Combination Sum Ii
Sure! Let's break down how to solve this coding challenge step by step with detailed explanations and pseudocode.
To solve this coding challenge, we need to find all unique combinations of a given list of numbers (
) that sum up to a specified target number (
). Each number in
can only be used once in each combination, and we should avoid duplicate combinations in the result.
candidates
target
candidates
# Explanation
- Problem Understanding :
-
You are given a list of integers called
candidates
target
-
The aim is to find all unique combinations in
candidates
target
- Each number may be used only once in each combination.
-
It is crucial to handle duplicate values in
candidates
- Constraints :
-
The length of
candidates
-
Each integer in
candidates
-
The
target
- Approach :
-
Sorting
: Sort the
candidates
- Backtracking : Use a backtracking approach to build combinations incrementally. This involves:
- Recursive Function : A function that will try to add each candidate to a current combination (path), then recurse to see if a valid combination can be formed.
- Base Case : If the current target becomes zero, the current combination is valid and should be added to the results.
- Skipping Duplicates : When considering a candidate, skip it if it's the same as the previously considered candidate (to avoid duplicates).
- Parameters :
-
start_index
candidates
-
current_path
-
remaining_target
- Check for Duplicates : Skip if the candidate is the same as the previous one to avoid duplicate combinations.
-
Break if Candidate Exceeds Target
: If the candidate is greater than the
remaining_target
- Recursive Call : Add the candidate to the current path and call the backtrack function with updated parameters.
- Initial Setup :
- Sort the list of candidates.
-
Initialize an empty list
result
- Backtracking Function definition :
-
The
backtrack
start_index
current_path
remaining_target
-
If
remaining_target
current_path
result
- Looping and Conditions :
-
We loop through each candidate starting from
start_index
-
Skip duplicates: If the current candidate is the same as the previous one and it's not the first time we are looking at it on this recursive level (i.e.,
current_index > start_index
-
Stop further exploration if the current candidate exceeds
remaining_target
-
Include the current candidate in
current_path
- Backtracking :
-
After recursive call, remove the last candidate from
current_path
- Returning the Result :
-
Finally, return the
result
# Detailed Steps in Pseudocode
1. Sort the `candidates` list
Sorting helps to streamline the process of skipping duplicates and breaking out of the loop early if a candidate exceeds the target during recursion.2. Initialize the result list
This will store all the valid combinations found.3. Define the backtracking function
This function will handle the recursive aspect of exploring potential combinations.4. Base Case
Ifremaining_target
5. Iterate over the candidates starting from `start_index`
For each candidate:Pseudocode
function combinationSum2(candidates, target):
# Step 1: Sort the candidates list to handle duplicates and make breaking conditions easier
sort(candidates)
# Step 2: Initialize the result list
result = []
# Step 3: Define the backtracking function
function backtrack(start_index, current_path, remaining_target):
# Step 4: Base Case - If remaining_target is 0, store a copy of current_path in result
if remaining_target == 0:
result.append(copy_of(current_path))
return
# Step 5: Iterate over candidates starting from start_index
for current_index from start_index to length of candidates:
# Skip duplicate candidates to avoid duplicate combinations
if current_index > start_index and candidates[current_index] == candidates[current_index - 1]:
continue
# Break if current candidate exceeds the remaining_target
if candidates[current_index] > remaining_target:
break
# Choose the current candidate by adding it to current_path
current_path.append(candidates[current_index])
# Recur with updated parameters: move to next candidate and reduce the target
backtrack(current_index + 1, current_path, remaining_target - candidates[current_index])
# Un-choose the current candidate to backtrack
current_path.pop()
# Initialize the backtracking with initial parameters
backtrack(0, [], target)
# Step 6: Return the result
return result