494. Target Sum

Question

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

Method 1: dfs

class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        if(nums == null || nums.length == 0) return 0;
        return dfs(nums, S, 0, 0, 0);
    }
    private int dfs(int[] nums, int S, int index, int cur, int start){
        if(index == nums.length){
            return cur == S ? 1 : 0;
        }else{
            return dfs(nums, S, start + 1, cur + nums[start], start + 1)
                    + dfs(nums, S, start + 1, cur - nums[start], start + 1);
        }
    }
}

Method 2: dp

class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        int sum = 0;
        for(int num : nums) sum += num;
        if(sum < S) return 0;
        int offset = sum;
        int[][] dp = new int[1 + nums.length][sum + offset + 1];
        dp[0][offset] = 1;
        for(int i = 1; i <= nums.length; i++){
            for(int j = 0; j < sum * 2 + 1; j++){
                dp[i][j] += (j - nums[i - 1] >= 0 ? dp[i - 1][j - nums[i - 1]]: 0) + (j + nums[i - 1] < 2 * sum + 1 ? dp[i - 1][j + nums[i - 1]]: 0);
            }
        }
        return dp[nums.length][S + offset];
    }
}

Second Time

Method 1: dp

 class Solution {
 	public int findTargetSumWays(int[] nums, int S) {
 		int sum = 0;
 		for(int num : nums)
 			sum += num;
 		if(sum < S || -sum > S) return 0;
 		int offset = sum;
 		int[][] dp = new int[nums.length + 1][sum * 2 + 1];
 		dp[0][offset] = 1;
 		for(int i = 1; i <= nums.length; i++){
 			for(int j = 0; j < sum * 2 + 1; j++){
 				dp[i][j] = (j - nums[i - 1] >= 0 ? dp[i - 1][j - nums[i - 1]]: 0)
 					+ (j + nums[i - 1] < sum * 2 + 1 ? dp[i - 1][j + nums[i - 1]]: 0);
 			}
 		}
 		return dp[nums.length][S + offset];
 	}
 }

Reference

花花酱 LeetCode 494. Target Sum

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