Given an integer n, you must transform it into 0 using the following operations any number of times: Change the rightmost (0th) bit in the binary representation of n. Change the ith bit in the binary representation of n if the (i-1)th bit is set to

minimumOneBitOperations.js

/**
 * Calculates the minimum number of operations to reduce a number `n` to 0
 * using a specific bitwise operation rule:
 * - You can flip the lowest 1-bit and all bits to its right.
 *
 * This function uses a recursive approach based on the properties of Gray code.
 *
 * @param {number} n - The input number.
 * @return {number} - Minimum number of operations to reduce `n` to 0.
 */
var minimumOneBitOperations = function(n) {
    // Base case: if n is 0 or 1, return n directly
    if (n <= 1) return n;

    // Find the position of the highest set bit in `n`
    let highestBit = 0;
    while ((1 << highestBit) <= n) {
        highestBit++;
    }

    // Compute the value of the highest bit (2^(highestBit - 1))
    const highestBitValue = 1 << (highestBit - 1);

    // Recursive step:
    // - Flip all bits below the highest bit (which gives (2^k - 1) operations)
    // - Then recursively reduce the remaining part: n - 2^(k-1)
    return ((1 << highestBit) - 1) - minimumOneBitOperations(n - highestBitValue);
};