Expression Add Operators

▶

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.

Line:

curr_str = num[0:1]  # '1'
path.append(curr_str)
backtrack(1, 1, 1)

💡 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.

Line:

path.append('+'); path.append('2')
backtrack(2, 3, 2)

💡 The expression now is '1+2' with evaluation 3.

fill_cells

Parse operand '3' at index 2-2 and add with '+' operator

At index 2, the substring '3' is parsed as the next operand. The '+' operator is appended before it, updating the expression and evaluation.

💡 Adding '+' again extends the expression and updates evaluation by adding the operand.

Line:

path.append('+'); path.append('3')
backtrack(3, 6, 3)

💡 The expression '1+2+3' evaluates to 6, matching the target.

compare

Check if expression '1+2+3' matches target

Since the entire string is consumed (index == len(num)), the algorithm checks if the evaluated value equals the target.

💡 This is the base case where a complete expression is evaluated for validity.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 The expression '1+2+3' is a valid solution and is recorded.

backtrack

Backtrack to index 2 and try '-' operator with operand '3'

After exploring '+3', the algorithm backtracks to index 2 and tries the '-' operator before operand '3'.

💡 Backtracking allows exploring alternative operator choices at the same position.

Line:

path.append('-'); path.append('3')
backtrack(3, 0, -3)

💡 The expression '1+2-3' evaluates to 0, which is not the target.

compare

Check if expression '1+2-3' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '1+2-3' is invalid and will not be added to results.

backtrack

Backtrack to index 2 and try '*' operator with operand '3'

After exploring '-' operator, the algorithm tries '*' operator before operand '3'. It updates evaluation considering operator precedence.

💡 Multiplication requires adjusting the evaluated value by removing last operand and adding multiplied result.

Line:

path.append('*'); path.append('3')
backtrack(3, 1 + 2 * 3, 2 * 3)

💡 Expression '1+2*3' evaluates to 7, not the target.

compare

Check if expression '1+2*3' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '1+2*3' is invalid and will not be added to results.

backtrack

Backtrack to index 1 and try operand '23' with '+' operator

Backtracking to index 1, the algorithm parses '23' as the next operand and tries '+' operator before it.

💡 Parsing longer operands explores different splits of the input string.

Line:

curr_str = num[1:3]  # '23'
path.append('+'); path.append('23')
backtrack(3, 24, 23)

💡 Expression '1+23' evaluates to 24, not the target.

compare

Check if expression '1+23' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '1+23' is invalid and will not be added to results.

backtrack

Backtrack to index 0 and try operand '12' as first operand

Backtracking to the root, the algorithm parses '12' as the first operand and starts a new expression path.

💡 Trying longer first operands explores different expression bases.

Line:

curr_str = num[0:2]  # '12'
path.append('12')
backtrack(2, 12, 12)

💡 Expression starts with '12' and evaluation 12.

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.

💡 Adding '+' extends the expression and updates evaluation by adding the operand.

Line:

path.append('+'); path.append('3')
backtrack(3, 15, 3)

💡 Expression '12+3' evaluates to 15, not the target.

compare

Check if expression '12+3' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '12+3' is invalid and will not be added to results.

backtrack

Backtrack to index 0 and try operand '123' as first operand

Backtracking to the root, the algorithm parses '123' as the first operand and starts a new expression path.

💡 Trying the longest possible first operand explores all splits.

Line:

curr_str = num[0:3]  # '123'
path.append('123')
backtrack(3, 123, 123)

💡 Expression starts with '123' and evaluation 123.

compare

Check if expression '123' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '123' is invalid and will not be added to results.

backtrack

Backtrack to index 0 and try operand '1' with '-' operator

Backtracking to the root, the algorithm tries operand '1' as first operand, then explores '-' operator with next operands.

💡 Exploring all operators after the first operand is essential to find all valid expressions.

Line:

path.append('1')
backtrack(1, 1, 1)
path.append('-'); path.append('2')
backtrack(2, -1, -2)

💡 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.

Line:

path.append('*'); path.append('3')
backtrack(3, -1 - (-2) + (-2) * 3, -6)

💡 Expression '1-2*3' evaluates to -5, not the target.

compare

Check if expression '1-2*3' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '1-2*3' is invalid and will not be added to results.

backtrack

Backtrack to index 1 and try operand '2' with '*' operator and then operand '3'

Backtracking to index 1, the algorithm tries operand '2' with '*' operator and then operand '3' with '*' operator, forming '1*2*3'.

💡 Exploring multiplication chains is key to finding all valid expressions.

Line:

path.append('*'); path.append('2')
backtrack(2, 2, 2)
path.append('*'); path.append('3')
backtrack(3, 6, 6)

💡 Expression '1*2*3' evaluates to 6, matching the target.

compare

Check if expression '1*2*3' matches target

At the end of the string, the algorithm checks if the evaluated value equals the target.

💡 Only expressions matching the target are recorded as answers.

