medium Combinations

Problem Statement

Given two integers n and k, return all possible combinations of k numbers chosen from the range [1, n].

Each combination must be a unique set of numbers, order of which does not matter.

Examples

• Example 1:

  • Input: n = 3, k = 2
  • Expected Output: [[1, 2], [1, 3], [2, 3]]
  • Justification: [[1, 2], [1, 3], [2, 3]] are all combinations of size 2, which we can create using [1, 2, 3].

• Example 2:

  • Input: n = 4, k = 1
  • Expected Output: [[1], [2], [3], [4]]
  • Justification: [[1], [2], [3], [4]] are all combinations of size 1, which we can create using `[1, 2, 3, 4].

• Example 3:

  • Input: n = 5, k = 3
  • Expected Output: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
  • Justification: We unique triplets using numbers 1 to 5, yielding ten different combinations.

Solution

To solve this problem, we can employ a backtracking approach. Backtracking is a recursive strategy for solving problems by exploring all possible solutions and backtrack whenever a solution that's currently being explored no longer seems viable. This approach is particularly effective for generating combinations because it allows us to build combinations one element at a time and backtrack as soon as we've either found a valid combination or need to make a different choice to continue exploring possibilities.

Backtracking works here by choosing a start number for a combination, then recursively choosing the next number, ensuring each number is greater than the previous to maintain uniqueness and avoid duplication. This method efficiently explores all potential combinations of k numbers within the given range, ensuring no possibilities are missed and no invalid combinations are included.

Step-by-step Algorithm

1. Initialize an empty list (or array) result to store all the combinations.
2. Define a helper function backtrack that takes the current combination (tempList), the starting number (start), 3. n, and k as parameters. Begin the process by calling backtrack with an empty tempList, start as 1, n, and k.
3. Check if the current combination's size equals k. If so, add a copy of tempList to result and return.
4. Iterate from the current start number to n:
   • Add the current number i to tempList.
   • Recursively call backtrack with i + 1 as the new start, to ensure each combination is unique and respects the ascending order.
   • Remove the last number added to tempList to backtrack, exploring other possible combinations by trying the next number in the sequence.
5. Once all combinations are explored, return the result.

Algorithm Walkthrough

Let's consider the input n = 5 and k = 3:

1. Initialize result as an empty list to store combinations.
2. Call backtrack with an empty list as tempList, start = 1, n = 5, and k = 3.
3. First Call to Backtrack:
   • tempList = [], start = 1
   • Loop from i = 1 to 5:
     • Add 1 to tempList and call backtrack with start = 2.
       • Second Call to Backtrack:
         • tempList = [1], start = 2
         • Loop from i = 2 to 5:
           • Add 2 to tempList and call backtrack with start = 3.
             • Third Call to Backtrack:
               • tempList = [1, 2], start = 3
               • Loop from i = 3 to 5:
                 • Add 3 to tempList making it [1, 2, 3] and since tempList.size() == k, add [1, 2, 3] to result.
                 • Backtrack: Remove 3, making tempList = [1, 2].
                 • Add 4 to tempList making it [1, 2, 4] and since tempList.size() == k, add [1, 2, 4] to result.
                 • Backtrack: Remove 4, making tempList = [1, 2].
                 • Add 5 to tempList making it [1, 2, 5] and since tempList.size() == k, add [1, 2, 5] to result.
                 • Backtrack: Remove 5, making tempList = [1, 2].
               • Backtrack to the second call: tempList = [1].
           • Next, i = 3 in the second call, add 3 to tempList making it [1, 3] and proceed similarly.
           • Continue this process until all combinations starting with 1 are explored.
     • Backtrack and try next numbers as the starting point, [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5] will be generated in subsequent calls.
4. After exploring all possible combinations starting from 1 to 5, the result will contain all unique combinations of size 3 from numbers 1 to 5.
5. Return result which now holds all the required combinations: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]].

Code

Complexity Analysis

Time Complexity

• : The primary factor affecting the time complexity is the number of combinations generated, denoted as C(n, k), where C(n, k) is the binomial coefficient representing the number of ways to choose k elements out of n options. For each combination, we perform operations that are proportional to k (to add elements to a temporary list or array). Therefore, the total time complexity is .

Space Complexity

• : The space required to store all combinations is directly proportional to the number of combinations times the size of each combination, which is k. This accounts for the space needed to store the output.

🎯 STRICT STANDOUT — Solution Combinations

0. Pattern family

Family: Backtracking / combinations C(n,k) with start-index discipline

1. Why / judgment (K3)

Combinations are unordered subsets of fixed size k. The standout discipline is the start index: always append from i..n and recurse with i+1 so each set is built in strictly ascending order. That kills permutations-of-the-same-set without a visited array. Judgment: if order mattered it would be permutations; if k were free it would be subsets; here size is locked to k.

2. Worked complexity / derivation (K11)

Output size Θ(C(n,k)·k) to write answers.
Search tree: at depth d, branching ≤ n−start+1; total internal nodes O(C(n,k)·k) in the
classic analysis (each combination prefix is charged once).
Time Θ(C(n,k)·k) dominated by copying each length-k combo into the answer.
Space O(k) recursion depth + O(C(n,k)·k) for output.
Naive generate-all-permutations-then-dedup: O(P(n,k)·k) — strictly worse; start-index is the fix.

3. Pattern + recognition + when-NOT (K12)

Name: COMBINATIONS BACKTRACK (start = i+1, stop at len==k)

Recognition: choose k from 1..n; order irrelevant; classic “pick / skip with ascending start”.

When-NOT: If order matters → permutations (swap or used[]). If any size → subsets (include/exclude). If reuse allowed → combination sum (start stays i). If only count C(n,k) → DP/Pascal, not DFS.

4. Edge hand-run (K13)

n=3,k=2 → [[1,2],[1,3],[2,3]] (not [2,1]).
n=1,k=1 → [[1]]. k=0 → [[]] convention; constraints usually k≥1.
n=4,k=4 → [[1,2,3,4]] one path. n=4,k=5 → [] prune when remaining < needed.

5. Interviewer follow-ups & drills

Q1. Why start=i+1 not start=i?
Model answer: start=i reuses the same number and produces multisets / perm-duplicates; combinations need each index once, ascending.

Q2. Prune early?
Model answer: If remaining numbers n−i+1 < k−len(path), return — cannot fill the combo.

Q3. How does this transfer to Combination Sum?
Model answer: Same DFS skeleton; change start to i (reuse) and stop on remaining target 0 / negative.