README_EN.md
| comments | difficulty | edit_url | tags | ||
|---|---|---|---|---|---|
| true | Medium | https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0039.Combination%20Sum/README_EN.md |
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39. Combination Sum
Description
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
Example 1:
Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.
Example 2:
Input: candidates = [2,3,5], target = 8 Output: [[2,2,2,2],[2,3,3],[3,5]]
Example 3:
Input: candidates = [2], target = 1 Output: []
Constraints:
1 <= candidates.length <= 302 <= candidates[i] <= 40- All elements of
candidatesare distinct. 1 <= target <= 40
Solutions
Solution 1: Sorting + Pruning + Backtracking
We can first sort the array to facilitate pruning.
Next, we design a function dfs(i, s), which means starting the search from index i with a remaining target value of s. Here, i and s are both non-negative integers, the current search path is t, and the answer is ans.
In the function dfs(i, s), we first check whether s is 0. If it is, we add the current search path t to the answer ans, and then return. If s \lt candidates[i], it means that the elements of the current index and the following indices are all greater than the remaining target value s, and the path is invalid, so we return directly. Otherwise, we start the search from index i, and the search index range is j \in [i, n), where n is the length of the array candidates. During the search, we add the element of the current index to the search path t, recursively call the function dfs(j, s - candidates[j]), and after the recursion ends, we remove the element of the current index from the search path t.
In the main function, we just need to call the function dfs(0, target) to get the answer.
The time complexity is O(2^n \times n), and the space complexity is O(n). Here, n is the length of the array candidates. Due to pruning, the actual time complexity is much less than O(2^n \times n).
Similar problems:
Python3
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
def dfs(i: int, s: int):
if s == 0:
ans.append(t[:])
return
if s < candidates[i]:
return
for j in range(i, len(candidates)):
t.append(candidates[j])
dfs(j, s - candidates[j])
t.pop()
candidates.sort()
t = []
ans = []
dfs(0, target)
return ans
Java
class Solution {
private List<List<Integer>> ans = new ArrayList<>();
private List<Integer> t = new ArrayList<>();
private int[] candidates;
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
this.candidates = candidates;
dfs(0, target);
return ans;
}
private void dfs(int i, int s) {
if (s == 0) {
ans.add(new ArrayList(t));
return;
}
if (s < candidates[i]) {
return;
}
for (int j = i; j < candidates.length; ++j) {
t.add(candidates[j]);
dfs(j, s - candidates[j]);
t.remove(t.size() - 1);
}
}
}
C++
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> ans;
vector<int> t;
function<void(int, int)> dfs = [&](int i, int s) {
if (s == 0) {
ans.emplace_back(t);
return;
}
if (s < candidates[i]) {
return;
}
for (int j = i; j < candidates.size(); ++j) {
t.push_back(candidates[j]);
dfs(j, s - candidates[j]);
t.pop_back();
}
};
dfs(0, target);
return ans;
}
};
Go
func combinationSum(candidates []int, target int) (ans [][]int) {
sort.Ints(candidates)
t := []int{}
var dfs func(i, s int)
dfs = func(i, s int) {
if s == 0 {
ans = append(ans, slices.Clone(t))
return
}
if s < candidates[i] {
return
}
for j := i; j < len(candidates); j++ {
t = append(t, candidates[j])
dfs(j, s-candidates[j])
t = t[:len(t)-1]
}
}
dfs(0, target)
return
}
TypeScript
function combinationSum(candidates: number[], target: number): number[][] {
candidates.sort((a, b) => a - b);
const ans: number[][] = [];
const t: number[] = [];
const dfs = (i: number, s: number) => {
if (s === 0) {
ans.push(t.slice());
return;
}
if (s < candidates[i]) {
return;
}
for (let j = i; j < candidates.length; ++j) {
t.push(candidates[j]);
dfs(j, s - candidates[j]);
t.pop();
}
};
dfs(0, target);
return ans;
}
Rust
impl Solution {
fn dfs(i: usize, s: i32, candidates: &Vec<i32>, t: &mut Vec<i32>, ans: &mut Vec<Vec<i32>>) {
if s == 0 {
ans.push(t.clone());
return;
}
if s < candidates[i] {
return;
}
for j in i..candidates.len() {
t.push(candidates[j]);
Self::dfs(j, s - candidates[j], candidates, t, ans);
t.pop();
}
}
pub fn combination_sum(mut candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
candidates.sort();
let mut ans = Vec::new();
Self::dfs(0, target, &candidates, &mut vec![], &mut ans);
ans
}
}
C#
public class Solution {
private List<IList<int>> ans = new List<IList<int>>();
private List<int> t = new List<int>();
private int[] candidates;
public IList<IList<int>> CombinationSum(int[] candidates, int target) {
Array.Sort(candidates);
this.candidates = candidates;
dfs(0, target);
return ans;
}
private void dfs(int i, int s) {
if (s == 0) {
ans.Add(new List<int>(t));
return;
}
if (s < candidates[i]) {
return;
}
for (int j = i; j < candidates.Length; ++j) {
t.Add(candidates[j]);
dfs(j, s - candidates[j]);
t.RemoveAt(t.Count - 1);
}
}
}
Solution 2: Sorting + Pruning + Backtracking(Another Form)
We can also change the implementation logic of the function dfs(i, s) to another form. In the function dfs(i, s), we first check whether s is 0. If it is, we add the current search path t to the answer ans, and then return. If i \geq n or s \lt candidates[i], the path is invalid, so we return directly. Otherwise, we consider two situations, one is not selecting the element of the current index, that is, recursively calling the function dfs(i + 1, s), and the other is selecting the element of the current index, that is, recursively calling the function dfs(i, s - candidates[i]).
