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Steps
setup
Start Backtracking at index 0
The algorithm begins at the start of the string with an empty expression path and zero evaluated value. It prepares to parse operands starting from index 0.
💡 Initializing the recursion sets the stage for exploring all possible expressions from the beginning of the input.
Line:backtrack(0, 0, 0)
💡 The recursion starts with no expression and zero evaluation, ready to build expressions from the first digit.
fill_cells
Parse operand '1' at index 0-0 and add as first operand
The algorithm parses the substring '1' as the first operand. Since this is the first operand, it is added to the path without an operator.
💡 The first operand sets the base of the expression; no operator is needed before it.
💡 The first operand initializes the expression and evaluation to 1.
fill_cells
Parse operand '2' at index 1-1 and add with '+' operator
From index 1, the substring '2' is parsed as the next operand. The '+' operator is appended before it, and the expression evaluation is updated accordingly.
💡 Adding '+' extends the expression and updates the evaluation by adding the new operand.
💡 Expression '1-2' evaluates to -1, not the target.
fill_cells
Parse operand '3' at index 2-2 and add with '*' operator
From index 2, the substring '3' is parsed as the next operand. The '*' operator is appended before it, updating the expression and evaluation with precedence.
💡 Multiplication changes evaluation by combining with last operand.
Expression Add Operators - Watch the Algorithm Execute, Step by Step
Watching this visualization helps you understand how backtracking incrementally builds expressions, prunes invalid paths, and collects valid answers without generating all permutations blindly.
Step 1/21
·Active fill★Answer cell
enteringindex=0evaluated=0last_operand=0
Call Path (current → root)
backtrack(0,0,0)
Remaining
parse operand from index 0 to 0parse operand from index 0 to 1parse operand from index 0 to 2
enteringindex=1evaluated=1last_operand=1
Call Path (current → root)
backtrack(1,1,1)
backtrack(0,0,0)
Remaining
parse operand from index 1 to 1parse operand from index 1 to 2
Partial: [1]
enteringindex=2evaluated=3last_operand=2
Call Path (current → root)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Remaining
parse operand from index 2 to 2
Partial: [1, +, 2]
enteringindex=3evaluated=6last_operand=3
Call Path (current → root)
backtrack(3,6,3)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, +, 3]
backtrackingindex=3evaluated=6last_operand=3
Call Path (current → root)
backtrack(3,6,3)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, +, 3]
Found 1: [1+2+3]
enteringindex=3evaluated=0last_operand=-3
Call Path (current → root)
backtrack(3,0,-3)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, -, 3]
Found 1: [1+2+3]
backtrackingindex=3evaluated=0last_operand=-3
Call Path (current → root)
backtrack(3,0,-3)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, -, 3]
Found 1: [1+2+3]
enteringindex=3evaluated=7last_operand=6
Call Path (current → root)
backtrack(3,7,6)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, *, 3]
Found 1: [1+2+3]
backtrackingindex=3evaluated=7last_operand=6
Call Path (current → root)
backtrack(3,7,6)
backtrack(2,3,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 2, *, 3]
Found 1: [1+2+3]
enteringindex=3evaluated=24last_operand=23
Call Path (current → root)
backtrack(3,24,23)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 23]
Found 1: [1+2+3]
backtrackingindex=3evaluated=24last_operand=23
Call Path (current → root)
backtrack(3,24,23)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, +, 23]
Found 1: [1+2+3]
enteringindex=2evaluated=12last_operand=12
Call Path (current → root)
backtrack(2,12,12)
backtrack(0,0,0)
Remaining
parse operand from index 2 to 2
Partial: [12]
Found 1: [1+2+3]
enteringindex=3evaluated=15last_operand=3
Call Path (current → root)
backtrack(3,15,3)
backtrack(2,12,12)
backtrack(0,0,0)
Partial: [12, +, 3]
Found 1: [1+2+3]
backtrackingindex=3evaluated=15last_operand=3
Call Path (current → root)
backtrack(3,15,3)
backtrack(2,12,12)
backtrack(0,0,0)
Partial: [12, +, 3]
