Physical Address
304 North Cardinal St.
Dorchester Center, MA 02124
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.
nums = [2, 1]
, you can add a '+'
before 2
and a '-'
before 1
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
public int findTargetSumWays(int[] nums, int s) {
Map<Integer, Integer> dp = new HashMap();
dp.put(0, 1);
for (int num : nums) {
Map<Integer, Integer> dp2 = new HashMap();
for (int tempSum : dp.keySet()) {
int key1 = tempSum + num;
dp2.put(key1, dp2.getOrDefault(key1, 0) + dp.get(tempSum));
int key2 = tempSum - num;
dp2.put(key2, dp2.getOrDefault(key2, 0) + dp.get(tempSum));
}
dp = dp2;
}
return dp.getOrDefault(s, 0);
}
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]
int countSubsets(vector<int>& nums, int n, int M)
{
int t[n+1][M+1];
for(int i=0; i<=n; i++)
{
for(int j=0; j<=M; j++)
{
if(i==0)
t[i][j]=0;
if(j==0)
t[i][j]=1;
}
}
//t[0][0] = 1;
for(int i=1; i<=n; i++)
{
for(int j=0; j<=M; j++)
{
if(nums[i-1]<=j)
t[i][j]=t[i-1][j-nums[i-1]]+t[i-1][j];
else
t[i][j]=t[i-1][j];
}
}
return t[n][M];
}
int findTargetSumWays(vector<int>& nums, int target)
{
target=abs(target);
int n=nums.size();
int sum=0;
for(int i=0; i<n; i++)
sum+=nums[i];
int M=(sum+target)/2;
if(sum<target||(sum+target)%2!=0)
return 0;
return countSubsets(nums, n, M);
}
In our experience, we suggest you solve this Target Sum LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.
If you are stuck anywhere between any coding problem, just visit Queslers to get the Target Sum LeetCode Solution
I hope this Target Sum LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.
This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.
Keep Learning!
More Coding Solutions >>