LeetCode #216 — MEDIUM

Combination Sum III

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used.
Each number is used at most once.

Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Example 1:

Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Example 2:

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Example 3:

Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Constraints:

2 <= k <= 9
1 <= n <= 60

Step 01

Brute Force Baseline

Problem summary: Find all valid combinations of k numbers that sum up to n such that the following conditions are true: Only numbers 1 through 9 are used. Each number is used at most once. Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking

Example 1

3
7

Example 2

3
9

Example 3

4
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #216: Combination Sum III
class Solution {
    private List<List<Integer>> ans = new ArrayList<>();
    private List<Integer> t = new ArrayList<>();
    private int k;

    public List<List<Integer>> combinationSum3(int k, int n) {
        this.k = k;
        dfs(1, n);
        return ans;
    }

    private void dfs(int i, int s) {
        if (s == 0) {
            if (t.size() == k) {
                ans.add(new ArrayList<>(t));
            }
            return;
        }
        if (i > 9 || i > s || t.size() >= k) {
            return;
        }
        t.add(i);
        dfs(i + 1, s - i);
        t.remove(t.size() - 1);
        dfs(i + 1, s);
    }
}

// Accepted solution for LeetCode #216: Combination Sum III
func combinationSum3(k int, n int) (ans [][]int) {
	t := []int{}
	var dfs func(i, s int)
	dfs = func(i, s int) {
		if s == 0 {
			if len(t) == k {
				ans = append(ans, slices.Clone(t))
			}
			return
		}
		if i > 9 || i > s || len(t) >= k {
			return
		}
		t = append(t, i)
		dfs(i+1, s-i)
		t = t[:len(t)-1]
		dfs(i+1, s)
	}
	dfs(1, n)
	return
}

# Accepted solution for LeetCode #216: Combination Sum III
class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        def dfs(i: int, s: int):
            if s == 0:
                if len(t) == k:
                    ans.append(t[:])
                return
            if i > 9 or i > s or len(t) >= k:
                return
            t.append(i)
            dfs(i + 1, s - i)
            t.pop()
            dfs(i + 1, s)

        ans = []
        t = []
        dfs(1, n)
        return ans

// Accepted solution for LeetCode #216: Combination Sum III
impl Solution {
    #[allow(dead_code)]
    pub fn combination_sum3(k: i32, n: i32) -> Vec<Vec<i32>> {
        let mut ret = Vec::new();
        let mut candidates = (1..=9).collect();
        let mut cur_vec = Vec::new();
        Self::dfs(n, k, 0, 0, &mut cur_vec, &mut candidates, &mut ret);
        ret
    }

    #[allow(dead_code)]
    fn dfs(
        target: i32,
        length: i32,
        cur_index: usize,
        cur_sum: i32,
        cur_vec: &mut Vec<i32>,
        candidates: &Vec<i32>,
        ans: &mut Vec<Vec<i32>>,
    ) {
        if cur_sum > target || cur_vec.len() > (length as usize) {
            // No answer for this
            return;
        }
        if cur_sum == target && cur_vec.len() == (length as usize) {
            // We get an answer
            ans.push(cur_vec.clone());
            return;
        }
        for i in cur_index..candidates.len() {
            cur_vec.push(candidates[i]);
            Self::dfs(
                target,
                length,
                i + 1,
                cur_sum + candidates[i],
                cur_vec,
                candidates,
                ans,
            );
            cur_vec.pop().unwrap();
        }
    }
}

// Accepted solution for LeetCode #216: Combination Sum III
function combinationSum3(k: number, n: number): number[][] {
    const ans: number[][] = [];
    const t: number[] = [];
    const dfs = (i: number, s: number) => {
        if (s === 0) {
            if (t.length === k) {
                ans.push(t.slice());
            }
            return;
        }
        if (i > 9 || i > s || t.length >= k) {
            return;
        }
        t.push(i);
        dfs(i + 1, s - i);
        t.pop();
        dfs(i + 1, s);
    };
    dfs(1, n);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^9 × 9)

Space

O(k)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.