This is a problem I brought along to the WICSxACM Technical Interview Prep workshop on 13 January 2020. It is also available at bit.ly/integer-replacement.

1. Problem Description
2. Hints
3. Solution

## Problem Description

Source: LeetCode 397

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

1. If $n$ is even, replace $n$ with $n/2$
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

1. 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?
2. 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. I chose to use C++ because it is the language I prefer to use in interviews. Python makes things too easy. IMHO interviewers will want to see your implementation skills, and C++ is the way to do that. Of course, this will vary on a case-by-case basis. Some companies mainly use Python and will want to interview you in Python, for instance.

### 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. These certainly work too.