Line:

if index == len(num):
  if evaluated == target:
    res.append(''.join(path))

💡 Expression '1*2*3' is a valid solution and is recorded.

from typing import List

def addOperators(num: str, target: int) -> List[str]:
    res = []
    path = []
    def backtrack(index: int, evaluated: int, last_operand: int):
        if index == len(num):  # STEP 5, 7, 9, 11, 14, 16, 19, 21
            if evaluated == target:
                res.append(''.join(path))
            return
        for i in range(index, len(num)):
            if i != index and num[index] == '0':
                break
            curr_str = num[index:i+1]  # STEP 2, 10, 12, 15
            curr = int(curr_str)
            length_before = len(path)
            if index == 0:
                path.append(curr_str)  # STEP 2, 12, 15
                backtrack(i+1, curr, curr)  # STEP 2, 12, 15
                path[length_before:] = []
            else:
                path.append('+'); path.append(curr_str)  # STEP 3, 4, 10, 13
                backtrack(i+1, evaluated + curr, curr)  # STEP 3, 4, 10, 13
                path[length_before:] = []

                path.append('-'); path.append(curr_str)  # STEP 6, 17
                backtrack(i+1, evaluated - curr, -curr)  # STEP 6, 17
                path[length_before:] = []

                path.append('*'); path.append(curr_str)  # STEP 8, 18, 20
                backtrack(i+1, evaluated - last_operand + last_operand * curr, last_operand * curr)  # STEP 8, 18, 20
                path[length_before:] = []
    backtrack(0, 0, 0)  # STEP 1
    return res

if __name__ == '__main__':
    print(addOperators("123", 6))  # Expected: ['1+2+3', '1*2*3']

📊

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.

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. What is the time complexity of the backtracking with bitmask optimization approach for counting beautiful arrangements of size n?

medium

A. O(n * n!)

B. O(n^2)

C. O(n * 2^n)

D. O(n!)

Solution

1. Step 1: Identify the search space

   The algorithm explores permutations of n numbers, which is n! in worst case.

2. Step 2: Consider pruning and bitmask usage

   Bitmask helps track used numbers efficiently, pruning invalid branches but the total number of recursive calls is proportional to n * n! because for each position, up to n choices are tried.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Backtracking explores permutations with n choices per position, leading to O(n * n!) complexity [OK]

Hint: Backtracking permutations with bitmask -> O(n * n!) time [OK]

Common Mistakes:

• Confusing bitmask with exponential 2^n complexity
• Assuming polynomial due to pruning

2. Identify the bug in the following code snippet for counting beautiful arrangements:

medium

A. Line with 'bit = 1 << num' bit shift

B. Line with 'if (used & bit) == 0' check

C. Line with 'if pos > n:' condition

D. Line with recursive call 'backtrack(pos + 1, used | bit)'

Solution

1. Step 1: Analyze bit shift operation

   Bitmask uses 0-based indexing, so bit for number num should be 1 << (num - 1), not 1 << num.

2. Step 2: Consequence of incorrect bit shift

   Incorrect bit shifts cause wrong bits to be set/checked, leading to missing or duplicate counts.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct bit shift is critical for accurate used-state tracking [OK]

Hint: Bit shifts must be zero-based for bitmask [OK]

Common Mistakes:

• Off-by-one bit shifts
• Forgetting to unmark used numbers

3. 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

1. Step 1: Identify number of combinations

   Each digit maps to up to 4 letters, so total combinations are up to 4^n.

2. 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.

3. 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.

4. Final Answer:

   Option A -> Option A

5. 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

4. What is the time complexity of the bitmask-optimized backtracking solution for the N-Queens problem, and why is the common misconception that it is O(n^3) incorrect?

medium

A. O(n^3) because there are three nested loops over rows, columns, and diagonals

B. O(2^n) because bitmasking iterates over all subsets of columns

C. O(n!) because each row places one queen and pruning reduces the search space factorially

D. O(n^2) because each queen placement checks all previous rows and columns

Solution

1. Step 1: Identify the branching factor per row

   Each row places exactly one queen, and pruning avoids invalid columns and diagonals, reducing choices drastically.

2. Step 2: Understand factorial growth

   Because queens cannot share columns, the number of ways to place queens is at most n!; pruning reduces this further but worst-case remains O(n!).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Bitmask pruning reduces search to permutations of columns [OK]

Hint: N-Queens search space is permutations, not polynomial loops [OK]

Common Mistakes:

• Assuming nested loops imply cubic time
• Confusing bitmask subsets with full subsets
• Ignoring pruning effect

5. Suppose the Word Search problem is modified so that the same cell can be used multiple times in forming the word. Which modification to the backtracking algorithm correctly handles this change?

hard

A. Add a global visited set to track cells used in all paths

B. Keep marking visited cells but reset them after each recursion

C. Remove the in-place marking of visited cells to allow reuse

D. Use dynamic programming to store partial results and avoid recomputation

Solution

1. Step 1: Understand new constraint

   Cells can be reused multiple times, so marking visited cells to prevent reuse is no longer needed.

2. Step 2: Modify algorithm accordingly

   Removing in-place marking allows revisiting cells, correctly reflecting the relaxed constraint.

3. Step 3: Evaluate other options

   Keeping marking or using global visited set prevents reuse incorrectly; DP is not suitable here.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Removing visited marking enables cell reuse as required [OK]

Hint: Relaxing constraints often means removing pruning steps [OK]

Common Mistakes:

• Keeping visited marking, causing false negatives
• Adding unnecessary global state