Backtracking Concept and Decision Tree Visualization in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start backtracking from the first position.

DSA C++

void backtrack(int pos) {
    if (pos == n) {
        // process solution
        return;
    }
    for (int choice = 0; choice < options; ++choice) {
        // make choice
        backtrack([1]);
        // undo choice
    }
}

int main() {
    backtrack(0);
    return 0;
}

Drag options to blanks, or click blank then click option'

A0

Bpos - 1

Cn

Dpos + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using pos - 1 causes infinite recursion backward.

Using 0 resets position and causes repeated work.

Using n skips all positions.

In backtracking, after making a choice at position pos, we move to the next position pos + 1 to continue exploring.

2fill in blank

medium

Complete the code to check if the current partial solution is valid before continuing.

DSA C++

bool isValid(int pos) {
    // check constraints for position pos
    return [1];
}

void backtrack(int pos) {
    if (pos == n) {
        // process solution
        return;
    }
    if (!isValid(pos)) return;
    for (int choice = 0; choice < options; ++choice) {
        // make choice
        backtrack(pos + 1);
        // undo choice
    }
}

Drag options to blanks, or click blank then click option'

Atrue

Bfalse

Cpos < n

Dpos == n

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false always stops the search.

Checking pos == n here is incorrect; that is the base case.

The isValid function should return true if the partial solution is valid so far, allowing backtracking to continue.

3fill in blank

hard

Fix the error in the backtracking recursive call to avoid infinite recursion.

DSA C++

void backtrack(int pos) {
    if (pos == n) {
        // process solution
        return;
    }
    for (int choice = 0; choice < options; ++choice) {
        // make choice
        backtrack([1]);
        // undo choice
    }
}

Drag options to blanks, or click blank then click option'

Apos + 1

Bpos

Cpos - 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Calling backtrack(pos) repeats the same call endlessly.

Calling backtrack(pos - 1) goes backward and may cause errors.

The recursive call must move forward by calling backtrack(pos + 1) to avoid infinite recursion at the same position.

4fill in blank

hard

Fill both blanks to generate all subsets of a set using backtracking.

DSA C++

void backtrack(int pos, vector<int>& subset) {
    if (pos == n) {
        result.push_back(subset);
        return;
    }
    // Exclude current element
    backtrack([1], subset);
    // Include current element
    subset.push_back(nums[pos]);
    backtrack([2], subset);
    subset.pop_back();
}

Drag options to blanks, or click blank then click option'

Apos + 1

Bpos

Cpos - 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using pos instead of pos + 1 causes infinite recursion.

Using pos - 1 goes backward and causes errors.

In both recursive calls, we move forward by increasing pos by 1 to explore subsets including or excluding the current element.

5fill in blank

hard

Fill all three blanks to build a decision tree for permutations using backtracking.

DSA C++

void backtrack(vector<int>& current, vector<bool>& used) {
    if (current.size() == n) {
        result.push_back(current);
        return;
    }
    for (int i = 0; i < n; ++i) {
        if (!used[i]) {
            used[i] = true;
            current.push_back(nums[[1]]);
            backtrack(current, used);
            current.pop_back();
            used[[2]] = false;
        }
    }
}

int main() {
    vector<int> current;
    vector<bool> used(n, false);
    backtrack(current, used);
    return 0;
}

Drag options to blanks, or click blank then click option'

Ai

Bpos

Cn

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using pos instead of i causes undefined variable errors.

Using n or 0 instead of i breaks the logic.

The index i is used to track which element is chosen and marked used in the permutation backtracking.

Practice

1. What is the main idea behind backtracking in algorithms?

easy

A. Sort data before processing

B. Try all possible options and undo choices when they lead to dead ends

C. Use loops to repeat actions

D. Store data in arrays for quick access

Solution

1. Step 1: Understand backtracking purpose

   Backtracking tries all options to find solutions by exploring choices step-by-step.

2. Step 2: Recognize undoing choices

   When a choice leads to no solution, backtracking reverses it to try others.

