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| 1 | +# Solution Explanation: Maximum XOR Queries |
| 2 | + |
| 3 | +This repository contains a solution for solving the **Maximum XOR Queries** problem in multiple programming languages: C++, Java, JavaScript, Python, and Go. Each solution is broken down step-by-step for ease of understanding. |
| 4 | + |
| 5 | +## Problem Summary |
| 6 | + |
| 7 | +Given an array of numbers (`nums`) and a parameter `maximumBit`, we are to find the maximum XOR result for each query by choosing an integer `k` that maximizes the XOR between `k` and the cumulative XOR of all elements in `nums`. The `k` must be within a range defined by `maximumBit`. |
| 8 | + |
| 9 | +## Steps to Solution |
| 10 | + |
| 11 | +The approach is the same across all languages but adapted to the syntax of each language. Here’s a language-by-language breakdown of how each solution works. |
| 12 | + |
| 13 | +--- |
| 14 | + |
| 15 | +### C++ Code Explanation |
| 16 | + |
| 17 | +1. **Initialize Variables**: |
| 18 | + - We start by defining `XORed`, an integer to store the cumulative XOR of all elements in `nums`. |
| 19 | + - We also define `max_k`, which is the maximum value achievable with `maximumBit` bits. This is calculated as `2^maximumBit - 1`. |
| 20 | + |
| 21 | +2. **Compute Initial XOR**: |
| 22 | + - Using a loop, XOR all elements in `nums` together to get an initial cumulative XOR (`XORed`) of the entire array. |
| 23 | + |
| 24 | +3. **Process Each Query in Reverse**: |
| 25 | + - We loop from the last element of `nums` to the first. |
| 26 | + - For each query, calculate `k` as `XORed ^ max_k`. This gives the integer `k` that maximizes the XOR. |
| 27 | + - Append `k` to our `answer` array. |
| 28 | + - Update `XORed` by removing the effect of the last element in `nums`. |
| 29 | + |
| 30 | +4. **Return the Result**: |
| 31 | + - The final `answer` array contains the results for all queries, and we return it. |
| 32 | + |
| 33 | +--- |
| 34 | + |
| 35 | +### Java Code Explanation |
| 36 | + |
| 37 | +1. **Initialize Variables**: |
| 38 | + - Define `XORed` to store the cumulative XOR of all elements in `nums`. |
| 39 | + - Calculate `max_k` as `(1 << maximumBit) - 1` to get the maximum value achievable within `maximumBit` bits. |
| 40 | + |
| 41 | +2. **Compute Initial XOR**: |
| 42 | + - Use a loop to XOR all elements in `nums`, updating `XORed` with each element. |
| 43 | + |
| 44 | +3. **Reverse Loop for Queries**: |
| 45 | + - Loop backward from the last element of `nums`. |
| 46 | + - For each query, calculate `k` as `XORed ^ max_k`. |
| 47 | + - Store `k` in the `answer` array. |
| 48 | + - Update `XORed` by XORing it with the current element to "remove" its effect. |
| 49 | + |
| 50 | +4. **Return the Final Answer**: |
| 51 | + - The `answer` array now contains all query results, which we return as our solution. |
| 52 | + |
| 53 | +--- |
| 54 | + |
| 55 | +### JavaScript Code Explanation |
| 56 | + |
| 57 | +1. **Initialize Variables**: |
| 58 | + - Define `XORed` to store the cumulative XOR of all numbers in `nums`. |
| 59 | + - Calculate `max_k` as `(1 << maximumBit) - 1` to get the highest possible value within `maximumBit` bits. |
| 60 | + |
| 61 | +2. **Compute Initial XOR**: |
| 62 | + - Loop over `nums` and calculate the cumulative XOR (`XORed`). |
| 63 | + |
| 64 | +3. **Process Each Query in Reverse**: |
| 65 | + - Start from the last element of `nums` and work backwards. |
| 66 | + - For each query, calculate `k` as `XORed ^ max_k`. |
| 67 | + - Store `k` in the `answer` array. |
| 68 | + - Update `XORed` by removing the effect of the last element. |
| 69 | + |
| 70 | +4. **Return the Final Array**: |
| 71 | + - Return `answer` as the array containing all query results. |
| 72 | + |
| 73 | +--- |
| 74 | + |
| 75 | +### Python Code Explanation |
| 76 | + |
| 77 | +1. **Initialize Variables**: |
| 78 | + - Define `XORed` to hold the cumulative XOR of all elements in `nums`. |
| 79 | + - Calculate `max_k` as `(1 << maximumBit) - 1` to get the largest value allowed by `maximumBit`. |
| 80 | + |
| 81 | +2. **Compute Cumulative XOR**: |
| 82 | + - Loop through each number in `nums` to calculate the cumulative XOR. |
| 83 | + |
| 84 | +3. **Process Each Query in Reverse**: |
| 85 | + - Start from the last element of `nums` and process each in reverse. |
| 86 | + - For each query, calculate `k` as `XORed ^ max_k` to get the maximum XOR. |
| 87 | + - Append `k` to the `answer` list. |
| 88 | + - Update `XORed` by XORing it with the last element in `nums`. |
| 89 | + |
| 90 | +4. **Return Results**: |
| 91 | + - `answer` now holds the maximum XOR result for each query, which we return as the solution. |
| 92 | + |
| 93 | +--- |
| 94 | + |
| 95 | +### Go Code Explanation |
| 96 | + |
| 97 | +1. **Initialize Variables**: |
| 98 | + - Define `XORed` to keep the cumulative XOR of all elements in `nums`. |
| 99 | + - Calculate `max_k` as `(1 << maximumBit) - 1` to get the maximum possible integer with `maximumBit` bits. |
| 100 | + |
| 101 | +2. **Compute Initial XOR**: |
| 102 | + - XOR all elements in `nums` to get the cumulative XOR. |
| 103 | + |
| 104 | +3. **Process Each Query in Reverse**: |
| 105 | + - Start from the end of `nums` and move backwards. |
| 106 | + - For each query, calculate `k` as `XORed ^ max_k`. |
| 107 | + - Add `k` to the `answer` slice. |
| 108 | + - Update `XORed` by removing the effect of the last element. |
| 109 | + |
| 110 | +4. **Return the Result Slice**: |
| 111 | + - Return `answer` as the final result, containing the XOR values for each query. |
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