This posting will detail how to solve the problem of reducing a given integer to 1 using C#. This problem is a common type in coding tests and is very useful for developing problem-solving skills and algorithm design abilities.
Problem Description
There are several operations to reduce the given integer x
to 1. Each operation is as follows:
- Subtracting 1 (
x => x - 1
) - Dividing by 2 (
x => x / 2
, possible whenx
is even) - Dividing by 3 (
x => x / 3
, possible whenx
is divisible by 3)
The goal is to reduce x
to 1 with the minimum number of operations. For example, if x = 10
, the following operations can be performed:
Example:
10 → 5 (divide 10 by 2) → 4 (subtract 1 from 5) → 2 (divide 4 by 2) → 1 (subtract 1 from 2)
Problem Solving Strategy
There are various approaches to solve this problem, but one of the most efficient methods is to use BFS (Breadth-First Search). BFS is advantageous for finding the shortest path and solves the problem in a step-by-step manner.
BFS Algorithm Overview
BFS is an algorithm that stores nodes to be explored in a queue and explores them in the order they were added. In this problem, each state (integer) is treated as a node, and the new states reachable through each operation are added to the queue. This allows us to find the minimum number of operations to reach 1.
Implementation
Now, let’s write a code to solve the problem in C#. The code below is a function that calculates the minimum number of operations to reduce the given integer x
to 1.
using System;
using System.Collections.Generic;
public class Solution {
public int MinOperations(int x) {
// A set to track visited nodes
HashSet visited = new HashSet();
// A queue for BFS exploration
Queue> queue = new Queue>();
// Initial state (x, 0) - current value and operation count
queue.Enqueue(new Tuple(x, 0));
visited.Add(x); // Mark the initial node as visited
while (queue.Count > 0) {
var current = queue.Dequeue();
int currentValue = current.Item1;
int operationsCount = current.Item2;
// If we have reached the goal state of 1
if (currentValue == 1) {
return operationsCount; // Return the minimum operation count
}
// Perform possible operations
// x - 1
if (currentValue - 1 >= 1 && !visited.Contains(currentValue - 1)) {
visited.Add(currentValue - 1);
queue.Enqueue(new Tuple(currentValue - 1, operationsCount + 1));
}
// x / 2
if (currentValue % 2 == 0 && !visited.Contains(currentValue / 2)) {
visited.Add(currentValue / 2);
queue.Enqueue(new Tuple(currentValue / 2, operationsCount + 1));
}
// x / 3
if (currentValue % 3 == 0 && !visited.Contains(currentValue / 3)) {
visited.Add(currentValue / 3);
queue.Enqueue(new Tuple(currentValue / 3, operationsCount + 1));
}
}
return -1; // In case it is not reachable
}
}
Code Explanation
Looking at the C# code above, the MinOperations
function takes x
as input and returns the minimum number of operations. This function performs all necessary operations to reduce the integer to 1 through BFS exploration.
Main Components
- Visit Record (visited): Tracks nodes (integers) that have already been explored.
- Queue (queue): Stores the current node and operation counts to proceed with BFS.
- Conditional Statements: Check if each operation results in a valid new state (>= 1) and whether it has been visited.
Time Complexity Analysis
The time complexity of this algorithm is O(x) in the worst case. Since BFS is performed for each integer, in the worst case, it may need to explore all possibilities. However, it generally performs quickly.
Conclusion
In this posting, we explored how to solve the problem of reducing an integer to 1 using BFS in C#. By applying this method, you can efficiently find the minimum number of operations, and it greatly helps in developing algorithmic thinking. Solving additional practice problems is also a good way to expand this knowledge.
Based on this problem, try challenging various modified versions. Keep pondering problem-solving to gain a deeper understanding!