grid47 Exploring patterns and algorithms
Oct 30, 2024
5 min read
Solution to LeetCode 77: Combinations Problem
You are given two integers n and k. Find all possible combinations of k numbers selected from the range [1, n], where the numbers are chosen without repetition. The answer should contain all possible combinations of k numbers in any order.
Problem
Approach
Steps
Complexity
Input: The input consists of two integers, `n` and `k`. `n` represents the upper bound of the range of numbers, and `k` represents the number of numbers to be selected.
Example: n = 5, k = 3
Constraints:
• 1 <= n <= 20
• 1 <= k <= n
Output: Return all possible combinations of `k` numbers from the range [1, n], in any order.
• The output should be a list of combinations, each containing exactly `k` numbers from the range [1, n].
Goal: Generate all possible combinations of `k` numbers from the range [1, n] using a backtracking approach.
Steps:
• Use a backtracking technique to explore all possible combinations by selecting a number from the range [1, n] and recursively adding it to the current combination.
• If the current combination has `k` elements, add it to the result list.
• Backtrack by removing the last selected element and exploring other possible combinations.
Goal: The problem constraints ensure the size of the input values.
Steps:
• 1 <= n <= 20
• 1 <= k <= n
Assumptions:
• The input integers `n` and `k` are valid, with `k <= n`.
• Input: n = 5, k = 3
• Explanation: The output consists of all possible combinations of 3 numbers selected from the range [1, 5]. There are a total of 10 combinations.
• Input: n = 3, k = 2
• Explanation: The output consists of all possible combinations of 2 numbers selected from the range [1, 3]. There are a total of 3 combinations.
Approach: The approach to solving this problem involves using backtracking to generate all combinations of `k` numbers from the range [1, n].
Observations:
• We need to generate combinations, not permutations, so order doesn't matter.
• Backtracking is a natural fit for generating combinations, as it allows us to explore all possibilities and backtrack when we reach a valid combination.
Steps:
• Define a helper function that takes the current combination and recursively adds elements from the range [1, n] to it.
• If the current combination has `k` elements, add it to the result list.
• Ensure that each combination is unique by incrementing the starting index for the next recursive call.
Empty Inputs:
• If `k` is greater than `n`, there are no valid combinations.
Large Inputs:
• For the upper bound of `n = 20`, ensure the solution handles the large number of possible combinations efficiently.
Special Values:
• If `k = 1`, the output will contain `n` combinations, each containing a single number from 1 to n.
Constraints:
• Ensure that the solution efficiently handles the constraints of `1 <= n <= 20`.
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> combinations;
vector<int> combination;
backtrack(combinations, combination, 1, n, k);
return combinations;
}
voidbacktrack(vector<vector<int>>& combinations, vector<int>& combination, int start, int n, int k) {
if (combination.size() == k) {
combinations.push_back(combination);
return;
}
for (int i = start; i <= n; ++i) {
combination.push_back(i);
backtrack(combinations, combination, i +1, n, k);
combination.pop_back();
}
}
1 : Function Declaration
vector<vector<int>> combine(int n, int k) {
Declares a function `combine` that takes two integers `n` and `k` as input and returns a vector of vectors representing all possible combinations.
2 : Variable Initialization
vector<vector<int>> combinations;
Initializes an empty vector `combinations` to store the generated combinations.
3 : Variable Initialization
vector<int> combination;
Initializes an empty vector `combination` to store the current combination being built.
4 : Function Call
backtrack(combinations, combination, 1, n, k);
Calls the `backtrack` function to recursively generate combinations.
5 : Return
return combinations;
Returns the `combinations` vector containing all the generated combinations.
6 : Function Declaration
voidbacktrack(vector<vector<int>>& combinations, vector<int>& combination, int start, int n, int k) {
Declares a recursive `backtrack` function to explore combinations.
7 : Conditional
if (combination.size() == k) {
Checks if the current `combination` has `k` elements.
8 : Vector Operation
combinations.push_back(combination);
Adds the current `combination` to the `combinations` vector.
9 : Return
return;
Returns from the recursive call as a complete combination is found.
10 : Conditional
if (i > n) return;
Checks if the current index `i` has exceeded the maximum value `n`.
11 : Loop Iteration
for (int i = start; i <= n; ++i) {
Iterates from `start` to `n` to explore possible numbers for the current position in the combination.
12 : Vector Operation
combination.push_back(i);
Adds the current number `i` to the `combination`.
13 : Recursive Call
backtrack(combinations, combination, i +1, n, k);
Recursively calls the `backtrack` function with the updated `combination` and `start` index.
14 : Vector Operation
combination.pop_back();
Removes the last added number from the `combination` to explore other possibilities.
Best Case: O(n choose k)
Average Case: O(n choose k)
Worst Case: O(n choose k)
Description: The time complexity is determined by the number of combinations that need to be generated, which is `n choose k`.
Best Case: O(k)
Worst Case: O(k)
Description: The space complexity is O(k) due to the temporary storage used to hold the current combination during recursion.
