494. Target Sum

Approach 1: Brute Force

Algorithm

The brute force approach is based on recursion. We need to try to put both the + and - symbols at every location in the given $nums$ array and find out the assignments which lead to the required result $S$.

For this, we make use of a recursive function calculate(nums, i, sum, S), which returns the assignments leading to the sum $S$, starting from the $i^{th}$ index onwards, provided the sum of elements up to the $i^{th}$ element is $sum$. This function appends a + sign and a - sign both to the element at the current index and calls itself with the updated $sum$ as $sum + nums[i]$ and $sum - nums[i]$ respectively along with the updated current index as $i+1$. Whenever we reach the end of the array, we compare the sum obtained with $S$. If they are equal, we increment the $count$ value to be returned.

Thus, the function call calculate(nums, 0, 0, S) returns the required number of assignments.

Implementation

public class Solution {
    int count = 0;
    
    public int findTargetSumWays(int[] nums, int S) {
        calculate(nums, 0, 0, S);
        return count;
    }
    
    public void calculate(int[] nums, int i, int sum, int S) {
        if (i == nums.length) {
            if (sum == S) {
                count++;
            }
        } else {
            calculate(nums, i + 1, sum + nums[i], S);
            calculate(nums, i + 1, sum - nums[i], S);
        }
    }
}

Complexity Analysis

• Time complexity: $O(2^n)$. Size of recursion tree will be $2^n$. $n$ refers to the size of $nums$ array.

• Space complexity: $O(n)$. The depth of the recursion tree can go up to $n$.
  
  

Approach 2: Recursion with Memoization

Algorithm

In the last approach, we can observe that a lot of redundant function calls were made with the same value of $i$ as the current index and the same value of $sum$ as the current sum, since the same values could be obtained through multiple paths in the recursion tree. In order to remove this redundancy, we make use of memoization as well to store the results which have been calculated earlier.

Thus, for every call to calculate(nums, i, sum, S), we store the result obtained in $memo[i][sum + total]$, where total stands for the sum of all the elements from the input array. The factor of total has been added as an offset to the $sum$ value to map all the $sum$s possible to positive integer range. By making use of memoization, we can get the result of each redundant function call in constant time.

Implementation

public class Solution {
    int total;
    
    public int findTargetSumWays(int[] nums, int S) {
        total = Arrays.stream(nums).sum();
        
        int[][] memo = new int[nums.length][2 * total + 1];
        for (int[] row : memo) {
            Arrays.fill(row, Integer.MIN_VALUE);
        }
        return calculate(nums, 0, 0, S, memo);
    }
    
    public int calculate(int[] nums, int i, int sum, int S, int[][] memo) {
        if (i == nums.length) {
            if (sum == S) {
                return 1;
            } else {
                return 0;
            }
        } else {
            if (memo[i][sum + total] != Integer.MIN_VALUE) {
                return memo[i][sum + total];
            }
            int add = calculate(nums, i + 1, sum + nums[i], S, memo);
            int subtract = calculate(nums, i + 1, sum - nums[i], S, memo);
            memo[i][sum + total] = add + subtract;
            return memo[i][sum + total];
        }
    }
}

Complexity Analysis

• Time complexity: $O(t \cdot n)$. The memo array of size $O(t \cdot n)$ has been filled just once. Here, $t$ refers to the sum of the $nums$ array and $n$ refers to the length of the $nums$ array.

• Space complexity: $O(t \cdot n)$. The depth of recursion tree can go up to $n$. The memo array contains $t \cdot n$ elements.
  
  

Approach 3: 2D Dynamic Programming

Algorithm

The idea behind this approach is as follows. Suppose we can find out the number of times a particular sum, say $sum_i$ is possible up to a particular index, say $i$, in the given $nums$ array, which is given by say $count_i$. Now, we can find out the number of times the sum $sum_i + nums[i]$ can occur easily as $count_i$. Similarly, the number of times the sum $sum_i - nums[i]$ occurs is also given by $count_i$.

Thus, if we know all the sums $sum_j$'s which are possible up to the $j^{th}$ index by using various assignments, along with the corresponding count of assignments, $count_j$, leading to the same sum, we can determine all the sums possible up to the $(j+1)^{th}$ index along with the corresponding count of assignments leading to the new sums.

Based on this idea, we make use of a $dp$ to determine the number of assignments which can lead to the given sum. $dp[i][j]$ refers to the number of assignments which can lead to a sum of $j$ up to the $i^{th}$ index. To determine the number of assignments which can lead to a sum of $sum + nums[i]$ up to the $(i+1)^{th}$ index, we can use $dp[i][sum + nums[i]] = dp[i][sum + nums[i]] + dp[i-1][sum]$. Similarly, $dp[i][sum - nums[i]] = dp[i][sum - nums[i]] + dp[i-1][sum]$. We iterate over the $dp$ array in a row-wise fashion, i.e., firstly, we obtain all the sums which are possible up to a particular index along with the corresponding count of assignments and then proceed for the next element (index) in the $nums$ array.

But, since the $sum$ can range from -total to total, where total equals to the sum of the nums array, we need to add an offset of total to the sum indices (column number) to map all the sums obtained to positive range only.

At the end, the value of $dp[n-1][S+total]$ gives us the required number of assignments. Here, $n$ refers to the number of elements in the $nums$ array.

Implementation

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

Complexity Analysis

• Time complexity: $O(t \cdot n)$. The dp array of size $O(t \cdot n)$ has been filled just once. Here, $t$ refers to the sum of the $nums$ array and $n$ refers to the length of the $nums$ array.

• Space complexity: $O(t \cdot n)$. dp array of size $t \cdot n$ is used.
  
  

Approach 4: 1D Dynamic Programming

Algorithm

If we look closely at the last solution, we can observe that to evaluate the current row of $dp$, only the values of the last row of $dp$ are needed. Thus, we can save some space by using a 1D DP array instead of a 2D DP array. The only change we need to make is that we have to create an array $next$ of the same size as $dp$ so that we can update it while scanning through $dp$ since it is not safe to mutate $dp$ when the iteration is in progress. After the iteration is completed, we set $dp$ equal to $next$ and create a new empty array $next$ before the next iteration starts, and so on.

Implementation

public class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        int total = Arrays.stream(nums).sum();
        int[] dp = new int[2 * total + 1];
        dp[nums[0] + total] = 1;
        dp[-nums[0] + total] += 1;
        
        for (int i = 1; i < nums.length; i++) {
            int[] next = new int[2 * total + 1];
            for (int sum = -total; sum <= total; sum++) {
                if (dp[sum + total] > 0) {
                    next[sum + nums[i] + total] += dp[sum + total];
                    next[sum - nums[i] + total] += dp[sum + total];
                }
            }
            dp = next;
        }
        
        return Math.abs(S) > total ? 0 : dp[S + total];
    }
}

Complexity Analysis

• Time complexity: $O(t \cdot n)$. Each of the $n$ dp arrays of size $t$ has been filled just once. Here, $t$ refers to the sum of the $nums$ array and $n$ refers to the length of the $nums$ array.

• Space complexity: $O(t)$. Two dp arrays of size $2 \cdot t + 1$ are used, therefore the space usage is $O(t)$.