Leetcode

Target Sum

May 26, 2022 by aAshish

Approach 1: 2D DP

  • Time:O(nk), where k = (\Sigma \texttt{nums[i]} + S) / 2
  • Space:O(nk)

C++

 
class Solution {
 public:
  int findTargetSumWays(vector<int>& nums, int S) {
    const int sum = accumulate(begin(nums), end(nums), 0);
    if (sum < S || (sum + S) & 1)
      return 0;

    return knapsack(nums, (sum + S) / 2);
  }

 private:
  int knapsack(const vector<int>& nums, int target) {
    const int n = nums.size();
    // dp[i][j] := # 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];
  }
};

JAVA

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

    return knapsack(nums, (sum + S) / 2);
  }

  private int knapsack(int[] nums, int target) {
    final int n = nums.length;
    // dp[i][j] := # 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];
  }
}

Python

 
class Solution:
  def findTargetSumWays(self, nums: List[int], S: int) -> int:
    summ = sum(nums)
    if summ < S or (summ + S) & 1:
      return 0

    def knapsack(target: int) -> int:
      # dp[i] := # 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 + S) // 2)

Approach 2: 1D DP

  • Time:O(nk)
  • Space:O(k)

C++

 
class Solution {
 public:
  int findTargetSumWays(vector<int>& nums, int S) {
    const int sum = accumulate(begin(nums), end(nums), 0);
    if (sum < S || (sum + S) & 1)
      return 0;

    return knapsack(nums, (sum + S) / 2);
  }

 private:
  int knapsack(const vector<int>& nums, int target) {
    // dp[i] := # 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];
  }
};

JAVA

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

    return knapsack(nums, (sum + S) / 2);
  }

  private int knapsack(int[] nums, int target) {
    // dp[i] := # 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];
  }
}