[Dynamic Programming] Target Sum - Medium

The Target Sum problem is a well-known challenge in dynamic programming. We get an array of integers and a target sum. Our goal is to find out if we can put a ‘+’ or ‘-’ sign in front of each integer so that the total equals the target sum. This problem is interesting because we can solve it well using dynamic programming. This method helps us break the big problem into smaller, easier problems.

In this article, we will look into the Target Sum problem. We will talk about what it is, the problem itself, and how we can use dynamic programming to solve it. We will also show Java, Python, and C++ code examples. We can also explore ways to make our solutions use less space. We will compare different methods too. Plus, we will point out common mistakes people make when they implement this and answer some frequently asked questions about the Target Sum problem.

  • Dynamic Programming Target Sum Problem Overview
  • Understanding the Problem Statement
  • Dynamic Programming Approach to Target Sum
  • Java Implementation of Target Sum with Dynamic Programming
  • Python Solution for Target Sum Using Dynamic Programming
  • C++ Code Example for Target Sum Problem
  • Optimizing Space Complexity in Target Sum Solutions
  • Comparative Analysis of Different Approaches to Target Sum
  • Common Mistakes in Implementing Target Sum
  • Frequently Asked Questions

Understanding the Problem Statement

The Target Sum problem is a well-known dynamic programming challenge. It asks us to find how many ways we can put plus and minus signs in front of a set of numbers to reach a specific target sum.

Problem Definition

We have an array of integers called nums and another integer called target. Our job is to find out how many different ways we can add and subtract the numbers in nums so that the total equals the target.

Constraints

  • The length of nums can be from 1 to 20.
  • The target can be any integer from -1000 to 1000.
  • Each number in nums can also be between -1000 and 1000.

Example

For example, if we have this input:

  • nums = [1, 1, 1, 1, 1]
  • target = 3

The output should be 5. This is because there are five different ways to put the signs to get the target sum:

  • +1 + 1 + 1 - 1 - 1
  • +1 + 1 - 1 + 1 - 1
  • +1 - 1 + 1 + 1 - 1
  • -1 + 1 + 1 + 1 - 1
  • +1 + 1 - 1 - 1 + 1

Mathematical Formulation

Let’s say we call the sum of the numbers with a positive sign P and the sum of the numbers with a negative sign N. We can write this relationship:

[ P - N = ]

Also, we know that:

[ P + N = ]

From these, we can find:

[ P = ]

This means we need to find groups of numbers in nums that add up to ( P ). We can change this problem into a subset sum problem. This is a common example of using dynamic programming.

By understanding this, we can start to create a dynamic programming solution. This will help us find the number of ways to reach the target sum efficiently.

Dynamic Programming Approach to Target Sum

The Target Sum problem is a well-known problem in dynamic programming. We have an array of integers and a target sum. Our task is to find out if we can put ‘+’ or ‘-’ signs in front of each number in the array. We want the total to equal the target sum.

Problem Breakdown

We have an array called nums and a number called target. We can find a subset of nums that gives us a sum of (total_sum - target) / 2. If we can find this subset, the other numbers will balance to reach the target sum.

Dynamic Programming Approach

  1. Identify the Total Sum: First we need to find the total sum of the array. If (total_sum - target) is odd or negative, we should return false. It means we cannot divide it into two valid subsets.

  2. Subset Sum DP Table: We will use a boolean DP table. Here, dp[i][j] shows if we can form a subset with sum j using the first i elements from the array.

  3. Initialization: We set dp[0][0] = true. This means we can always get a sum of 0 with 0 elements. For every j, we set dp[0][j] = false. We cannot create a positive sum with 0 elements.

  4. Filling the DP Table: We will go through each element and each possible sum. We will update the table to decide if we should include the current element.

