[LeetCode] 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:
  1. The length of the given array is positive and will not exceed 20.
  2. The sum of elements in the given array will not exceed 1000.
  3. Your output answer is guaranteed to be fitted in a 32-bit integer.
Thought process:
DFS exhaust all possible combinations.

Solution 1 (recursion):
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public class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        return dfs(nums, S, 0, 0);
    }
    
    private int dfs(int[] nums, int S, int sum, int index) {
        if (index == nums.length) {
            return S == sum ? 1 : 0;
        }
        
        int plus = dfs(nums, S, sum + nums[index], index + 1);
        int minus = dfs(nums, S, sum - nums[index], index + 1);
        return plus + minus;
    }
}

Time complexity: O(2^n).

 The above solution makes many redundant calls to the same dfs(index, sum). I can use a 2D array to memoize the results and query them before making recursive calls. Since the sum of elements will not exceed 1000, the range of sums is [-1000, 1000]. So there are 2001 numbers in total.

 Solution 2 (recursion with memoization);
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class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        int[][] memo = new int[nums.length][2001];
        for (int i = 0; i < memo.length; i++) {
            Arrays.fill(memo[i], Integer.MIN_VALUE);
        }
        return findTargetSumWays(nums, S, 0, 0, memo);
    }
    
    private int findTargetSumWays(int[] nums, int S, int index, int sum, int[][] memo) {
        if (index == nums.length) {
            if (sum == S) {
                return 1;
            }
            return 0;
        }
        
        int column = sum + 1000;
        if (memo[index][column] != Integer.MIN_VALUE) {
            return memo[index][column];
        }
        int plus = findTargetSumWays(nums, S, index + 1, sum + nums[index], memo);
        int minus = findTargetSumWays(nums, S, index + 1, sum - nums[index], memo);
        memo[index][column] = plus + minus;
        return plus + minus;
    }
}

Time complexity:
Say the range of sums is s. The overall time complexity is O(ns).

 There is also a dynamic programming solution.
  1. Sub-problem: find out how many ways to assign symbols to a sub-array of nums to make the sum equal to a (a < S).
  2. Function: f[i][j] = f[i - 1][j - nums[i]] + f[i - 1][j + nums[i]].
  3. Initialization: f[0][j] (where j = nums[j]) = 1.
  4. Answer: f[nums.length][S].

Solution 3 (DP):

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class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        if (nums.length == 0) {
            return 0;
        }
        
        int[] f = new int[2001];
        f[nums[0] + 1000] = 1;
        f[- nums[0] + 1000] += 1;
        
        for (int i = 1; i < nums.length; i++) {
            int[] next = new int[2001];
            for (int sum = -1000; sum <= 1000; sum++) {
                if (f[sum + 1000] > 0) {
                    next[sum + nums[i] + 1000] += f[sum + 1000];
                    next[sum - nums[i] + 1000] += f[sum + 1000];
                }
            }
            f = next;
        }
        
        return S >= -1000 && S <= 1000 ? f[S + 1000] : 0;
    }
}

Time complexity: O(ns).
Location: Philadelphia, PA, USA

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