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494. Target Sum 👍

Approach 1: 2D DP

  • Time: $O(nk)$, where $k = (\Sigma \texttt{nums[i]} + \texttt{target}) / 2$
  • Space: $O(nk)$
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class Solution {
 public:
  int findTargetSumWays(vector<int>& nums, int target) {
    const int sum = accumulate(nums.begin(), nums.end(), 0);
    if (sum < abs(target) || (sum + target) % 2 == 1)
      return 0;
    return knapsack(nums, (sum + target) / 2);
  }

 private:
  int knapsack(const vector<int>& nums, int target) {
    const int n = nums.size();
    // dp[i][j] := the number of ways to sum to j by nums[0..i)
    vector<vector<int>> dp(n + 1, vector<int>(target + 1));
    dp[0][0] = 1;

    for (int i = 1; i <= n; ++i) {
      const int num = nums[i - 1];
      for (int j = 0; j <= target; ++j)
        if (j < num)
          dp[i][j] = dp[i - 1][j];
        else
          dp[i][j] = dp[i - 1][j] + dp[i - 1][j - num];
    }

    return dp[n][target];
  }
};

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class Solution {
  public int findTargetSumWays(int[] nums, int target) {
    final int sum = Arrays.stream(nums).sum();
    if (sum < Math.abs(target) || (sum + target) % 2 == 1)
      return 0;
    return knapsack(nums, (sum + target) / 2);
  }

  private int knapsack(int[] nums, int target) {
    final int n = nums.length;
    // dp[i][j] := the number of ways to sum to j by nums[0..i)
    int[][] dp = new int[n + 1][target + 1];
    dp[0][0] = 1;

    for (int i = 1; i <= n; ++i) {
      final int num = nums[i - 1];
      for (int j = 0; j <= target; ++j)
        if (j < num)
          dp[i][j] = dp[i - 1][j];
        else
          dp[i][j] = dp[i - 1][j] + dp[i - 1][j - num];
    }

    return dp[n][target];
  }
}

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class Solution:
  def findTargetSumWays(self, nums: list[int], target: int) -> int:
    summ = sum(nums)
    if summ < abs(target) or (summ + target) % 2 == 1:
      return 0

    def knapsack(target: int) -> int:
      # dp[i] := the number of ways to sum to i by nums so far
      dp = [1] + [0] * summ

      for num in nums:
        for j in range(summ, num - 1, -1):
          dp[j] += dp[j - num]

      return dp[target]

    return knapsack((summ + target) // 2)

Approach 2: 1D DP

  • Time: $O(nk)$
  • Space: $O(k)$
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class Solution {
 public:
  int findTargetSumWays(vector<int>& nums, int target) {
    const int sum = accumulate(nums.begin(), nums.end(), 0);
    if (sum < abs(target) || (sum + target) % 2 == 1)
      return 0;
    return knapsack(nums, (sum + target) / 2);
  }

 private:
  int knapsack(const vector<int>& nums, int target) {
    // dp[i] := the number of ways to sum to i by nums so far
    vector<int> dp(target + 1);
    dp[0] = 1;

    for (const int num : nums)
      for (int i = target; i >= num; --i)
        dp[i] += dp[i - num];

    return dp[target];
  }
};

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class Solution {
  public int findTargetSumWays(int[] nums, int target) {
    final int sum = Arrays.stream(nums).sum();
    if (sum < Math.abs(target) || (sum + target) % 2 == 1)
      return 0;
    return knapsack(nums, (sum + target) / 2);
  }

  private int knapsack(int[] nums, int target) {
    // dp[i] := the number of ways to sum to i by nums so far
    int[] dp = new int[target + 1];
    dp[0] = 1;

    for (final int num : nums)
      for (int i = target; i >= num; --i)
        dp[i] += dp[i - num];

    return dp[target];
  }
}

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class Solution:
  def findTargetSumWays(self, nums: list[int], target: int) -> int:
    summ = sum(nums)
    if summ < abs(target) or (summ + target) % 2 == 1:
      return 0

    def knapsack(nums: list[int], target: int) -> int:
      # dp[i] := the number of ways to sum to i by nums so far
      dp = [0] * (target + 1)
      dp[0] = 1

      for num in nums:
        for i in range(target, num - 1, -1):
          dp[i] += dp[i - num]

      return dp[target]

    return knapsack(nums, (summ + target) // 2)