Integer Replacement

This is a problem I brought along to the WICSxACM Technical Interview Prep workshop on January 13, 2020.

Problem Description ¶

Source: LeetCode 397

Given a positive integer $n$ , you can perform the following operations:

If $n$ is even, replace $n$ with $n/2$
If $n$ is odd, replace $n$ with either $n + 1$ or $n - 1$

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

Hints ¶

The initial temptation is to go for a fancy, math-based $O(1)$ solution, but there are more general approaches that will increase your chances of actually coming up with a solution. What are some other approaches you can think of?
How can you think of the problem as a graph? (For this workshop, I know there have been quite a few other graph problems, so this will be fresh in your mind.)

Solution ¶

I have implemented one possible solution to the problem in C++.

Approach 1: BFS ¶

This solution was suggested in the Hints. The idea is to think of the numbers as a graph, where each number is a node and there are edges connecting each node $x$ to $x/2$ if $x$ is even, or to $x + 1$ and $x - 1$ if $x$ is odd. We can then perform a BFS (Breadth-First Search) on the graph to find the length of the shortest path from $n$ to $1$ .

Time complexity: $O(n)$ – we can only encounter numbers in the range $[1, 2n]$ since if you reach $2n$ , you wasted a bunch of steps because $2n$ can just be divided by $2$ to get $n$ . Since we never visit the same number twice, the time complexity is $O(2n) = O(n)$
Space complexity: $O(n)$ – same reason as above.

class Solution {
 public:
  int integerReplacement(int n) {
    // Queue for BFS -- each pair holds a number `x` and the number
    // of steps it is from `n`. We use `long long` to account for the
    // edge case when `n == INT_MAX`. Note that as we pop nodes from
    // and add nodes to this queue, the number of steps gradually
    // increases. Furthermore, note that we are guaranteed to visit all nodes
    // that are `k` steps from `n` before any nodes that are `k + 1` steps
    // from `n`.
    queue<pair<long long, int>> q;

    // Visited set -- holds numbers that have already been encountered.
    // Numbers enter this set when they are first encountered by the
    // algorithm. We do not want to revisit these numbers because we only
    // care about the minimum number of steps to visit them.
    unordered_set<long long> visited;

    // The queue starts with `n` at 0 steps, and `n` is visited to begin with.
    q.push({n, 0});
    visited.insert(n);

    // BFS loop.
    while (!q.empty()) {
      // Retrieve the next `x` and `steps` from the queue.
      long long x;
      int steps;
      tie(x, steps) = q.front();  // See std::tie
      q.pop();

      // We solved the problem -- return the number of steps.
      if (x == 1) return steps;

      if (x % 2 == 0) {
        // If x is even, the only node connected to it will be `x/2`
        if (visited.find(x / 2) == visited.end()) {
          // Make sure to record that `x/2` is visited.
          visited.insert(x / 2);
          // Add `x/2` with one more step than `x`.
          q.push({x / 2, steps + 1});
        }
      } else {
        // Same as above, except we now have `x + 1` and `x - 1`
        if (visited.find(x + 1) == visited.end()) {
          visited.insert(x + 1);
          q.push({x + 1, steps + 1});
        }
        if (visited.find(x - 1) == visited.end()) {
          visited.insert(x - 1);
          q.push({x - 1, steps + 1});
        }
      }
    }

    // If for some reason no solution was found.
    return -1;
  }
};

Alternate Approaches ¶

If you look at the discussions on the original LeetCode problem, you will see references to dynamic programming (DP) and recursion.