Generate All Combinations Sum K in DSA Python - Execution Trace

Concept Flow - Generate All Combinations Sum K

Start with empty combination

↓

Choose next number from candidates

↓

Add number to combination

↓

Check sum of combination

↓

Continue adding

↓

If sum == K

↓

Save current combination

↓

Backtrack

↓

Try next number

↓

End

Start with an empty list, add numbers one by one, check if sum is less than, equal to, or greater than K. If equal, save combination. If less, continue adding. If greater, backtrack and try next number.

Execution Sample

DSA Python

def comb_sum_k(candidates, k):
    result = []
    def backtrack(start, path, total):
        if total == k:
            result.append(path[:])
            return
        if total > k:
            return
        for i in range(start, len(candidates)):
            path.append(candidates[i])
            backtrack(i + 1, path, total + candidates[i])
            path.pop()
    backtrack(0, [], 0)
    return result

# Example:
# comb_sum_k([2,3,6,7], 7) -> [[7]]

This code finds all unique combinations of numbers from candidates that sum to k by exploring choices recursively and backtracking.

Execution Table

Step	Operation	Current Combination	Sum	Action	Visual State
1	Start backtrack(0, [], 0)	[]	0	Begin with empty combination	[]
2	Add 2	[2]	2	Sum < 7, continue	[2]
3	Add 3	[2, 3]	5	Sum < 7, continue	[2 -> 3]
4	Add 6	[2, 3, 6]	11	Sum > 7, backtrack	[2 -> 3 -> 6]
5	Remove 6	[2, 3]	5	Backtrack to try next	[2 -> 3]
6	Add 7	[2, 3, 7]	12	Sum > 7, backtrack	[2 -> 3 -> 7]
7	Remove 7	[2, 3]	5	Backtrack to try next	[2 -> 3]
8	Remove 3	[2]	2	Backtrack to try next	[2]
9	Add 6	[2, 6]	8	Sum > 7, backtrack	[2 -> 6]
10	Remove 6	[2]	2	Backtrack to try next	[2]
11	Add 7	[2, 7]	9	Sum > 7, backtrack	[2 -> 7]
12	Remove 7	[2]	2	Backtrack to try next	[2]
13	Remove 2	[]	0	Backtrack to try next	[]
14	Add 3	[3]	3	Sum < 7, continue	[3]
15	Add 6	[3, 6]	9	Sum > 7, backtrack	[3 -> 6]
16	Remove 6	[3]	3	Backtrack to try next	[3]
17	Add 7	[3, 7]	10	Sum > 7, backtrack	[3 -> 7]
18	Remove 7	[3]	3	Backtrack to try next	[3]
19	Remove 3	[]	0	Backtrack to try next	[]
20	Add 6	[6]	6	Sum < 7, continue	[6]
21	Add 7	[6, 7]	13	Sum > 7, backtrack	[6 -> 7]
22	Remove 7	[6]	6	Backtrack to try next	[6]
23	Remove 6	[]	0	Backtrack to try next	[]
24	Add 7	[7]	7	Sum == 7, save combination	[7]
25	Remove 7	[]	0	Backtrack to try next	[]
26	End	[]	0	All combinations tried	[]

💡 All candidates explored, backtracked fully, no more combinations

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 14	After Step 20	After Step 24	Final
path	[]	[2]	[2, 3]	[2, 3, 6]	[2, 3]	[3]	[6]	[7]	[]
total	0	2	5	11	5	3	6	7	0
start	0	1	2	3	2	1	2	3	end
result	[]	[]	[]	[]	[]	[]	[]	[[7]]	[[7]]

Key Moments - 3 Insights

Why do we backtrack (remove last number) after sum exceeds K?

Because when sum > K (see steps 4, 6, 9), continuing deeper won't find valid combinations. We remove last number to try other options (see steps 5, 7, 10).

Why do we save the combination only when sum == K?

Only when sum equals K (step 24) do we have a valid combination. If sum < K, we keep adding numbers; if sum > K, we backtrack.

Why do we increase 'start' index when recursing?

To avoid reusing the same number and to ensure combinations are unique and ordered, we move start forward (see code line backtrack(i + 1, ...)).

Visual Quiz - 3 Questions

Look at the execution table at step 24. What is the current combination?

A [7]

B [6, 7]

C [2, 3, 7]

D [3, 6]

💡 Hint

Check the 'Current Combination' column at step 24 in execution_table.

At which step does the sum first exceed K (7)?

A Step 3

B Step 4

C Step 6

D Step 9

💡 Hint

Look for the first 'Sum > 7' action in the execution_table.

If we allowed reuse of the same number (no i+1 in recursion), how would the combinations change?

A More combinations with repeated numbers

B Fewer combinations

C No change

D Algorithm would fail

💡 Hint

Consider how moving start index affects reuse of numbers in variable_tracker and code.

Concept Snapshot

Generate All Combinations Sum K:
- Use backtracking to build combinations.
- Add numbers while sum < K.
- Save combination if sum == K.
- Backtrack if sum > K.
- Move start index forward to avoid duplicates.
- Explore all unique combinations recursively.

Full Transcript

This visualization shows how to generate all unique combinations of numbers that sum to a target K. We start with an empty list and add numbers one by one from the candidates. Each time we add a number, we check the sum. If the sum is less than K, we continue adding. If the sum equals K, we save the current combination. If the sum exceeds K, we remove the last number and try the next candidate. We move forward in the candidate list to avoid duplicates. The execution table tracks each step, showing the current combination, sum, and actions like adding or removing numbers. The variable tracker shows how the path and total sum change over time. Key moments clarify why backtracking is needed and why we save combinations only when sum equals K. The quiz tests understanding of these steps and the effect of changing the recursion start index.