Found 1: [1+2+3]
enteringindex=3evaluated=123last_operand=123
Call Path (current → root)
backtrack(3,123,123)
backtrack(0,0,0)
Partial: [123]
Found 1: [1+2+3]
backtrackingindex=3evaluated=123last_operand=123
Call Path (current → root)
backtrack(3,123,123)
backtrack(0,0,0)
Partial: [123]
Found 1: [1+2+3]
enteringindex=2evaluated=-1last_operand=-2
Call Path (current → root)
backtrack(2,-1,-2)
backtrack(1,1,1)
backtrack(0,0,0)
Remaining
parse operand from index 2 to 2
Partial: [1, -, 2]
Found 1: [1+2+3]
enteringindex=3evaluated=-5last_operand=-6
Call Path (current → root)
backtrack(3,-5,-6)
backtrack(2,-1,-2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, -, 2, *, 3]
Found 1: [1+2+3]
backtrackingindex=3evaluated=-5last_operand=-6
Call Path (current → root)
backtrack(3,-5,-6)
backtrack(2,-1,-2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, -, 2, *, 3]
Found 1: [1+2+3]
enteringindex=3evaluated=6last_operand=6
Call Path (current → root)
backtrack(3,6,6)
backtrack(2,2,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, *, 2, *, 3]
Found 1: [1+2+3]
backtrackingindex=3evaluated=6last_operand=6
Call Path (current → root)
backtrack(3,6,6)
backtrack(2,2,2)
backtrack(1,1,1)
backtrack(0,0,0)
Partial: [1, *, 2, *, 3]
Found 2: [1+2+3] [1*2*3]
Key Takeaways
✓ Backtracking incrementally builds expressions by choosing operands and operators step-by-step, allowing efficient exploration of all valid expressions.
This insight is hard to see from code alone because the recursion and backtracking logic is intertwined and subtle.
✓ Handling multiplication requires careful evaluation updates to respect operator precedence, by adjusting the evaluated value using the last operand.
Understanding this adjustment is crucial and often confusing without seeing the evaluation state change visually.
✓ Backtracking prunes invalid paths early, such as skipping operands with leading zeros, which prevents unnecessary recursion and improves efficiency.
This pruning is subtle in code but becomes clear when watching the recursion tree skip certain branches.
Practice
(1/5)
1. You need to count the number of permutations of numbers from 1 to n such that for each position i, either the number at position i is divisible by i or i is divisible by the number. Which algorithmic approach best guarantees an efficient solution?
easy
A. Sorting numbers and checking divisibility in a single pass
B. Greedy algorithm that places numbers based on divisibility from left to right
C. Dynamic programming with memoization over subsets
D. Backtracking with pruning to explore valid permutations only
Solution
Step 1: Understand problem constraints
The problem requires counting permutations with divisibility conditions at each position, which is a combinatorial search problem.
Step 2: Identify suitable algorithm
Backtracking with pruning efficiently explores only valid permutations, avoiding full enumeration, unlike greedy or sorting approaches which fail to guarantee correctness.
Final Answer:
Option D -> Option D
Quick Check:
Backtracking prunes invalid branches early, ensuring correctness and efficiency [OK]
Hint: Backtracking prunes invalid permutations early [OK]
Common Mistakes:
Assuming greedy can solve divisibility constraints
Thinking sorting suffices for permutation constraints
2. You need to find the k-th lexicographically smallest permutation of numbers from 1 to n. Which approach guarantees an optimal solution without generating all permutations explicitly?
easy
A. Use a greedy algorithm that swaps elements to reach the k-th permutation directly.
B. Compute factorial values to determine each digit's position and build the permutation directly.
C. Use dynamic programming to count permutations and reconstruct the k-th sequence.
D. Generate all permutations using backtracking and return the k-th one.
Solution
Step 1: Understand the problem constraints
The problem requires finding the k-th permutation without enumerating all permutations, which is infeasible for large n.
Step 2: Identify the approach that uses factorial number system
Using precomputed factorials, we can determine the index of each digit in the permutation directly, avoiding full enumeration.
