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Introduction to Backtracking

What is Backtracking?

Backtracking is an algorithmic technique for solving problems recursively by trying to build a solution incrementally, one piece at a time, removing those solutions that fail to satisfy the constraints of the problem at any point in time.

It's like exploring a maze - if you reach a dead end, you backtrack to the previous junction and try a different path.

Key Characteristics

  • Recursive Nature: Backtracking is inherently recursive, building solutions step by step
  • Constraint Satisfaction: It checks if the current state satisfies all constraints before proceeding
  • Pruning: It abandons a path as soon as it determines that it cannot lead to a valid solution
  • Exhaustive Search: It explores all possible solutions but in an optimized way

When to Use Backtracking

Backtracking is particularly useful for:

  • Combinatorial Problems: Finding all possible combinations or permutations with constraints
  • Constraint Satisfaction Problems: Sudoku, N-Queens, Map Coloring
  • Decision Problems: Problems where you need to find if a solution exists
  • Optimization Problems: Finding the best solution among many possible ones

Basic Structure of Backtracking

A typical backtracking algorithm follows this structure:

function backtrack(state, choices): if isGoalReached(state): recordSolution(state) return for choice in choices: if isValid(state, choice): makeChoice(state, choice) backtrack(state, remainingChoices) undoChoice(state, choice) // Backtrack!

The key is the "undoChoice" step - this is where we backtrack when a path doesn't lead to a solution.

Simple Example: Finding Subsets

Let's look at a simple example of finding all subsets of a set [1, 2, 3]:

function findSubsets(nums): result = [] backtrack([], 0, nums, result) return result function backtrack(current, start, nums, result): // Add the current subset to our result result.push([...current]) // Try adding each remaining number for i from start to nums.length - 1: // Include nums[i] current.push(nums[i]) // Recursively find subsets with nums[i] included backtrack(current, i + 1, nums, result) // Backtrack - remove nums[i] to try next number current.pop()

This will generate all possible subsets: [], [1], [2], [3], [1,2], [1,3], [2,3], [1,2,3]

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