Question

2915. Length of the Longest Subsequence That Sums to Target

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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:
  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]

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];
  }
};

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];
}

Solution 2

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]

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];
  }
};

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];
}