Java Implementation

public boolean canPartition(int[] nums, int target) {
    int totalSum = 0;
    for (int num : nums) totalSum += num;
    if (totalSum < target || (totalSum - target) % 2 != 0) return false;

    int subsetSum = (totalSum - target) / 2;
    boolean[] dp = new boolean[subsetSum + 1];
    dp[0] = true;

    for (int num : nums) {
        for (int j = subsetSum; j >= num; j--) {
            dp[j] = dp[j] || dp[j - num];
        }
    }
    return dp[subsetSum];
}

Python Implementation

def canPartition(nums, target):
    total_sum = sum(nums)
    if total_sum < target or (total_sum - target) % 2 != 0:
        return False

    subset_sum = (total_sum - target) // 2
    dp = [False] * (subset_sum + 1)
    dp[0] = True

    for num in nums:
        for j in range(subset_sum, num - 1, -1):
            dp[j] = dp[j] or dp[j - num]

    return dp[subset_sum]

C++ Implementation

bool canPartition(vector<int>& nums, int target) {
    int totalSum = accumulate(nums.begin(), nums.end(), 0);
    if (totalSum < target || (totalSum - target) % 2 != 0) return false;

    int subsetSum = (totalSum - target) / 2;
    vector<bool> dp(subsetSum + 1, false);
    dp[0] = true;

    for (int num : nums) {
        for (int j = subsetSum; j >= num; j--) {
            dp[j] = dp[j] || dp[j - num];
        }
    }
    return dp[subsetSum];
}

This dynamic programming method solves the Target Sum problem well. It uses the idea of subset sum. This way, we can save time and space while solving the problem.

Java Implementation of Target Sum with Dynamic Programming

In this section, we show a Java way to solve the Target Sum problem using dynamic programming. The problem is simple. We have an array of numbers and a target sum. We need to check if we can put ‘+’ or ‘-’ signs in front of the numbers to make their sum equal to the target.

Java Code Implementation

import java.util.Arrays;

public class TargetSum {
    public boolean canPartition(int[] nums, int target) {
        int sum = Arrays.stream(nums).sum();
        if (sum < target || (sum - target) % 2 != 0) {
            return false;
        }
        int subsetSum = (sum - target) / 2;
        return isSubsetSum(nums, subsetSum);
    }

    private boolean isSubsetSum(int[] nums, int subsetSum) {
        boolean[] dp = new boolean[subsetSum + 1];
        dp[0] = true;

        for (int num : nums) {
            for (int j = subsetSum; j >= num; j--) {
                dp[j] = dp[j] || dp[j - num];
            }
        }
        return dp[subsetSum];
    }

    public static void main(String[] args) {
        TargetSum ts = new TargetSum();
        int[] nums = {1, 1, 2, 3};
        int target = 1;
        System.out.println(ts.canPartition(nums, target));
    }
}

Explanation of the Code

  • canPartition Method: This method checks if we can reach the target by splitting the array into two parts.
    • First, it calculates the total of all numbers in the array.
    • If the total is less than the target or if the difference between total and target is odd, it returns false.
    • It finds the subset sum we need with the formula (sum - target) / 2.
  • isSubsetSum Method: This method checks if there is a part of the array that adds up to the subsetSum.
    • It uses a boolean array dp. Here, dp[j] tells if we can reach a sum of j.
    • It goes through each number in the nums array. It updates the dp array backward so it doesn’t change values we still need.
  • Main Method: This is where the program starts. Here, we call the canPartition method with a sample array and a target.

Properties

  • Time Complexity: O(N * S), where N is the number of elements in the array and S is the target subset sum.
  • Space Complexity: O(S), for the boolean array used in dynamic programming.

This Java code uses dynamic programming ideas to solve the Target Sum problem. It gives a clear and simple solution. If you want to learn more, you can look at related dynamic programming problems like Partition Equal Subset Sum. This helps to understand the ideas better.

Python Solution for Target Sum Using Dynamic Programming

We solve the Target Sum problem by using Dynamic Programming in Python. First, we explain the problem. We have an array of integers and a target sum. Our task is to find how many ways we can put ‘+’ or ‘-’ signs in front of each integer so that their total equals the target.

Problem Breakdown

  1. We can see this problem as a subset sum problem. We need to find subsets that can make a certain sum.
  2. Let S1 be the sum of numbers with a ‘+’ sign and S2 be the sum of numbers with a ‘-’ sign. We want S1 - S2 = target. We can change this to S1 = (total_sum + target) / 2. Here, total_sum is the sum of all numbers in the array.

