;` ` ` `int` `B = 2;` ` ` `int` `C = 10;` ` ` `cout << maximumTripletXOR(A, B, C);` `}` |

## Java

`// Java Program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to calculate` `// maximum triplet XOR form 3 ranges` `static` `int` `maximumTripletXOR(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` ` ` `// Initialize a variable` ` ` `// to store the answer` ` ` `int` `ans = ` `0` `;` ` ` `for` `(` `int` `i = ` `30` `; i >= ` `0` `; i--) {` ` ` `// create minimum number` ` ` `// that have a set bit` ` ` `// at ith position` ` ` `int` `cur = ` `1` `<< i;` ` ` `// Check for each number` ` ` `// and try to greedily choose` ` ` `// the bit if possible` ` ` `// If A >= cur then A also have a` ` ` `// set bit at ith position` ` ` `if` `(A >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from A` ` ` `A -= cur;` ` ` `}` ` ` `// Check for B` ` ` `else` `if` `(B >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from B` ` ` `B -= cur;` ` ` `}` ` ` `// Check for C` ` ` `else` `if` `(C >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from C` ` ` `C -= cur;` ` ` `}` ` ` `// If any of the above conditions` ` ` `// is not satisfied then` ` ` `// there is no way to turn that bit ON.` ` ` `}` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `A = ` `6` `;` ` ` `int` `B = ` `2` `;` ` ` `int` `C = ` `10` `;` ` ` `System.out.print(maximumTripletXOR(A, B, C));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python Program to implement` `# the above approach` `# Function to calculate` `# maximum triplet XOR form 3 ranges` `def` `maximumTripletXOR(A, B, C):` ` ` ` ` `# Initialize a variable` ` ` `# to store the answer` ` ` `ans ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(` `30` `, ` `-` `1` `,` `-` `1` `):` ` ` ` ` `# create minimum number` ` ` `# that have a set bit` ` ` `# at ith position` ` ` `cur ` `=` `1` `<< i` ` ` ` ` `# Check for each number` ` ` `# and try to greedily choose` ` ` `# the bit if possible` ` ` `# If A >= cur then A also have a` ` ` `# set bit at ith position` ` ` `if` `(A >` `=` `cur):` ` ` ` ` `# Increment the answer` ` ` `ans ` `+` `=` `cur` ` ` ` ` `# Subtract this value from A` ` ` `A ` `-` `=` `cur` ` ` `# Check for B` ` ` `elif` `(B >` `=` `cur):` ` ` ` ` `# Increment the answer` ` ` `ans ` `+` `=` `cur` ` ` ` ` `# Subtract this value from B` ` ` `B ` `-` `=` `cur` ` ` ` ` `# Check for C` ` ` `elif` `(C >` `=` `cur):` ` ` `# Increment the answer` ` ` `ans ` `+` `=` `cur` ` ` ` ` `# Subtract this value from C` ` ` `C ` `-` `=` `cur` ` ` ` ` `# If any of the above conditions` ` ` `# is not satisfied then` ` ` `# there is no way to turn that bit ON.` ` ` ` ` `return` `ans` ` ` `# Driver Code` `A ` `=` `6` `B ` `=` `2` `C ` `=` `10` `print` `(maximumTripletXOR(A, B, C))` `# This code is contributed by shivanisinghss2110` |

## C#

`// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{` `// Function to calculate` `// maximum triplet XOR form 3 ranges` `static` `int` `maximumTripletXOR(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` ` ` `// Initialize a variable` ` ` `// to store the answer` ` ` `int` `ans = 0;` ` ` `for` `(` `int` `i = 30; i >= 0; i--) {` ` ` `// create minimum number` ` ` `// that have a set bit` ` ` `// at ith position` ` ` `int` `cur = 1 << i;` ` ` `// Check for each number` ` ` `// and try to greedily choose` ` ` `// the bit if possible` ` ` `// If A >= cur then A also have a` ` ` `// set bit at ith position` ` ` `if` `(A >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from A` ` ` `A -= cur;` ` ` `}` ` ` `// Check for B` ` ` `else` `if` `(B >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from B` ` ` `B -= cur;` ` ` `}` ` ` `// Check for C` ` ` `else` `if` `(C >= cur) {` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from C` ` ` `C -= cur;` ` ` `}` ` ` `// If any of the above conditions` ` ` `// is not satisfied then` ` ` `// there is no way to turn that bit ON.` ` ` `}` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `A = 6;` ` ` `int` `B = 2;` ` ` `int` `C = 10;` ` ` `Console.Write(maximumTripletXOR(A, B, C));` `}` `}` `// This code is contributed by shivanisinghss2110` |

## Javascript

`<script>` `// JavaScript program to implement` `// the above approach` `// Function to calculate maximum ` `// triplet XOR form 3 ranges` `function` `maximumTripletXOR(A, B, C)` `{` ` ` ` ` `// Initialize a variable` ` ` `// to store the answer` ` ` `let ans = 0;` ` ` `for` `(let i = 30; i >= 0; i--)` ` ` `{` ` ` ` ` `// Create minimum number` ` ` `// that have a set bit` ` ` `// at ith position` ` ` `let cur = 1 << i;` ` ` `// Check for each number` ` ` `// and try to greedily choose` ` ` `// the bit if possible` ` ` `// If A >= cur then A also have a` ` ` `// set bit at ith position` ` ` `if` `(A >= cur) ` ` ` `{` ` ` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from A` ` ` `A -= cur;` ` ` `}` ` ` `// Check for B` ` ` `else` `if` `(B >= cur)` ` ` `{` ` ` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from B` ` ` `B -= cur;` ` ` `}` ` ` `// Check for C` ` ` `else` `if` `(C >= cur)` ` ` `{` ` ` ` ` `// Increment the answer` ` ` `ans += cur;` ` ` `// Subtract this value from C` ` ` `C -= cur;` ` ` `}` ` ` `// If any of the above conditions` ` ` `// is not satisfied then` ` ` `// there is no way to turn that bit ON.` ` ` `}` ` ` `return` `ans;` `}` `// Driver Code` `let A = 6;` `let B = 2;` `let C = 10;` `document.write(maximumTripletXOR(A, B, C));` `// This code is contributed by Potta Lokesh` `</script>` |

**Output**

15

**Time Complexity: **O(1)

**Auxiliary Space: **O(1)