Final Answer:
Option B -> Option B
Quick Check:
Factorial-based direct computation is optimal and avoids timeouts [OK]
Hint: Factorials let you jump directly to the k-th permutation [OK]
Common Mistakes:
Thinking greedy swaps can find k-th permutation efficiently
Assuming DP is needed here
Trying to generate all permutations for large n
3. Consider the following snippet from an optimized Sudoku solver using backtracking with heuristics. Given the board below, what is the length of candidates for the first empty cell chosen after sorting empties by candidate count?
Board:
[['5', '3', '.', '.', '7', '.', '.', '.', '.'],
['6', '.', '.', '1', '9', '5', '.', '.', '.'],
['.', '9', '8', '.', '.', '.', '.', '6', '.'],
['8', '.', '.', '.', '6', '.', '.', '.', '3'],
['4', '.', '.', '8', '.', '3', '.', '.', '1'],
['7', '.', '.', '.', '2', '.', '.', '.', '6'],
['.', '6', '.', '.', '.', '.', '2', '8', '.'],
['.', '.', '.', '4', '1', '9', '.', '.', '5'],
['.', '.', '.', '.', '8', '.', '.', '7', '9']]
Code snippet:
def get_candidates(r, c):
b = (r//3)*3 + c//3
return [ch for ch in '123456789' if ch not in rows[r] and ch not in cols[c] and ch not in boxes[b]]
empties.sort(key=lambda x: len(get_candidates(x[0], x[1])))
first_empty = empties[0]
candidates_len = len(get_candidates(first_empty[0], first_empty[1]))
What is the value of candidates_len?
easy
A. 2
B. 3
C. 4
D. 5
Solution
Step 1: Identify first empty cell after sorting
Check empties and compute candidates for each; the cell with fewest candidates is chosen first.
Step 2: Calculate candidates for first empty cell (0,2)
Row 0 has {'5','3','7'}, column 2 has {'8'}, box 0 has {'5','3','6','9','8'}. Candidates are digits not in any of these sets: '1','2','4'. So length is 3.
Final Answer:
Option B -> Option B
Quick Check:
Counting candidates for (0,2) yields 3 [OK]
Hint: Count candidates by excluding row, column, box digits [OK]
Common Mistakes:
Forgetting box constraints
Miscounting candidates for first empty cell
4. What is the time complexity of the optimal iterative approach using a queue to generate all letter combinations for a digit string of length n, assuming each digit maps to at most 4 letters?
medium
A. O(n * 4^n) because we process each digit and expand combinations exponentially
B. O(2^n) because each digit doubles the number of combinations
C. O(4^n * n) because there are up to 4^n combinations and each combination is built with n concatenations
D. O(n^2) because we process n digits and each combination takes n steps
Solution
Step 1: Identify number of combinations
Each digit maps to up to 4 letters, so total combinations are up to 4^n.
Step 2: Analyze work per combination
Each combination is built by concatenating letters for n digits, so each combination requires O(n) time to build.
Step 3: Simplify complexity expression
O(4^n * n) and O(n * 4^n) are equivalent; however, O(n * 4^n) because we process each digit and expand combinations exponentially is the standard notation emphasizing processing each digit and expanding combinations exponentially.
Final Answer:
Option A -> Option A
Quick Check:
Time is exponential in n with linear cost per combination [OK]
Hint: Exponential combinations times linear build cost [OK]
Common Mistakes:
Confusing exponential base (2 vs 4)
Ignoring concatenation cost per combination
5. Suppose the problem is modified so that numbers can be reused multiple times in the arrangement. Which change to the backtracking algorithm correctly adapts to this variant?
hard
A. Keep the 'used' bitmask but reset bits after each recursive call to allow reuse.
B. Remove the 'used' bitmask and allow all numbers at every position, checking divisibility only.
C. Use a frequency array to track counts of each number and decrement/increment during recursion.
D. Modify the divisibility condition to allow partial matches and keep the bitmask unchanged.
Solution
Step 1: Understand reuse implication
Allowing reuse means numbers are no longer unique per position, so tracking used numbers is unnecessary.
Step 2: Adapt algorithm
Removing the 'used' bitmask and checking only divisibility conditions at each position correctly counts valid arrangements with reuse.
Final Answer:
Option B -> Option B
Quick Check:
Without uniqueness constraint, 'used' tracking is redundant [OK]
Hint: Reuse means no need to track used numbers [OK]