700

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The 700th LC problem that I solved reminds me of the Collatz Conjecture. Here it is https://leetcode.com/problems/integer-replacement/

397. Integer Replacement
Medium

Given a positive integer n and you can do operations as follow:

  1. If n is even, replace n with n/2.
  2. If n is odd, you can replace n with either n + 1 or n - 1.

What is the minimum number of replacements needed for n to become 1?

Example 1:

Input:
8

Output:
3

Explanation:
8 -> 4 -> 2 -> 1

Example 2:

Input:
7

Output:
4

Explanation:
7 -> 8 -> 4 -> 2 -> 1
or
7 -> 6 -> 3 -> 2 -> 1

The idea is to implement the rules mentioned in the problem statement, but with the following observations and optimizations: 
1) Use a static hash table to keep track of the results. That's your cache
2) Use long instead of int to avoid overflows
3) Last but not least, optimize for when N is a power of 2. If N is a power of 2, the solution if Log(N,2). To check whether a number is a power of 2, simply count the number of bits. If the sum is 1, you have a power of 2.

That's it. Code is below. Cheers friends, ACC.




static Hashtable irCache = new Hashtable();
public int IntegerReplacement(int n)
{
    return IntegerReplacementLong(n);
}
private int IntegerReplacementLong(long n)
{
    if (irCache.ContainsKey(n))
    {
        return (int)irCache[n];
    }

    if (n == 1)
    {
        irCache.Add(n, 0);
        return 0;
    }

    //Power of 2
    if (IsPowerOfTwo(n))
    {
        irCache.Add(n, (int)Math.Log(n, 2));
        return (int)Math.Log(n, 2);
    }

    if (n % 2 == 0)
    {
        int val = IntegerReplacementLong(n / 2);
        if (!irCache.ContainsKey(n))
        {
            irCache.Add(n, val + 1);
        }
        return val + 1;
    }
    else
    {
        int valMinus = IntegerReplacementLong(n - 1);
        int valPlus = Int32.MaxValue;

        if (valMinus != 0) valPlus = IntegerReplacementLong(n + 1);

        int min = Math.Min(valPlus, valMinus);
        if (!irCache.ContainsKey(n))
        {
            irCache.Add(n, min + 1);
        }
        return min + 1;
    }
}

private bool IsPowerOfTwo(long n)
{
    int bits = 0;
    while (n > 0)
    {
        bits += (int)(n % 2);
        if (bits > 1) return false;
        n /= 2;
    }

    return bits == 1;
}

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