LeetCode Q 494 - Target Sum

You are given a list of non-negative integers, a1, a2, …, an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.
Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:
Input: nums is [1, 1, 1, 1, 1], S is 3. ; Output: 5
Explanation:
-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
There are 5 ways to assign symbols to make the sum of nums be target 3.
Note:

The length of the given array is positive and will not exceed 20.
The sum of elements in the given array will not exceed 1000.
Your output answer is guaranteed to be fitted in a 32-bit integer.

Solution :

Solution 1 : DFS / backtracking

Time Complexity: O(2^n)

Code:

private int res;
public int findTargetSumWays(int[] nums, int S) {
  if (nums == null || nums.length == 0) return 0;
  backtrack(nums, S, 0, 0);
  return res;
}

private void backtrack(int[] nums, int S, int index, int currSum) {
  if (index == nums.length) {
    if (currSum == S) res++;
    return;
  }

  //explore
  backtrack(nums, S, index + 1, currSum + nums[index]);
  backtrack(nums, S, index + 1, currSum - nums[index]);

  return;
}

Solution 2 : DP

dp[i][j] refers to the number of assignments which can lead to a sum of j upto the ith index.

State Transfer Function
dp[i][sum+nums[i]] = dp[i][sum+nums[i]]+dp[i-1][sum] dp[i][sum-nums[i]] = dp[i][sum-nums[i]]+dp[i-1][sum]

Note: we add an offset sum to the second index to avoid negative index value.

Time Complexity: O(kn)
Time Complexity: O(kn)

Code:

public int findTargetSumWays(int[] nums, int S) {
  if (nums == null || nums.length == 0) return 0;

  int sum = 0;
  for (int num: nums) sum += num;

  int[][] dp = new int[nums.length][2 * sum + 1];

  dp[0][sum + nums[i]] = 1;
  dp[0][sum - nums[i]] += 1; // = is not correct!

  for (int i = 1; i < nums.length; i++) {
    for (int s = 0; s < 2 * sum + 1; s++) {
      if (dp[i - 1][s] != 0) {
        int n = nums[i];
        dp[i][s + n] += dp[i - 1][s];
        dp[i][s - n] += dp[i - 1][s];
      }
    }
  }

  return dp[nums.length - 1][sum + S];
}

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