Combination Sum

LeetCode: 39. Combination SumDifficulty: Medium Pattern: Backtracking with Repetition

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Problem Statement

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

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Examples

Example 1

Input: candidates = [2, 3, 6, 7], target = 7
Output: [[2, 2, 3], [7]]
Explanation:
- 2 + 2 + 3 = 7
- 7 = 7
Both are valid combinations.

Example 2

Input: candidates = [2, 3, 5], target = 8
Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]

Example 3

Input: candidates = [2], target = 1
Output: []
Explanation: No combination sums to 1.

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Constraints

• 1 <= candidates.length <= 30
• 2 <= candidates[i] <= 40
• All elements of candidates are distinct
• 1 <= target <= 40

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Backtracking Tree Visualization

Key Insight: Reuse Elements

Unlike standard combinations, the same element can be used multiple times.

For candidates = [2, 3], target = 7:

                    []  (remaining=7)
                  /    \
              [2]       [3]
            (rem=5)   (rem=4)
           /    \        \
        [2,2]  [2,3]    [3,3]
       (rem=3) (rem=2)  (rem=1)
       /    \    \
    [2,2,2] [2,2,3] [2,3,3]
   (rem=1)  (rem=-1) (rem=-1)
     |        X        X
   [2,2,2,2]
   (rem=-1)
     X

Backtrack! No valid continuation...

Continue with [2,3] path -> [2,2,3] works!

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Backtracking Template

def combinationSum(candidates, target):
    """
    Template for combination sum with reuse allowed.

    Key difference: Use index `i` (not `i+1`) in recursive call
    to allow reusing the same element.
    """
    result = []

    def backtrack(start, path, remaining):
        # BASE CASE: found valid combination
        if remaining == 0:
            result.append(path[:])
            return

        # PRUNING: exceeded target
        if remaining < 0:
            return

        for i in range(start, len(candidates)):
            # CHOOSE
            path.append(candidates[i])

            # EXPLORE: use `i` (not i+1) to allow reuse
            backtrack(i, path, remaining - candidates[i])

            # UNCHOOSE
            path.pop()

    backtrack(0, [], target)
    return result

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Solutions

Solution

Python

def combinationSum(candidates: list, target: int) -> list:
    """
    Find all combinations that sum to target (reuse allowed).

    Time Complexity: O(n^(target/min)) where min is smallest candidate
    Space Complexity: O(target/min) for recursion depth
    """
    result = []

    def backtrack(start, path, remaining):
        if remaining == 0:
            result.append(path[:])
            return

        if remaining < 0:
            return

        for i in range(start, len(candidates)):
            path.append(candidates[i])
            # Use i (not i+1) to allow reusing same element
            backtrack(i, path, remaining - candidates[i])
            path.pop()

    backtrack(0, [], target)
    return result


# Test
print(combinationSum([2, 3, 6, 7], 7))
# Output: [[2, 2, 3], [7]]

Java

public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> result = new ArrayList<>();
    backtrack(result, new ArrayList<>(), candidates, 0, target);
    return result;
}

private void backtrack(List<List<Integer>> result,
                       List<Integer> path,
                       int[] candidates,
                       int start,
                       int remaining) {
    if (remaining == 0) {
        result.add(new ArrayList<>(path)); // deep copy
        return;
    }
    if (remaining < 0) return;
    for (int i = start; i < candidates.length; i++) {
        path.add(candidates[i]);
        // Pass i (not i+1) to allow reusing the same element
        backtrack(result, path, candidates, i, remaining - candidates[i]);
        path.remove(path.size() - 1);
    }
}

Complexity: Time O(n^(T/M)) · Space O(T/M)

• Time: In the worst case, we branch n ways at each level, and the tree depth is target/min(candidates). Each valid combination takes O(T/M) to copy.
• Space: Maximum recursion depth is target divided by the smallest candidate (how many times we can reuse the smallest element).

Solution 2: Backtracking with Sorting and Pruning

def combinationSum_optimized(candidates: list, target: int) -> list:
    """
    Optimized with sorting and early termination.