Dynamic Programming Approach

  1. First, we calculate the total_sum of the array.
  2. Then, we find target_sum as (total_sum + target) / 2.
  3. Now, we use dynamic programming to count how many ways we can reach target_sum with the numbers in the array.

Python Code Implementation

def findTargetSumWays(nums, target):
    total_sum = sum(nums)
    if (total_sum + target) % 2 != 0 or total_sum < target:
        return 0
    
    target_sum = (total_sum + target) // 2
    dp = [0] * (target_sum + 1)
    dp[0] = 1  # There is one way to get a sum of 0: use no elements.
    
    for num in nums:
        for j in range(target_sum, num - 1, -1):
            dp[j] += dp[j - num]
    
    return dp[target_sum]

# Example usage
nums = [1, 1, 1, 1, 1]
target = 3
print(findTargetSumWays(nums, target))  # Output: 5

Explanation of the Code

  • We start with a list dp. Here, dp[i] means how many ways we can get the sum i.
  • We go through each number in nums. We update the dp list in reverse. This is to avoid changing values that we still need to use.
  • At the end, we return dp[target_sum]. This shows the total number of ways to reach the target sum.

This method works well to find the number of ways to reach the target sum. It uses dynamic programming with a time complexity of O(n * target_sum) and a space complexity of O(target_sum). If you want to learn more about dynamic programming, you can read about the Coin Change problem.

C++ Code Example for Target Sum Problem

The Target Sum problem is about finding how many ways we can put symbols to make the sum equal to a target value. We usually solve this using dynamic programming. Here is a C++ code that shows the dynamic programming method for the Target Sum problem.

C++ Implementation

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int num : nums) sum += num;
        
        if (target > sum || (sum + target) % 2 != 0) return 0;

        int s = (sum + target) / 2;
        vector<int> dp(s + 1, 0);
        dp[0] = 1;

        for (int num : nums) {
            for (int j = s; j >= num; j--) {
                dp[j] += dp[j - num];
            }
        }

        return dp[s];
    }
};

int main() {
    Solution solution;
    vector<int> nums = {1, 1, 1, 1, 1};
    int target = 3;
    cout << "Number of ways to achieve target sum: " << solution.findTargetSumWays(nums, target) << endl;
    return 0;
}

Explanation

  • Input: We have a vector of integers called nums and an integer named target.
  • Output: We want to find the number of ways to assign + or - to each number in nums so that their sum equals target.
  • Dynamic Programming Array: The DP array dp holds the number of ways to get sums from 0 to s. Here s is half of the needed adjusted sum.
  • Time Complexity: O(n * s). Here n is how many elements are in nums and s is the subset sum.
  • Space Complexity: O(s) because we use the DP array.

This code gives a good solution for the Target Sum problem using ideas from dynamic programming. If you want to learn more about dynamic programming, you can look at problems like the Coin Change problem.

Optimizing Space Complexity in Target Sum Solutions

When we solve the Target Sum problem with dynamic programming, space complexity is important. This is true, especially when we have large datasets. Usually, we create a 2D array to hold our results. This can use a lot of space. But we can make space complexity better. We can reduce it to O(S), where S is the target sum. We can do this by using a 1D array instead of a 2D one.

Space-Optimized Dynamic Programming Approach

  1. Use a 1D Array: We can use one array to save results for the current target sum. This makes space complexity much lower.

  2. Iterate Backwards: When we update the array, we should go backwards. This way, we do not use the same element more than once when we calculate.

