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Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

  1. Recursion with Memoization - Complex approach
  2. Dynamic Programming (0/1 Knapsack) - Complex approach
  3. 2D Dynamic Programming - Complex approach

Approach 1: Recursion with Memoization

The key insight for solving this problem is to use recursion with memoization to explore all possible combinations of '+' and '-' signs.

We can define a recursive function calculate(index, sum) that returns the number of ways to achieve the target sum starting from the given index with the current sum.

For each index, we have two choices: add the current number or subtract it. We recursively explore both options and sum up the results.

To avoid redundant calculations, we use memoization to store the results of subproblems.

Algorithm:

  1. Define a recursive function calculate(index, sum) that returns the number of ways to achieve the target sum starting from the given index with the current sum.
  2. Base case: If index == nums.length, return 1 if sum == target, otherwise return 0.
  3. Recursive case: Return calculate(index + 1, sum + nums[index]) + calculate(index + 1, sum - nums[index]).
  4. Use memoization to avoid redundant calculations by storing the results of subproblems in a hash map.
  5. Call calculate(0, 0) to start the recursion from the beginning with an initial sum of 0.

Time Complexity:

O(n * sum)

Where n is the length of the nums array and sum is the sum of all elements in the array. With memoization, each subproblem is computed only once, and there are n * sum subproblems.

Space Complexity:

O(n * sum)

We use a hash map to store the results of subproblems, and there are n * sum subproblems.

Approach 2: Dynamic Programming (0/1 Knapsack)

We can transform this problem into a subset sum problem by observing that we're essentially partitioning the array into two subsets: one with '+' signs and one with '-' signs.

If we denote the sum of the '+' subset as P and the sum of the '-' subset as N, then we have:

  • P + N = sum(nums) (the sum of all elements)
  • P - N = target (the target sum)

Solving these equations, we get:

  • 2P = sum(nums) + target
  • P = (sum(nums) + target) / 2

So, we need to find the number of subsets with sum P. This is a classic 0/1 knapsack problem that can be solved using dynamic programming.

Note that P must be a non-negative integer for this approach to work. If sum(nums) + target is odd or target > sum(nums), there are no valid subsets.

Algorithm:

  1. Calculate the sum of all elements in the nums array.
  2. Calculate P = (sum + target) / 2.
  3. If P is not a non-negative integer (i.e., sum + target is odd or target > sum), return 0.
  4. Initialize a 1D DP array dp of size P + 1, where dp[j] represents the number of ways to form a sum of j.
  5. Set dp[0] = 1, as there is one way to form a sum of 0 (by not selecting any element).
  6. For each number num in nums:
  7. a. For j from P down to num:
  8. i. Update dp[j] += dp[j - num].
  9. Return dp[P] as the number of ways to form a sum of P.

Time Complexity:

O(n * sum)

Where n is the length of the nums array and sum is the sum of all elements in the array. We need to compute dp[j] for each j from 0 to P, and for each j, we consider all n elements.

Space Complexity:

O(sum)

We use a 1D DP array of size P + 1, which is proportional to the sum of all elements in the array.

Approach 3: 2D Dynamic Programming

Another approach is to use a 2D dynamic programming array to directly solve the original problem without transforming it.

Let's define dp[i][j] as the number of ways to achieve a sum of j using the first i elements of the array.

For each element nums[i-1], we have two choices: add it or subtract it. So, the recurrence relation is:

dp[i][j] = dp[i-1][j-nums[i-1]] + dp[i-1][j+nums[i-1]]

However, since the sum can be negative, we need to shift the indices to ensure they are non-negative. We can use an offset equal to the sum of all elements in the array.

Algorithm:

  1. Calculate the sum of all elements in the nums array.
  2. Initialize a 2D DP array dp of size (n+1) x (2*sum+1), where dp[i][j] represents the number of ways to achieve a sum of j-sum using the first i elements.
  3. Set dp[0][sum] = 1, as there is one way to form a sum of 0 using 0 elements.
  4. For i from 1 to n:
  5. a. For j from 0 to 2*sum:
  6. i. If j - nums[i-1] >= 0, add dp[i-1][j-nums[i-1]] to dp[i][j].
  7. ii. If j + nums[i-1] <= 2*sum, add dp[i-1][j+nums[i-1]] to dp[i][j].
  8. Return dp[n][sum+target] as the number of ways to achieve the target sum.

Time Complexity:

O(n * sum)

Where n is the length of the nums array and sum is the sum of all elements in the array. We need to compute dp[i][j] for each i from 0 to n and each j from 0 to 2*sum.

Space Complexity:

O(n * sum)

We use a 2D DP array of size (n+1) x (2*sum+1), which is proportional to n * sum.

Pseudocode

solution.pseudo
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function findTargetSumWays(nums, target):
    memo = new HashMap<String, Integer>()
 
    function calculate(index, sum):
        if index == nums.length:
            if sum == target:
                return 1
            else:
                return 0
 
        key = index + "," + sum
        if key in memo:
            return memo[key]
 
        // Add the current number
        add = calculate(index + 1, sum + nums[index])
 
        // Subtract the current number
        subtract = calculate(index + 1, sum - nums[index])
 
        memo[key] = add + subtract
        return memo[key]
 
    return calculate(0, 0)
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