Target Sum Leetcode Problem 494 [Python Solution]

In the Target Sum problem, you are given an integer array nums and an integer target.

The task is to build an expression out of the numbers in nums by adding either a ‘+’ or ‘-‘ symbol before each integer in nums and then concatenate all the integers.

The goal is to determine the number of different expressions that can be built, which evaluate to the given 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 equal to 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

Understanding the Constraints

Before we dive into the solution, let’s understand the constraints:

1. The input array nums can have at most 20 elements, making it a relatively small input size.
2. The values in the array nums range from 0 to 1000, and the sum of all elements in the array also ranges from 0 to 1000.
3. The target can have values between -1000 and 1000.

Target Sum LeetCode Problem Solution

Now, let’s discuss the Python solution for the Target Sum problem.

Bruteforce Approach

We can start by implementing a brute-force recursive solution.

This approach explores all possible combinations of adding or subtracting each number in the array to determine the total that equals the target.

The pseudocode for the brute-force approach is as follows:

def findWays(nums, target, index, total):
    # Base case
    if index == len(nums):
        return 1 if total == target else 0
    
    # Recursive case
    ways_add = findWays(nums, target, index + 1, total + nums[index])
    ways_subtract = findWays(nums, target, index + 1, total - nums[index])
    
    return ways_add + ways_subtract


result = findWays(nums, target, 0, 0)

This recursive approach explores all possible combinations, but it can be very slow for larger inputs due to its exponential time complexity.

An Efficient Approach with Caching

To optimize the solution, we can use dynamic programming and caching to avoid redundant calculations.

We’ll create a hash map (dp) to store the number of ways to reach a certain state (index, total) to prevent recalculating the same subproblem.

Here’s the efficient Python solution:

def findTargetSumWays(self, nums: List[int], target: int) -> int:
        # Create a cache to store computed results
        dp = {}

        def backtrack(index, total):
            # Base case
            if index == len(nums):
                return 1 if total == target else 0

            # Check if this state is already computed
            if (index, total) in dp:
                return dp[(index, total)]

            # Calculate the number of ways by adding and subtracting
            ways_add = backtrack(index + 1, total + nums[index])
            ways_subtract = backtrack(index + 1, total - nums[index])

            # Store the result in the cache
            dp[(index, total)] = ways_add + ways_subtract

            return dp[(index, total)]

        # Start the recursion with initial values
        return backtrack(0, 0)

This efficient approach utilizes caching to store previously computed subproblems, significantly reducing the time complexity.

Time and Space Complexity

Time Complexity:
The efficient approach using caching has a time complexity of O(n * t), where n is the length of the nums array, and t is the total sum of the array.

This is a substantial improvement over the exponential time complexity of the brute-force approach.

Space Complexity:
The space complexity of this solution is O(n * t) due to the storage of results in the dp cache.

Reasoning Behind Our Approach

The efficient approach optimizes the brute-force solution by using caching to avoid redundant calculations.

It uses dynamic programming to keep track of the number of ways to reach a particular state (index, total) and returns the total count of ways to reach the target value.

By storing and reusing previously computed results, the solution significantly improves the time complexity, making it suitable for larger input sizes.

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Conclusion

In this blog post, we’ve explored the Target Sum problem, a LeetCode problem with a medium difficulty level.

We discussed the problem’s overview, provided Python solutions for both a brute-force approach and an efficient approach using dynamic programming and caching.

The efficient approach offers a significant improvement in time complexity and is the recommended solution for solving the Target Sum problem.

It utilizes a hash map to store previously calculated results, avoiding redundant calculations and making it suitable for larger inputs.

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Link to the LeetCode problem

Happy coding!