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494. 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

Solution:

class Solution {
    int count = 0;
    public int findTargetSumWays(int[] nums, int target) {
        int index = 0;
        int sum = 0;
        helper(nums, index, sum, target);
        return count;

    }

    private void helper(int[] nums, int index, int sum, int target){
        if (index == nums.length){
            if (sum == target){
                count++;
            }
        }else{
            helper(nums, index + 1, sum + nums[index], target);
            helper(nums, index + 1, sum - nums[index], target);
        }
    }
}



// TC: O(branch^level) = O(2^n)
// SC: O(n)

/*

                    1 1 1 1 1

index i           1 
               +/      \-
              1+1      1-1
              /\ 


index 

index: means current nums[index]


*/

public class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        int total = Arrays.stream(nums).sum();
        int[][] dp = new int[nums.length][2 * total + 1];
        dp[0][nums[0] + total] = 1;
        dp[0][-nums[0] + total] += 1;

        for (int i = 1; i < nums.length; i++) {
            for (int sum = -total; sum <= total; sum++) {
                if (dp[i - 1][sum + total] > 0) {
                    dp[i][sum + nums[i] + total] += dp[i - 1][sum + total];
                    dp[i][sum - nums[i] + total] += dp[i - 1][sum + total];
                }
            }
        }

        return Math.abs(S) > total ? 0 : dp[nums.length - 1][S + total];
    }
}