Target Sum – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given an integer array nums
and an integer target
.
You want to build an expression out of nums by adding one of the symbols '+'
and '-'
before each integer in nums and then concatenate all the integers.
- For example, if
nums = [2, 1]
, you can add a'+'
before2
and a'-'
before1
and concatenate them to build the expression"+2-1"
.
Return the number of different expressions that you can build, which evaluates to target
.
Example 1:
Input: nums = [1,1,1,1,1], target = 3 Output: 5 Explanation: There are 5 ways to assign symbols to make the sum of nums be target 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 +1 + 1 + 1 + 1 - 1 = 3
Example 2:
Input: nums = [1], target = 1 Output: 1
Constraints:
1 <= nums.length <= 20
0 <= nums[i] <= 1000
0 <= sum(nums[i]) <= 1000
-1000 <= target <= 1000
C++ Target Sum LeetCode Solution
class Solution {
public:
int findTargetSumWays(vector<int>& nums, int s) {
int sum = accumulate(nums.begin(), nums.end(), 0);
return sum < s || (s + sum) & 1 ? 0 : subsetSum(nums, (s + sum) >> 1);
}
int subsetSum(vector<int>& nums, int s) {
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];
}
};
Java Target Sum LeetCode Solution
public int findTargetSumWays(int[] nums, int s) {
int sum = 0;
for (int n : nums)
sum += n;
return sum < s || (s + sum) % 2 > 0 ? 0 : subsetSum(nums, (s + sum) >>> 1);
}
public int subsetSum(int[] nums, int s) {
int[] dp = new int[s + 1];
dp[0] = 1;
for (int n : nums)
for (int i = s; i >= n; i--)
dp[i] += dp[i - n];
return dp[s];
}
Python 3 Target Sum LeetCode Solution
def findTargetSumWays(self, A, S):
count = collections.Counter({0: 1})
for x in A:
step = collections.Counter()
for y in count:
step[y + x] += count[y]
step[y - x] += count[y]
count = step
return count[S]
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