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# Target Sum LeetCode Solution – Queslers

## Problem – Target Sum

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 `'+'` 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`

### Target Sum LeetCode Solution in Java

``````    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);
}
``````

### Target Sum LeetCode Solution in Python

``````    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]
``````

### Target Sum LeetCode Solution in C++

``````    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);
}
``````
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