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);
    }  

Target Sum LeetCode Solution Review:

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

Find on LeetCode

Conclusion:

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 >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions