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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[] 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[j-x]+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];
}

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