Generate All Combinations Sum K in DSA Go - Execution Trace

Concept Flow - Generate All Combinations Sum K

Start with empty combination

↓

Choose next number from candidates

↓

Add number to current combination

↓

Check sum of combination

↓

Save

↓

Try next candidate

↓

End when all candidates tried or sum exceeded

Start with an empty list, try adding each candidate number, check if sum equals K to save or less to continue, backtrack when sum exceeds or all candidates are tried.

Execution Sample

DSA Go

func combineSumK(candidates []int, k int) [][]int {
  var res [][]int
  var comb []int
  var backtrack func(start, target int)
  backtrack = func(start, target int) {
    if target == 0 {
      res = append(res, append([]int{}, comb...))
      return
    }
    for i := start; i < len(candidates); i++ {
      if candidates[i] > target {
        break
      }
      comb = append(comb, candidates[i])
      backtrack(i+1, target-candidates[i])
      comb = comb[:len(comb)-1]
    }
  }
  sort.Ints(candidates)
  backtrack(0, k)
  return res
}

This Go function finds all unique combinations of numbers from candidates that sum to k.

Execution Table

Step	Operation	Current Combination	Remaining Sum	Action	Visual State
1	Start backtrack(0, 7)	[]	7	Begin with empty combination	[]
2	Choose candidates[0]=2	[2]	5	Add 2 to combination	[2]
3	Choose candidates[1]=3	[2,3]	2	Add 3 to combination	[2,3]
4	candidates[2]=6 > 2	[2,3]	2	Prune: too large to add, backtrack	[2,3]
5	Backtrack remove 3	[2]	5	Remove last number	[2]
6	candidates[2]=6 > 5	[2]	5	Prune: too large to add, backtrack	[2]
7	Backtrack remove 2	[]	7	Remove last number	[]
8	Choose candidates[1]=3	[3]	4	Add 3 to combination	[3]
9	candidates[2]=6 > 4	[3]	4	Prune: too large to add, backtrack	[3]
10	Backtrack remove 3	[]	7	Remove last number	[]
11	Choose candidates[2]=6	[6]	1	Add 6 to combination	[6]
12	candidates[3]=7 > 1	[6]	1	Prune: too large to add, backtrack	[6]
13	Choose candidates[3]=7	[7]	0	Sum equals target, save	[7]
14	Backtrack remove 7	[]	7	Remove last number	[]
15	End	[]	7	All candidates tried	[]

💡 All candidates tried or sum exceeded, backtracking ends.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 5	After Step 9	After Step 13	Final
comb	[]	[2]	[2,3]	[2]	[3]	[7]	[]
remaining sum	7	5	2	5	4	0	7
res (results)	[]	[]	[]	[]	[]	[[7]]	[[7]]

Key Moments - 3 Insights

Why do we backtrack (remove last number) after exploring a path?

Because after exploring all possibilities down that path (including prunings), we must remove it to try other candidates. See steps 5, 7, and 10 in execution_table where we remove numbers to explore new combinations.

Why do we prune (skip) candidates larger than the remaining sum?

Because adding it would exceed K, and since candidates are sorted ascending, later candidates are larger, making it impossible to sum exactly to K. This is shown in steps 4, 6, 9, and 12.

How do we know when to save a combination?

When the remaining sum reaches zero, meaning the combination sums exactly to K. Step 13 shows saving the combination [7].

Visual Quiz - 3 Questions

Look at the execution_table at Step 3, what is the current combination?

A [2, 3]

B [3, 6]

C [7]

D [2]

💡 Hint

Check the 'Current Combination' column at Step 3 in execution_table.

At which step does the algorithm save a valid combination?

A Step 9

B Step 5

C Step 13

D Step 2

💡 Hint

Look for 'Sum equals target, save' in the 'Action' column of execution_table.

If the candidate 7 was removed from the list, what would happen to the results?

A Only combinations summing to 7 without 7 would be found

B No combinations would be found

C The algorithm would run infinitely

D The combination [7] would still be found

💡 Hint

Check execution_table and variable_tracker; the only saved combination is [7] at Step 13. Without 7, none sum to 7.

Concept Snapshot

Generate All Combinations Sum K:
- Use backtracking to build combinations
- Add candidates if sum <= K
- Save combination if sum == K
- Remove last added number to try others
- Sort candidates to optimize pruning

Full Transcript

This visualization shows how to generate all combinations of numbers that sum to a target K using backtracking. We start with an empty combination and try adding each candidate number if it <= remaining sum. If remaining sum == 0, save it. Prune if candidate > remaining (due to sorting). Backtrack by removing last after exploring. The execution table tracks steps, combination, remaining sum, actions (add/prune/remove). Key moments explain backtracking, pruning, saving.

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 sum of all numbers from 1 to K

D. The product of all numbers that divide K

Solution

1. Step 1: Understand the problem goal

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

2. Step 2: Clarify what combinations mean

   Combinations mean groups of numbers where order does not matter, and each number can be used once or multiple times depending on problem rules.

