# 2915. Length of the Longest Subsequence That Sums to Target

## Description

You are given a 0-indexed 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 $f[i][j]=f[i-1][j]$. If we select $x$, then $f[i][j]=f[i-1][j-x]+1$, where $j\ge x$. Therefore, we have the state transition equation:

$f[i][j]=\max\{f[i-1][j],f[i-1][j-x]+1\}$

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 $f[i-1][\cdot]$, 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 n = nums.size();
int[][] f = new int[n + 1][target + 1];
final int inf = 1 << 30;
for (int[] g : f) {
Arrays.fill(g, -inf);
}
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
int x = nums.get(i - 1);
for (int j = 0; j <= target; ++j) {
f[i][j] = f[i - 1][j];
if (j >= x) {
f[i][j] = Math.max(f[i][j], f[i - 1][j - x] + 1);
}
}
}
return f[n][target] <= 0 ? -1 : f[n][target];
}
}

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

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


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

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