3. Final Answer:

   Try all possible options and undo choices when they lead to dead ends -> Option B

4. Quick Check:

   Backtracking = Try and undo choices [OK]

Hint: Backtracking means try, undo if fails, then try next [OK]

Common Mistakes:

• Confusing backtracking with sorting
• Thinking backtracking only uses loops
• Assuming backtracking stores all data upfront

2. Which of the following is the correct syntax to start a backtracking function in C++?

easy

A. backtrack() -> void { }

B. int backtrack[] = {0};

C. void backtrack(int step) { /* code */ }

D. function backtrack(int step) { }

Solution

1. Step 1: Identify valid C++ function syntax

   In C++, functions are declared with return type, name, and parameters inside parentheses.

2. Step 2: Check each option

   void backtrack(int step) { /* code */ } uses correct C++ syntax: return type void, function name, parameter list.

3. Final Answer:

   void backtrack(int step) { /* code */ } -> Option C

4. Quick Check:

   C++ function syntax = void backtrack(int step) { /* code */ } [OK]

Hint: C++ functions start with return type and parentheses [OK]

Common Mistakes:

• Using JavaScript or Python syntax in C++
• Missing return type in function declaration
• Using arrow notation which is invalid in C++

3. Consider this backtracking code snippet in C++ that prints all subsets of {1, 2}:

void backtrack(vector<int>& subset, int start) {
  if (start == 2) {
    for (int x : subset) cout << x << ' ';
    cout << '\n';
    return;
  }
  backtrack(subset, start + 1);
  subset.push_back(start + 1);
  backtrack(subset, start + 1);
  subset.pop_back();
}

int main() {
  vector<int> subset;
  backtrack(subset, 0);
  return 0;
}

What is the output?

medium

A. \n \n1 \n1 2 \n

B. \n\n\n\n\n\n\n

C. \n\n\n\n1 2 \n

D. \n\n\n2 \n1 \n1 2 \n

Solution

1. Step 1: Trace recursive calls without adding elements

   First call prints empty subset (no elements), so prints blank line.

2. Step 2: Trace recursive calls adding elements

   Then adds 1, prints '1 ', then adds 2, prints '1 2 ', and also prints '2 ' when only 2 is added.

3. Final Answer:

   \n \n1 \n1 2 \n -> Option A

4. Quick Check:

   All subsets printed = \n \n1 \n1 2 \n [OK]

Hint: Backtracking prints subsets by including/excluding elements [OK]

Common Mistakes:

• Forgetting to pop elements after recursion
• Mixing order of adding elements
• Missing empty subset output

4. The following backtracking code tries to generate all binary strings of length 2 but has a bug:

void backtrack(string& s, int pos) {
  if (pos == 2) {
    cout << s << '\n';
    return;
  }
  s[pos] = '0';
  backtrack(s, pos + 1);
  s[pos] = '1';
  backtrack(s, pos + 1);
}

int main() {
  string s = "";
  backtrack(s, 0);
  return 0;
}

What is the bug?

medium

A. String s is empty and cannot assign characters by index

B. The base case condition is incorrect

C. The recursion does not increment position

D. The output statements are missing

Solution

1. Step 1: Check string initialization

   String s is empty (length 0), so s[pos] = '0' causes out-of-range error.

2. Step 2: Understand string indexing requirement

   To assign by index, string must have length at least 2, e.g., initialized as "00".

3. Final Answer:

   String s is empty and cannot assign characters by index -> Option A

4. Quick Check:

   Empty string indexing causes error [OK]

Hint: Initialize string length before assigning by index [OK]

Common Mistakes:

• Ignoring string length before assignment
• Thinking base case is wrong
• Missing recursion increment

5. You want to use backtracking to solve the N-Queens problem for N=4. Which of these best describes the decision tree visualization for this problem?

hard

A. The tree has only one branch because queens are placed sequentially without choices

B. Each node stores the entire chessboard state and branches by swapping queens

C. Each level represents removing queens from the board to find solutions

D. Each level represents placing a queen in a row, branching by possible columns without conflicts

Solution

1. Step 1: Understand N-Queens backtracking

   Backtracking places queens row by row, choosing columns that don't conflict.

2. Step 2: Visualize decision tree levels

   Each level corresponds to a row; branches represent valid column choices for that row.

3. Final Answer:

   Each level represents placing a queen in a row, branching by possible columns without conflicts -> Option D

4. Quick Check:

   Decision tree levels = rows, branches = columns [OK]

Hint: Decision tree levels = rows, branches = valid columns [OK]

Common Mistakes:

• Thinking tree branches represent swapping queens
• Assuming only one branch per level
• Confusing placing with removing queens