Practice

1. What does the problem "Generate All Combinations Sum K" ask you to find?

easy

A. All unique combinations of numbers that add up to K

B. The largest number less than K

C. The count of numbers equal to K

D. The smallest number greater than K

Solution

1. Step 1: Understand the problem goal

   The problem asks to find all unique sets of numbers that sum exactly to K.

2. Step 2: Clarify uniqueness and sum condition

   Each combination must add up to K without duplicates in the result.

3. Final Answer:

   All unique combinations of numbers that add up to K -> Option A

4. Quick Check:

   Combinations sum to K = All unique combinations of numbers that add up to K [OK]

Hint: Focus on combinations that sum exactly to K [OK]

Common Mistakes:

• Confusing combinations with permutations
• Looking for single numbers equal to K
• Ignoring the sum condition

2. Which of the following Python function signatures is correct to start solving "Generate All Combinations Sum K"?

easy

A. def combinationSumK(nums, k): return int

B. def combination_sum_k(k: int, nums: list[int]) -> list[int]:

C. def combination_sum_k(nums: int, k: int) -> list[int]:

D. def combination_sum_k(nums: list[int], k: int) -> list[list[int]]:

Solution

1. Step 1: Check parameter types and return type

   The function should take a list of integers and an integer k, returning a list of lists of integers.

2. Step 2: Match signature to problem needs

   def combination_sum_k(nums: list[int], k: int) -> list[list[int]]: matches the expected input and output types correctly.

3. Final Answer:

   def combination_sum_k(nums: list[int], k: int) -> list[list[int]]: -> Option D

4. Quick Check:

   Correct signature with list inputs and list of lists output [OK]

Hint: Use list[int] for input and list[list[int]] for output [OK]

Common Mistakes:

• Wrong parameter types
• Incorrect return type
• Swapping parameter order

3. What is the output of the following code?

def comb_sum(nums, k):
    result = []
    def backtrack(start, path, total):
        if total == k:
            result.append(path[:])
            return
        if total > k:
            return
        for i in range(start, len(nums)):
            path.append(nums[i])
            backtrack(i + 1, path, total + nums[i])
            path.pop()
    backtrack(0, [], 0)
    return result

print(comb_sum([2,3,6,7], 7))

medium

A. [[2, 2, 3]]

B. [[2, 3, 6]]

C. [[7]]

D. [[3, 7]]

Solution

1. Step 1: Trace backtracking with input [2,3,6,7] and target 7

   Check combinations starting from index 0, adding numbers without repetition.

2. Step 2: Identify valid combinations summing to 7

   Only [7] sums exactly to 7 without exceeding or repeating indices.

3. Final Answer:

   [[7]] -> Option C

4. Quick Check:

   Only [7] sums to 7 in given list [OK]

Hint: Backtrack with index increment to avoid duplicates [OK]

Common Mistakes:

• Allowing repeated elements
• Not stopping when total > k
• Appending path without copying

4. Find the error in this code snippet for generating combinations summing to k:

def comb_sum(nums, k):
    result = []
    def backtrack(start, path, total):
        if total == k:
            result.append(path[:])
            return
        if total > k:
            return
        for i in range(start, len(nums)):
            path.append(nums[i])
            backtrack(i, path, total + nums[i])
            path.pop()
    backtrack(0, [], 0)
    return result

medium

A. Appending path without copying causes all results to be the same

B. Using backtrack(i, ...) allows repeated elements incorrectly

C. Missing base case for total < k

D. Not initializing result as a global variable

Solution

1. Step 1: Analyze recursive call backtrack(i, ...)

   Using i instead of i+1 allows reuse of the same element multiple times.

2. Step 2: Understand problem requires unique combinations without repeats

   To avoid duplicates, index must increase to i+1 in recursive calls.

3. Final Answer:

   Using backtrack(i, path, total + nums[i]) allows repeated elements incorrectly -> Option B

4. Quick Check:

   Index must increase to avoid repeats [OK]

Hint: Increment index in recursion to prevent repeats [OK]

Common Mistakes:

• Not copying path when appending
• Allowing repeated elements by wrong index
• Ignoring base cases

5. Given the list [1, 2, 3, 4, 5] and target sum 5, which of the following is the correct set of all unique combinations that sum to 5?

hard

A. [[1, 4], [2, 3], [5]]

B. [[1, 2, 2], [5]]

C. [[1, 1, 3], [2, 3]]

D. [[5, 0], [2, 3]]

Solution

1. Step 1: List all combinations of numbers from [1,2,3,4,5] that sum to 5

   Possible sums: 1+4=5, 2+3=5, 5=5.

2. Step 2: Verify uniqueness and validity

   Combinations like [1,2,2] or [1,1,3] are invalid as 2 or 1 repeated without duplicates allowed.

3. Final Answer:

   [[1, 4], [2, 3], [5]] -> Option A

4. Quick Check:

   Unique combos summing to 5 = [[1, 4], [2, 3], [5]] [OK]

Hint: Sum unique sets without repeating elements [OK]

Common Mistakes:

• Including repeated elements not in input
• Adding zero which is not in list
• Ignoring uniqueness of combinations