3. Final Answer:

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

4. Quick Check:

   Combinations sum K = unique sets adding to K [OK]

Hint: Focus on sums that exactly equal K, ignoring order [OK]

Common Mistakes:

• Confusing combinations with permutations
• Looking for a single sum instead of all combinations
• Ignoring the target sum K

2. Which of the following Go function signatures is correct for a function that generates all combinations summing to K from a slice of integers?

easy

A. func combinationSum(nums []int, target int) int

B. func combinationSum(nums int, target []int) []int

C. func combinationSum(nums []int, target int) []int

D. func combinationSum(nums []int, target int) [][]int

Solution

1. Step 1: Identify input parameters

   The function should take a slice of integers and a target integer sum.

2. Step 2: Identify return type

   It should return a slice of slices of integers, representing all combinations.

3. Final Answer:

   func combinationSum(nums []int, target int) [][]int -> Option D

4. Quick Check:

   Input slice and target int, output slice of int slices [OK]

Hint: Return type must hold multiple combinations ([][]int) [OK]

Common Mistakes:

• Swapping parameter types
• Returning a single slice instead of slice of slices
• Returning an int instead of combinations

3. What is the output of the following Go code snippet?

func main() {
    nums := []int{2, 3, 6, 7}
    target := 7
    result := combinationSum(nums, target)
    fmt.Println(result)
}

func combinationSum(nums []int, target int) [][]int {
    var res [][]int
    var backtrack func(start, target int, path []int)
    backtrack = func(start, target int, path []int) {
        if target == 0 {
            temp := make([]int, len(path))
            copy(temp, path)
            res = append(res, temp)
            return
        }
        for i := start; i < len(nums); i++ {
            if nums[i] > target {
                continue
            }
            backtrack(i, target-nums[i], append(path, nums[i]))
        }
    }
    backtrack(0, target, []int{})
    return res
}

medium

A. [[2 2 3] [7]]

B. [[2 3 6] [7]]

C. [[3 3 3] [7]]

D. [[2 2 2 3] [7]]

Solution

1. Step 1: Understand backtracking logic

   The function tries combinations starting from each index, allowing reuse of numbers, and collects paths summing to target.

2. Step 2: Trace combinations for target 7

   Possible sums are [2,2,2,3] and [7] because 2+2+2+3=9 is incorrect, so check carefully: 2+2+2+3=9 (too big), so that is invalid. The actual combinations are [2,2,3] (2+2+3=7) and [7].

3. Step 3: Verify options

   [[2 2 2 3] [7]] shows [2 2 2 3] which sums to 9, invalid. [[2 2 3] [7]] shows [2 2 3] and [7], which is correct.

4. Final Answer:

   [[2 2 3] [7]] -> Option A

5. Quick Check:

   Backtracking finds [2 2 3] and [7] sums [OK]

Hint: Look for combinations that sum exactly to 7 using given nums [OK]

Common Mistakes:

• Including invalid sums like [2 3 6]
• Ignoring reuse of numbers
• Confusing order of elements

4. Identify the bug in this Go function for generating combinations summing to K:

func combinationSum(nums []int, target int) [][]int {
    var res [][]int
    var backtrack func(start, target int, path []int)
    backtrack = func(start, target int, path []int) {
        if target == 0 {
            res = append(res, path)
            return
        }
        for i := start; i < len(nums); i++ {
            if nums[i] > target {
                continue
            }
            backtrack(i, target-nums[i], append(path, nums[i]))
        }
    }
    backtrack(0, target, []int{})
    return res
}

medium

A. The target check should be target < 0 instead of target == 0

B. The loop should start from 0 every time, not from start

C. Appending path directly causes shared slice reference issues

D. The function should return a single slice, not slice of slices

Solution

1. Step 1: Check how path is appended to result

   Appending path directly stores the reference, causing all results to point to the same underlying array.

2. Step 2: Fix by copying path before appending

   Make a copy of path slice before appending to res to avoid shared references.

3. Final Answer:

   Appending path directly causes shared slice reference issues -> Option C

4. Quick Check:

   Copy path before append to avoid shared slice bugs [OK]

Hint: Always copy slices before appending to result [OK]

Common Mistakes:

• Appending slice references directly
• Incorrect loop start index
• Wrong base case condition

5. Given a list of integers with duplicates, how would you modify the combination sum algorithm to generate all unique combinations summing to K without duplicates? Choose the best approach.

hard

A. Use a map to store sums and ignore duplicates after generation

B. Sort the list and skip duplicate numbers during recursion

C. Allow duplicates freely and filter results at the end

D. Increase target by 1 to avoid duplicates

Solution

1. Step 1: Sort the input list

   Sorting helps group duplicates together so they can be skipped during recursion.

2. Step 2: Skip duplicates in recursion

   When iterating, skip numbers equal to the previous one at the same recursion level to avoid duplicate combinations.

3. Final Answer:

   Sort the list and skip duplicate numbers during recursion -> Option B

4. Quick Check:

   Sorting + skipping duplicates = unique combinations [OK]

Hint: Sort input and skip duplicates while recursing [OK]

Common Mistakes:

• Filtering duplicates after generating all combinations
• Not sorting input before recursion
• Changing target incorrectly