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2915. Length of the Longest Subsequence That Sums to Target
Description
You are given a 0indexed array of integers nums
, and an integer target
.
Return the length of the longest subsequence of nums
that sums up to target
. If no such subsequence exists, return 1
.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [1,2,3,4,5], target = 9 Output: 3 Explanation: There are 3 subsequences with a sum equal to 9: [4,5], [1,3,5], and [2,3,4]. The longest subsequences are [1,3,5], and [2,3,4]. Hence, the answer is 3.
Example 2:
Input: nums = [4,1,3,2,1,5], target = 7 Output: 4 Explanation: There are 5 subsequences with a sum equal to 7: [4,3], [4,1,2], [4,2,1], [1,1,5], and [1,3,2,1]. The longest subsequence is [1,3,2,1]. Hence, the answer is 4.
Example 3:
Input: nums = [1,1,5,4,5], target = 3 Output: 1 Explanation: It can be shown that nums has no subsequence that sums up to 3.
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 1000
1 <= target <= 1000
Solutions
Solution 1: Dynamic Programming
We define $f[i][j]$ as the length of the longest subsequence that selects several numbers from the first $i$ numbers and the sum of these numbers is exactly $j$. Initially, $f[0][0]=0$, and all other positions are $\infty$.
For $f[i][j]$, we consider the $i$th number $x$. If we do not select $x$, then $f[i][j]=f[i1][j]$. If we select $x$, then $f[i][j]=f[i1][jx]+1$, where $j\ge x$. Therefore, we have the state transition equation:
\[f[i][j]=\max\{f[i1][j],f[i1][jx]+1\}\]The final answer is $f[n][target]$. If $f[n][target]\le0$, there is no subsequence with a sum of $target$, return $1$.
The time complexity is $O(n\times target)$, and the space complexity is $O(n\times target)$. Here, $n$ is the length of the array, and $target$ is the target value.
We notice that the state of $f[i][j]$ is only related to $f[i1][\cdot]$, so we can optimize the first dimension and reduce the space complexity to $O(target)$.

class Solution { public int lengthOfLongestSubsequence(List<Integer> nums, int target) { int[] f = new int[target + 1]; final int inf = 1 << 30; Arrays.fill(f, inf); f[0] = 0; for (int x : nums) { for (int j = target; j >= x; j) { f[j] = Math.max(f[j], f[j  x] + 1); } } return f[target] <= 0 ? 1 : f[target]; } }

class Solution { public: int lengthOfLongestSubsequence(vector<int>& nums, int target) { int f[target + 1]; memset(f, 0x3f, sizeof(f)); f[0] = 0; for (int x : nums) { for (int j = target; j >= x; j) { f[j] = max(f[j], f[j  x] + 1); } } return f[target] <= 0 ? 1 : f[target]; } };

class Solution: def lengthOfLongestSubsequence(self, nums: List[int], target: int) > int: f = [0] + [inf] * target for x in nums: for j in range(target, x  1, 1): f[j] = max(f[j], f[j  x] + 1) return 1 if f[1] <= 0 else f[1]

func lengthOfLongestSubsequence(nums []int, target int) int { f := make([]int, target+1) for i := range f { f[i] = (1 << 30) } f[0] = 0 for _, x := range nums { for j := target; j >= x; j { f[j] = max(f[j], f[jx]+1) } } if f[target] <= 0 { return 1 } return f[target] }

function lengthOfLongestSubsequence(nums: number[], target: number): number { const f: number[] = Array(target + 1).fill(Infinity); f[0] = 0; for (const x of nums) { for (let j = target; j >= x; j) { f[j] = Math.max(f[j], f[j  x] + 1); } } return f[target] <= 0 ? 1 : f[target]; }