Solution Explanation for LeetCode 77. Combinations

This problem asks us to find all possible combinations of k numbers chosen from the range [1, n]. The solution uses backtracking, a recursive approach to explore all possible combinations. Two variations of the backtracking approach are presented.

Approach 1: Backtracking (Two Choices at Each Step)

This approach considers each number from 1 to n. For each number, we have two choices: either include it in the current combination or exclude it. This leads to a recursive branching process.

Algorithm:

1. dfs(i) function: This recursive function takes an integer i as input, representing the starting index for the next number to consider.
2. Base Cases:
   • If the length of the current combination (t) equals k, we've found a valid combination, so we add it to the result (ans) and return.
   • If i exceeds n, we've exhausted all numbers, so we return.
3. Recursive Steps:
   • Include: Add i to the current combination (t), recursively call dfs(i + 1) to explore combinations starting from the next number, and then remove i from t (backtracking).
   • Exclude: Recursively call dfs(i + 1) without adding i to t, exploring combinations that don't include i.

Code (Python):

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        def dfs(i: int):
            if len(t) == k:
                ans.append(t[:])  # Append a copy to avoid modification
                return
            if i > n:
                return
            t.append(i)
            dfs(i + 1)
            t.pop()
            dfs(i + 1)
 
        ans = []
        t = []
        dfs(1)
        return ans

Time Complexity: O(k * C(n, k)), where C(n, k) is the number of combinations (n choose k). The algorithm explores all combinations, and for each combination, we spend O(k) time to copy it into the result.

Space Complexity: O(k) due to the recursion depth and the t list storing the current combination. The space for ans is O(C(n, k) * k) in the worst case, but this is usually considered separately from the algorithm's space complexity.

Approach 2: Backtracking (Iterating Through Choices)

This approach is similar to Approach 1 but slightly more efficient because it avoids the redundant recursive calls. Instead of two recursive calls for each number (include/exclude), it iterates through the possible numbers to include in the combination.

Algorithm:

1. dfs(i) function: This recursive function takes an integer i as input, representing the starting index for the next number to consider.
2. Base Cases: Same as Approach 1.
3. Recursive Steps:
   • It iterates from i to n. For each number j, it adds j to the current combination (t), recursively calls dfs(j + 1) to explore further combinations, and then removes j from t (backtracking).

Code (Python):

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        def dfs(i: int):
            if len(t) == k:
                ans.append(t[:])
                return
            if i > n:
                return
            for j in range(i, n + 1):
                t.append(j)
                dfs(j + 1)
                t.pop()
 
        ans = []
        t = []
        dfs(1)
        return ans

Time Complexity: O(k * C(n, k)). The analysis is the same as in Approach 1.

Space Complexity: O(k), same as Approach 1.

Both approaches have the same time complexity. Approach 2 might be slightly more efficient in practice due to reduced function calls, but the difference is often negligible. The choice between the two approaches is largely a matter of preference and coding style. Other languages follow similar logic and complexity.