Example Implementation in Java

public class TargetSum {
    public int findTargetSumWays(int[] nums, int target) {
        int sum = 0;
        for (int num : nums) sum += num;

        if (target > sum || (sum + target) % 2 != 0) return 0;

        int newTarget = (sum + target) / 2;
        int[] dp = new int[newTarget + 1];
        dp[0] = 1;

        for (int num : nums) {
            for (int j = newTarget; j >= num; j--) {
                dp[j] += dp[j - num];
            }
        }
        return dp[newTarget];
    }
}

Example Implementation in Python

def findTargetSumWays(nums, target):
    total_sum = sum(nums)
    
    if target > total_sum or (total_sum + target) % 2 != 0:
        return 0

    new_target = (total_sum + target) // 2
    dp = [0] * (new_target + 1)
    dp[0] = 1

    for num in nums:
        for j in range(new_target, num - 1, -1):
            dp[j] += dp[j - num]
    
    return dp[new_target]

Example Implementation in C++

#include <vector>
using namespace std;

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int num : nums) sum += num;

        if (target > sum || (sum + target) % 2 != 0) return 0;

        int newTarget = (sum + target) / 2;
        vector<int> dp(newTarget + 1, 0);
        dp[0] = 1;

        for (int num : nums) {
            for (int j = newTarget; j >= num; j--) {
                dp[j] += dp[j - num];
            }
        }
        return dp[newTarget];
    }
};

Key Takeaways

  • By using a 1D array and going backwards, we can lower the space complexity of the Target Sum problem from O(N * S) to O(S).
  • This method keeps the efficiency of dynamic programming while using less memory.

For more on dynamic programming optimizations, check out Dynamic Programming Coin Change Problem.

Comparative Analysis of Different Approaches to Target Sum

The Target Sum problem can have different methods. Each method has its own pros and cons in terms of complexity and efficiency. Here we will compare the most common approaches.

1. Recursive Approach

  • Time Complexity: O(2^n)
  • Space Complexity: O(n) (due to call stack)
  • Description: The simple recursive method checks all combinations of adding or subtracting each number to find the target sum.
public int findTargetSumWays(int[] nums, int S) {
    return findWays(nums, 0, S);
}

private int findWays(int[] nums, int index, int S) {
    if (index == nums.length) return S == 0 ? 1 : 0;
    return findWays(nums, index + 1, S + nums[index]) +
           findWays(nums, index + 1, S - nums[index]);
}

2. Dynamic Programming (DP) Approach

  • Time Complexity: O(n * sum)
  • Space Complexity: O(sum) (1D DP array)
  • Description: This method uses a DP table to keep track of how many ways we can reach each possible sum. It builds from smaller problems.
def findTargetSumWays(nums, S):
    total = sum(nums)
    if total < S or (total + S) % 2 == 1: return 0
    target = (total + S) // 2
    dp = [0] * (target + 1)
    dp[0] = 1

    for num in nums:
        for i in range(target, num - 1, -1):
            dp[i] += dp[i - num]
    return dp[target]

3. Memoization

  • Time Complexity: O(n * sum)
  • Space Complexity: O(n * sum) (for the memoization table)
  • Description: This is a top-down approach like the recursive method but it uses a cache to store results we already computed.
class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        unordered_map<string, int> memo;
        return dfs(nums, 0, S, memo);
    }

    int dfs(vector<int>& nums, int index, int S, unordered_map<string, int>& memo) {
        if (index == nums.size()) return S == 0 ? 1 : 0;
        string key = to_string(index) + "," + to_string(S);
        if (memo.find(key) != memo.end()) return memo[key];
        memo[key] = dfs(nums, index + 1, S + nums[index], memo) +
                     dfs(nums, index + 1, S - nums[index], memo);
        return memo[key];
    }
};

4. Iterative DP with Space Optimization

  • Time Complexity: O(n * sum)
  • Space Complexity: O(sum) (optimized space)
  • Description: This is like the DP approach but uses one array to keep track of combinations. We update it in reverse order to avoid overwriting.
def findTargetSumWays(nums, S):
    total = sum(nums)
    if total < S or (total + S) % 2 == 1: return 0
    target = (total + S) // 2
    dp = [0] * (target + 1)
    dp[0] = 1

    for num in nums:
        for i in range(target, num - 1, -1):
            dp[i] += dp[i - num]
    return dp[target]

Summary of Comparisons

  • Recursive: It is simple but not good for big inputs; it has high time complexity.
  • Dynamic Programming: It works better; it is good for larger inputs with reasonable sum limits.
  • Memoization: It mixes recursive style with better efficiency; it is good to avoid deep recursion limits.
  • Iterative DP: It uses less space; best when we have memory limits.