    Sorting allows us to break early when candidate > remaining.

    Time Complexity: O(n^(target/min)) but with better pruning
    Space Complexity: O(target/min)
    """
    result = []
    candidates.sort()  # Sort for pruning optimization

    def backtrack(start, path, remaining):
        if remaining == 0:
            result.append(path[:])
            return

        for i in range(start, len(candidates)):
            # Early termination: if current candidate > remaining,
            # all subsequent candidates will also be > remaining
            if candidates[i] > remaining:
                break

            path.append(candidates[i])
            backtrack(i, path, remaining - candidates[i])
            path.pop()

    backtrack(0, [], target)
    return result


# Test
print(combinationSum_optimized([2, 3, 6, 7], 7))
# Output: [[2, 2, 3], [7]]

Complexity: Time O(n^(T/M)) · Space O(T/M)

• Time: Same worst-case complexity, but sorting enables early termination (break instead of continue) when candidate exceeds remaining.
• Space: Recursion depth is still bounded by target/min(candidates); sorting is O(n log n) which is dominated by the exponential work.

Solution 3: Dynamic Programming

def combinationSum_dp(candidates: list, target: int) -> list:
    """
    DP approach: Build combinations bottom-up.

    dp[i] = list of all combinations that sum to i

    Time Complexity: O(target * n * average_combination_length)
    Space Complexity: O(target * num_combinations)
    """
    dp = [[] for _ in range(target + 1)]
    dp[0] = [[]]  # One way to sum to 0: empty combination

    for candidate in candidates:
        for i in range(candidate, target + 1):
            for combination in dp[i - candidate]:
                dp[i].append(combination + [candidate])

    return dp[target]


# Test
print(combinationSum_dp([2, 3, 6, 7], 7))
# Output: [[2, 2, 3], [7]]

Complexity: Time O(T * n * L) · Space O(T * C)

• Time: For each value from 1 to target, we iterate through n candidates, and extending combinations takes O(L) where L is average combination length.
• Space: dp array stores all combinations for each sum from 0 to target; C is the total number of combinations across all sums.

Solution 4: BFS Approach

from collections import deque

def combinationSum_bfs(candidates: list, target: int) -> list:
    """
    BFS approach using a queue.

    Time Complexity: O(n^(target/min))
    Space Complexity: O(n^(target/min)) for queue
    """
    result = []
    candidates.sort()
    queue = deque([(0, [], target)])  # (start_idx, path, remaining)

    while queue:
        start, path, remaining = queue.popleft()

        if remaining == 0:
            result.append(path)
            continue

        for i in range(start, len(candidates)):
            if candidates[i] > remaining:
                break
            queue.append((i, path + [candidates[i]], remaining - candidates[i]))

    return result


# Test
print(combinationSum_bfs([2, 3, 6, 7], 7))

Complexity: Time O(n^(T/M)) · Space O(n^(T/M))

• Time: Similar to backtracking; BFS explores the same search space level by level instead of depth-first.
• Space: The queue can hold all partial combinations at the current BFS level, which can be exponential in size.

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Complexity Analysis

Approach	Time	Space	Notes
Basic Backtracking	O(n^(T/M))	O(T/M)	T=target, M=min candidate
With Sorting	O(n^(T/M))	O(T/M)	Better pruning
DP	O(T * n * L)	O(T * C)	L=avg length, C=num combinations

Understanding the Time Complexity

The worst case is when we can reuse the smallest element many times:

• If min(candidates) = 2 and target = 10
• We could have combinations like [2,2,2,2,2]
• Maximum depth of recursion: target / min(candidates)
• At each level, we try up to n candidates

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Pruning Visualization

Without sorting (candidates = [7, 3, 2], target = 5):
- Try 7: 7 > 5, continue...
- Try 3: [3], remaining = 2
  - Try 7: 7 > 2, continue...
  - Try 3: 3 > 2, continue...
  - Try 2: [3,2], remaining = 0, SUCCESS!
- Try 2: [2], remaining = 3
  - ...continues exploring...

