# Largest possible value of M not exceeding N having equal Bitwise OR and XOR between them

Given an integer N, the task is to find the largest number M, where (M < N), such that N(XOR)M is equal to N(OR)M i.e. (N ^ M) = (N | M).

Examples:

Input: N = 5
Output:
5 ^ 4 = 1 and 5 | 4 = 5. Therefore, XOR and OR between them are not equal.
5 ^ 3 = 6 and 5 | 3 = 7. Therefore, XOR and OR between them are not equal.
5 ^ 2 = 7 and 5 | 2 = 7. Therefore, XOR and OR between them are equal.

Input: N = 14
Output:

Approach:
To get the required number M, traverse all the bits of N from its Least Significant Bit (LSB) to Most Significant Bit (MSB). Two cases arise here:

1. If the ith bit of N is 1 then:
• If the ith bit of M is set to 1, then N^M will not be equal to N|M as (1^1 = 0) and (1|1 = 1).
• If the ith bit is set of M to 0, then N^M will be equal to N|M as (1^0 = 1) and (1|0 = 1).
• So if the ith bit of N is 1, set the ith bit of M to 0.
2. If the ith bit of N is 0 then:
• If the ith bit of M is set to 1, then N^M will be equal to N|M as (0^1 = 1) and (0|1 = 1).
• If we set the ith bit of M to 0, then N^M will be equal to N|M as (0^0 = 0) and (0|0 = 0).
• So, if the ith bit of M is set to either 0 or 1, N^M will always be equal to N|M.
• As the largest value of M which is less than N has to be found out, always set the ith bit of M to 1.

Illustration:

• N = 5
• 32-bit representation of 5 = 00000000000000000000000000000101
• LSB index of 5 = 31
• MSB index of 5 = 29
• Traversing from LSB to MSB i.e. from 31 to 29:
• For index 31, N[31] = 1. So M[31] should be set to 0.
• For index 30, N[30] = 0. So M[30] should be set to 1.
• For index 29, N[29] = 1. So M[29] should be set to 0.
• Thus the 32-bit representation of M is 00000000000000000000000000000010, which is equal to 2 in decimal representation.

Below is the implementation of the above approach:

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find required ` `// number M ` `int` `equalXORandOR(``int` `n) ` `{ ` `    ``// Initialising m ` `    ``int` `m = 0; ` ` `  `    ``// Finding the index of the ` `    ``// most significant bit of N ` `    ``int` `MSB = (``int``)log2(n); ` ` `  `    ``// Calculating required number ` `    ``for` `(``int` `i = 0; i <= MSB; i++) { ` ` `  `        ``if` `(!(n & (1 << i))) { ` `            ``m += (1 << i); ` `        ``} ` `    ``} ` ` `  `    ``return` `m; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 14; ` `    ``cout << equalXORandOR(n); ` `    ``return` `0; ` `} `

 `// Java program to implement ` `// the above approach ` `class` `GFG{ ` ` `  `// Function to find required ` `// number M ` `static` `int` `equalXORandOR(``int` `n) ` `{ ` `     `  `    ``// Initialising m ` `    ``int` `m = ``0``; ` ` `  `    ``// Finding the index of the ` `    ``// most significant bit of N ` `    ``int` `MSB = (``int``)Math.log(n); ` ` `  `    ``// Calculating required number ` `    ``for``(``int` `i = ``0``; i <= MSB; i++) ` `    ``{ ` `        ``if` `((n & (``1` `<< i)) <= ``0``) ` `        ``{ ` `            ``m += (``1` `<< i); ` `        ``} ` `    ``} ` `    ``return` `m; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``14``; ` `     `  `    ``System.out.print(equalXORandOR(n)); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

 `# Python3 program to implement  ` `# the above approach  ` `from` `math ``import` `log2 ` ` `  `# Function to find required ` `# number M ` `def` `equalXORandOR(n): ` ` `  `    ``# Initialising m ` `    ``m ``=` `0` ` `  `    ``# Finding the index of the ` `    ``# most significant bit of N ` `    ``MSB ``=` `int``(log2(n)) ` ` `  `    ``# Calculating required number ` `    ``for` `i ``in` `range``(MSB ``+` `1``): ` `        ``if``(``not``(n & (``1` `<< i))): ` `            ``m ``+``=` `(``1` `<< i) ` ` `  `    ``return` `m ` ` `  `# Driver Code ` `n ``=` `14` ` `  `# Function call ` `print``(equalXORandOR(n)) ` ` `  `# This code is contributed by Shivam Singh`

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find required ` `// number M ` `static` `int` `equalXORandOR(``int` `n) ` `{ ` `     `  `    ``// Initialising m ` `    ``int` `m = 0; ` ` `  `    ``// Finding the index of the ` `    ``// most significant bit of N ` `    ``int` `MSB = (``int``)Math.Log(n); ` ` `  `    ``// Calculating required number ` `    ``for``(``int` `i = 0; i <= MSB; i++) ` `    ``{ ` `        ``if` `((n & (1 << i)) <= 0) ` `        ``{ ` `            ``m += (1 << i); ` `        ``} ` `    ``} ` `    ``return` `m; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 14; ` `     `  `    ``Console.Write(equalXORandOR(n)); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey`

Output:
```1
```

Time Complexity: O(log2 N)
Auxiliary Space: O(1)

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