LeetCode in Net

494. Target Sum

Medium

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

using System;
using System.Collections.Generic;

public class Solution {
    public int FindTargetSumWays(int[] nums, int s) {
        int sum = 0;
        s = Math.Abs(s);
        foreach (int num in nums) {
            sum += num;
        }
        // Invalid s, just return 0
        if (s > sum || (sum + s) % 2 != 0) {
            return 0;
        }
        int[][] dp = new int[(sum + s) / 2 + 1][];
        for (int i = 0; i < dp.Length; i++) {
            dp[i] = new int[nums.Length + 1];
        }
        dp[0][0] = 1;
        // empty knapsack must be processed specially
        for (int i = 0; i < nums.Length; i++) {
            if (nums[i] == 0) {
                dp[0][i + 1] = dp[0][i] * 2;
            } else {
                dp[0][i + 1] = dp[0][i];
            }
        }
        for (int i = 1; i < dp.Length; i++) {
            for (int j = 0; j < nums.Length; j++) {
                dp[i][j + 1] += dp[i][j];
                if (nums[j] <= i) {
                    dp[i][j + 1] += dp[i - nums[j]][j];
                }
            }
        }
        return dp[(sum + s) / 2][nums.Length];
    }
}

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