With sorting (candidates = [2, 3, 7], target = 5):
- Try 2: [2], remaining = 3
  - Try 2: [2,2], remaining = 1
    - Try 2: 2 > 1, BREAK! (skip 3, 7 too)
  - Try 3: [2,3], remaining = 0, SUCCESS!
  - Try 7: 7 > 3, BREAK!
- Try 3: [3], remaining = 2
  - Try 3: 3 > 2, BREAK! (skip 7 too)
- Try 7: 7 > 5, BREAK!

Sorting allows `break` instead of `continue`!

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Common Mistakes

1. Using i+1 Instead of i

# WRONG - doesn't allow reusing elements
def wrong(candidates, target):
    def backtrack(start, path, remaining):
        for i in range(start, len(candidates)):
            path.append(candidates[i])
            backtrack(i + 1, path, remaining - candidates[i])  # i+1 is wrong!
            path.pop()

# CORRECT - use i to allow reuse
def correct(candidates, target):
    def backtrack(start, path, remaining):
        for i in range(start, len(candidates)):
            path.append(candidates[i])
            backtrack(i, path, remaining - candidates[i])  # i allows reuse!
            path.pop()

2. Forgetting Pruning Base Case

# INEFFICIENT - continues even when remaining < 0
def inefficient(candidates, target):
    def backtrack(start, path, remaining):
        if remaining == 0:
            result.append(path[:])
            return
        # Missing: if remaining < 0: return
        for i in range(start, len(candidates)):
            path.append(candidates[i])
            backtrack(i, path, remaining - candidates[i])  # Goes negative!
            path.pop()

# CORRECT - prune when remaining < 0
def correct(candidates, target):
    def backtrack(start, path, remaining):
        if remaining == 0:
            result.append(path[:])
            return
        if remaining < 0:  # Pruning!
            return

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Comparison: Combination Problems

Problem	Reuse Allowed?	Duplicates in Input?	Index in Recursion
Combination Sum	Yes	No	i
Combination Sum II	No	Yes	i + 1 + skip duplicates
Combination Sum III	No	No	i + 1
Subsets	No	No	i + 1

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Related Problems

Combination Sum II (LeetCode 40)

Problem: Find unique combinations that sum to target. Each number used once, input may have duplicates.

Key Insight: Two changes: use i+1 (no reuse) and skip duplicates at same level.

Approach: Sort + skip duplicates + track remaining sum + use i+1.

Complexity: O(2^n) time, O(n) space

Link: Local Solution

Combination Sum III (LeetCode 216)

Problem: Find k numbers from 1-9 that sum to n. Each number used at most once.

Key Insight: Fixed pool (1-9), fixed count (k), no duplicates in input.

Approach: Backtracking with two termination conditions: count == k AND sum == n.

Complexity: O(C(9,k) * k) time, O(k) space

Link: LeetCode 216

Coin Change (LeetCode 322)

Problem: Find minimum number of coins needed to make amount.

Key Insight: Optimization problem (min count) vs enumeration (all combinations).

Approach: DP with dp[amount] = min coins. dp[i] = min(dp[i], dp[i-coin] + 1).

Complexity: O(amount * n) time, O(amount) space

Link: LeetCode 322

Coin Change 2 (LeetCode 518)

Problem: Count the number of combinations that sum to amount.

Key Insight: Counting problem vs listing problem. Use DP instead of backtracking.

Approach: DP counting. Process coins in outer loop to avoid duplicate counts.

Complexity: O(amount * n) time, O(amount) space

Link: LeetCode 518

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Interview Tips

1. Clarify reuse: "Can I use the same element multiple times?"
2. Explain the key insight: Use i instead of i+1 for reuse
3. Mention pruning: Sort + early termination optimization
4. Draw the tree: Shows understanding of branching factor
5. Discuss complexity: Explain why it's exponential

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Key Takeaways

1. Use i in recursive call to allow element reuse
2. Two base cases: remaining == 0 (success) and remaining < 0 (prune)
3. Sorting enables better pruning with early break
4. Start index prevents duplicates like [2,3] and [3,2]
5. Complexity is exponential in target/min(candidates)

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Last updated: January 2026