494. Target Sum

494. Target Sum (Medium)

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 
-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3
There are 5 ways to assign symbols to make the sum of nums be target 3.

Note:

1. The length of the given array is positive and will not exceed 20.
2. The sum of elements in the given array will not exceed 1000.
3. Your output answer is guaranteed to be fitted in a 32-bit integer.

Related Topics:
Dynamic Programming, Depth-first Search

Similar Questions:

• Expression Add Operators (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/target-sum
// Author: github.com/lzl124631x
// Time: O(2^N)
// Space: O(2^N)
class Solution {
public:
  int findTargetSumWays(vector<int>& nums, int S) {
    unordered_map<int, int> ans;
    ans[0] = 1;
    for (int n : nums) {
      unordered_map<int, int> newAns;
      for (auto p : ans) {
        int sum = p.first, cnt = p.second;
        newAns[sum + n] += cnt;
        newAns[sum - n] += cnt;
      }
      ans = newAns;
    }
    return ans[S];
  }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/target-sum/
// Author: github.com/lzl124631x
// Time: O(2^N)
// Space: O(N)
typedef long long LL;
class Solution {
    int dfs(vector<int> &nums, int start, LL target) {
        if (start == nums.size()) return target == 0;
        return dfs(nums, start + 1, target - nums[start])
            + dfs(nums, start + 1, target + nums[start]);
    }
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        return dfs(nums, 0, S);
    }
};

Solution 3. DFS + Memo

// OJ: https://leetcode.com/problems/target-sum/
// Author: github.com/lzl124631x
// Time: O(2^N)
// Space: O(2^N)
typedef long long LL;
class Solution {
    vector<unordered_map<int, int>> memo;
    int dfs(vector<int> &nums, int start, LL target) {
        if (start == nums.size()) return target == 0;
        if (memo[start].count(target)) return memo[start][target];
        return memo[start][target] = dfs(nums, start + 1, target - nums[start])
            + dfs(nums, start + 1, target + nums[start]);
    }
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        memo = vector<unordered_map<int, int>>(nums.size());
        return dfs(nums, 0, S);
    }
};

Solution 4. DP

// OJ: https://leetcode.com/problems/target-sum
// Author: github.com/lzl124631x
// Time: O(NS)
// Space: O(S)
// Ref: https://discuss.leetcode.com/topic/76243/java-15-ms-c-3-ms-o-ns-iterative-dp-solution-using-subset-sum-with-explanation
class Solution {
private:
  int subsetSum(vector<int> &nums, int S) {
    vector<int> dp(S + 1, 0);
    dp[0] = 1;
    for (int n : nums)
      for (int i = S; i >= n; --i) dp[i] += dp[i - n];
    return dp[S];
  }
public:
  int findTargetSumWays(vector<int>& nums, int S) {
    int sum = accumulate(nums.begin(), nums.end(), 0);
    return sum < S || (sum + S) % 2 ? 0 : subsetSum(nums, (sum + S) / 2);
  }
};