The time complexity is O(2^n \times n), and the space complexity is O(n). Here, n is the length of the array candidates. Due to pruning, the actual time complexity is much less than O(2^n \times n).
Python3
class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
def dfs(i: int, s: int):
if s == 0:
ans.append(t[:])
return
if i >= len(candidates) or s < candidates[i]:
return
dfs(i + 1, s)
t.append(candidates[i])
dfs(i, s - candidates[i])
t.pop()
candidates.sort()
t = []
ans = []
dfs(0, target)
return ans
Java
class Solution {
private List<List<Integer>> ans = new ArrayList<>();
private List<Integer> t = new ArrayList<>();
private int[] candidates;
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
this.candidates = candidates;
dfs(0, target);
return ans;
}
private void dfs(int i, int s) {
if (s == 0) {
ans.add(new ArrayList(t));
return;
}
if (i >= candidates.length || s < candidates[i]) {
return;
}
dfs(i + 1, s);
t.add(candidates[i]);
dfs(i, s - candidates[i]);
t.remove(t.size() - 1);
}
}
C++
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> ans;
vector<int> t;
function<void(int, int)> dfs = [&](int i, int s) {
if (s == 0) {
ans.emplace_back(t);
return;
}
if (i >= candidates.size() || s < candidates[i]) {
return;
}
dfs(i + 1, s);
t.push_back(candidates[i]);
dfs(i, s - candidates[i]);
t.pop_back();
};
dfs(0, target);
return ans;
}
};
Go
func combinationSum(candidates []int, target int) (ans [][]int) {
sort.Ints(candidates)
t := []int{}
var dfs func(i, s int)
dfs = func(i, s int) {
if s == 0 {
ans = append(ans, slices.Clone(t))
return
}
if i >= len(candidates) || s < candidates[i] {
return
}
dfs(i+1, s)
t = append(t, candidates[i])
dfs(i, s-candidates[i])
t = t[:len(t)-1]
}
dfs(0, target)
return
}
TypeScript
function combinationSum(candidates: number[], target: number): number[][] {
candidates.sort((a, b) => a - b);
const ans: number[][] = [];
const t: number[] = [];
const dfs = (i: number, s: number) => {
if (s === 0) {
ans.push(t.slice());
return;
}
if (i >= candidates.length || s < candidates[i]) {
return;
}
dfs(i + 1, s);
t.push(candidates[i]);
dfs(i, s - candidates[i]);
t.pop();
};
dfs(0, target);
return ans;
}
Rust
impl Solution {
fn dfs(i: usize, s: i32, candidates: &Vec<i32>, t: &mut Vec<i32>, ans: &mut Vec<Vec<i32>>) {
if s == 0 {
ans.push(t.clone());
return;
}
if i >= candidates.len() || s < candidates[i] {
return;
}
Self::dfs(i + 1, s, candidates, t, ans);
t.push(candidates[i]);
Self::dfs(i, s - candidates[i], candidates, t, ans);
t.pop();
}
pub fn combination_sum(mut candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
candidates.sort();
let mut ans = Vec::new();
Self::dfs(0, target, &candidates, &mut vec![], &mut ans);
ans
}
}
C#
public class Solution {
private List<IList<int>> ans = new List<IList<int>>();
private List<int> t = new List<int>();
private int[] candidates;
public IList<IList<int>> CombinationSum(int[] candidates, int target) {
Array.Sort(candidates);
this.candidates = candidates;
dfs(0, target);
return ans;
}
private void dfs(int i, int s) {
if (s == 0) {
ans.Add(new List<int>(t));
return;
}
if (i >= candidates.Length || s < candidates[i]) {
return;
}
dfs(i + 1, s);
t.Add(candidates[i]);
dfs(i, s - candidates[i]);
t.RemoveAt(t.Count - 1);
}
}
PHP
class Solution {
/**
* @param integer[] $candidates
* @param integer $target
* @return integer[][]
*/
function combinationSum($candidates, $target) {
$result = [];
$currentCombination = [];
$startIndex = 0;
sort($candidates);
$this->findCombinations($candidates, $target, $startIndex, $currentCombination, $result);
return $result;
}
function findCombinations($candidates, $target, $startIndex, $currentCombination, &$result) {
if ($target === 0) {
$result[] = $currentCombination;
return;
}
for ($i = $startIndex; $i < count($candidates); $i++) {
$num = $candidates[$i];
if ($num > $target) {
break;
}
$currentCombination[] = $num;
$this->findCombinations($candidates, $target - $num, $i, $currentCombination, $result);
array_pop($currentCombination);
}
}
}