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We can first sort the array to facilitate pruning.
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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 $ans$.
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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 $t$.
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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 $dfs(i, s - candidates[i])$.
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In the main function, we just need to call the function $dfs(0, target)$ to get the answer.
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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 $O(2^n \times n)$.
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 $dfs(i, s - candidates[i])$.
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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 $O(2^n \times n)$.
We can also change the implementation logic of the function $dfs(i, s)$ to another form. If we choose the current number, we add the current number to the search path $t$, then recursively call the function $dfs(i + 1, s - candidates[i])$, and after the recursion ends, we remove the current number from the search path $t$. If we do not choose the current number, we can skip all numbers that are the same as the current number, then recursively call the function $dfs(j, s)$, where $j$ is the index of the first number that is different from the current number.
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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 $O(2^n \times n)$.
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