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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0494_target_sum;

// #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Backtracking
// #Big_O_Time_O(n*(sum+s))_Space_O(n*(sum+s))
// #2022_07_21_Time_9_ms_(79.99%)_Space_45.2_MB_(32.79%)

/**
 * 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`
**/
public class Solution {
    public int findTargetSumWays(int[] nums, int s) {
        int sum = 0;
        s = Math.abs(s);
        for (int num : 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][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];
    }
}