Each method has its strengths. The choice of which one to use depends on the problem’s limits, like the input size and the target sum. For more insights into dynamic programming, we can look at related articles on topics like Dynamic Programming Coin Change or Dynamic Programming Longest Increasing Subsequence.

Common Mistakes in Implementing Target Sum

When we implement the Target Sum problem with dynamic programming, we can make some common mistakes. These mistakes can give us wrong answers or slow solutions. Here are some mistakes we often see:

  1. Incorrect Base Case Handling:
    • We must define the base cases for dynamic programming correctly. If we set up the DP array wrong, the results will be wrong too. Usually, the base case shows that we can reach a sum of 0 with an empty subset.
    dp[0] = true; // We can always get a sum of 0
  2. Not Considering All Possible Sums:
    • If we do not check all possible sums when updating the DP array, we can miss some solutions. We should always make sure that the DP loop checks all sums from the current target down to the current number’s value.
    for num in nums:
    for j in range(target, num - 1, -1):
        dp[j] = dp[j] or dp[j - num]
  3. Mismanagement of Space Complexity:
    • If we use a 2D array for a problem that can be solved with 1D, we waste space. We can save space by using one array and going backwards.
    vector<bool> dp(target + 1, false);
  4. Ignoring Edge Cases:
    • If we do not handle edge cases like negative numbers or when we cannot reach the target sum, we can get runtime errors or wrong outputs. We should check these cases at the start.
  5. Overlapping Subproblems:
    • If we do not use memoization correctly in recursive solutions, we can do the same calculations many times. We need to store and reuse the results we already computed.
    memo = {}
def findTargetSumWays(i, target):
    if (i, target) in memo:
        return memo[(i, target)]
    # computation logic
    memo[(i, target)] = result
    return result
  6. Misunderstanding Input Constraints:
    • If we do not follow the problem’s rules, like limits on the input array size or the target sum value, our algorithm can fail or run slow.
  7. Incorrect DP Transition Logic:
    • Mistakes in the transition logic while updating the DP table can give us wrong results. We should check the logic that combines results from earlier states very carefully.
    for (int i = 0; i < nums.length; i++) {
    for (int j = sum; j >= nums[i]; j--) {
        dp[j] = dp[j] || dp[j - nums[i]];
    }
}

By knowing these common mistakes, we can write better and faster solutions for the Target Sum problem with dynamic programming. For more reading on related dynamic programming topics, we can check articles on Dynamic Programming - Coin Change and Dynamic Programming - Partition Equal Subset Sum.

Frequently Asked Questions

1. What is the Target Sum problem in dynamic programming?

The Target Sum problem is about finding ways to add plus or minus signs to a set of numbers. We want them to add up to a specific target. We can solve this problem well with dynamic programming. This method helps us avoid doing the same calculations again by keeping track of results we get. For more details on dynamic programming basics, check our Dynamic Programming Overview.

2. How can dynamic programming be applied to solve the Target Sum problem?

We can use dynamic programming for the Target Sum problem by making a two-dimensional array. This array will hold the results of smaller problems. We fill this array based on if each number helps us reach the target sum positively or negatively. This way is much faster than simple recursive solutions. So, we often choose this method for the problem.

3. What are the time and space complexities of the Target Sum dynamic programming solution?

The time complexity for the dynamic programming solution is O(n * sum). Here, n is the number of elements and sum is the target sum. The space complexity is also O(n * sum) if we use a two-dimensional array. But we can make it smaller to O(sum) by using a one-dimensional array. Many dynamic programming solutions show how to do this.

4. Can the Target Sum problem be solved using recursion?

Yes, we can also use recursion for the Target Sum problem. Each number can be added with a positive or negative sign. But this simple method can take too much time for big inputs. So, using dynamic programming is much better. It makes our solution faster and more efficient.

5. What are some common mistakes when implementing the Target Sum problem?

Some common mistakes in the Target Sum problem are not setting up the dynamic programming table correctly. People may not understand the base cases well or forget to consider negative sums. We need to pay close attention to these details. This helps us make a correct and efficient solution. For more similar problems, you can look at the Coin Change problem.