What is the time complexity of Target Sum?

The optimal solution has a time complexity of O(n * S).

What data structures are used to solve Target Sum?

This problem uses Dynamic Programming, Backtracking. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Target Sum?

Insight 1: The core idea is to transform the problem into a subset sum problem. Instead of finding combinations of '+/-' signs, we can divide the numbers into two groups: P (positive) and N (negative). The goal is to find the number of ways to choose numbers for the P group such that sum(P) - sum(N) = target. Insight 2: We can further simplify the equation: sum(P) - sum(N) = target. Since sum(P) + sum(N) = sum(nums) = total, adding the two equations gives us 2 * sum(P) = target + total, or sum(P) = (target + total) / 2. Therefore, the problem is equivalent to finding the number of subsets whose sum equals (target + total) / 2.

Which companies ask LeetCode 494. Target Sum in interviews?

This problem is frequently asked at Myntra, Pinterest, ServiceNow.

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Target Sum - LeetCode 494 Solution

Target Sum - Complete Solution Guide

Target Sum is LeetCode problem 494, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 Exampl

Detailed Explanation

The problem asks us to find the number of ways to assign '+' or '-' signs to each number in a given integer array `nums` such that the sum of the resulting expression equals a given target value. For example, given `nums = [1, 1, 1, 1, 1]` and `target = 3`, we need to find how many combinations of plus and minus signs result in a sum of 3. The constraints state that the length of the array is at most 20, the elements are non-negative and less than or equal to 1000, and the absolute value of the target is at most 1000.

Solution Approach

The provided code uses a dynamic programming approach to solve the equivalent subset sum problem. First, it calculates the total sum of the numbers in the input array. It then checks if a solution is even possible by verifying two conditions: (1) if the absolute value of the target is greater than the total sum, there's no way to reach the target; and (2) if (target + total) is not even, (target + total) / 2 will not be an integer, so there's no solution. If both conditions are met, the code calculates the required subset sum (target + total) / 2 and uses a 1D dynamic programming array `dp` to find the number of subsets that sum up to this value.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the `nums` array.
Step 2: Check for invalid cases: if `abs(target) > total` or `(total + target)` is odd, return 0 because there are no possible solutions.
Step 3: Calculate the required subset sum: `subset = (total + target) / 2`.
Step 4: Initialize a 1D DP array `dp` of size `subset + 1` with all elements set to 0, except `dp[0] = 1` (representing the base case where an empty subset has a sum of 0).
Step 5: Iterate through the `nums` array. For each number `num`, iterate backward through the `dp` array from `subset` down to `num`. Update `dp[s]` by adding `dp[s - num]` to it. This represents the number of ways to form the sum `s` either without including `num` (which is `dp[s]`) or by including `num` (which is `dp[s - num]`). Iterating backwards ensures that we are considering each number only once when building subsets.
Step 6: After iterating through all the numbers, `dp[subset]` will contain the number of subsets that sum up to the calculated `subset` value. Return `dp[subset]`.

Key Insights

Insight 1: The core idea is to transform the problem into a subset sum problem. Instead of finding combinations of '+/-' signs, we can divide the numbers into two groups: P (positive) and N (negative). The goal is to find the number of ways to choose numbers for the P group such that sum(P) - sum(N) = target.
Insight 2: We can further simplify the equation: sum(P) - sum(N) = target. Since sum(P) + sum(N) = sum(nums) = total, adding the two equations gives us 2 * sum(P) = target + total, or sum(P) = (target + total) / 2. Therefore, the problem is equivalent to finding the number of subsets whose sum equals (target + total) / 2.
Insight 3: Dynamic programming is suitable for solving the subset sum problem. We can use a 1D DP array to store the number of ways to achieve each possible sum from 0 up to (target + total) / 2. Iterate through the nums array, and for each number, update the DP array by considering the possibility of including that number in the subset.

Complexity Analysis

Time Complexity: O(n * S)

Space Complexity: O(S)

Topics

This problem involves: Dynamic Programming, Backtracking.

Companies

Asked at: Myntra, Pinterest, ServiceNow.

Target Sum | LeetCode 494

Medium

Dynamic Programming Backtracking

Time: O(n * S)Space: O(S)

Solution Code

Available in:

class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int total = 0;
        for (int n : nums) total += n;
        if (Math.abs(target) > total || ((total + target) & 1) == 1) return 0;
        int subset = (total + target) / 2;
        int[] dp = new int[subset + 1];
        dp[0] = 1;
        for (int num : nums) {
            for (int s = subset; s >= num; s--) {
                dp[s] += dp[s - num];
            }
        }
        return dp[subset];
    }
}

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Problem Info

Description

Problem 494: Target Sum - Detailed explanation and solution approaches.

Topics

Dynamic Programming Backtracking

Companies

MyntraPinterestServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:494

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Calculate the total sum of the `nums` array.
2
Step 2: Check for invalid cases: if `abs(target) > total` or `(total + target)` is odd, return 0 because there are no possible solutions.
3
Step 3: Calculate the required subset sum: `subset = (total + target) / 2`.
4
Step 4: Initialize a 1D DP array `dp` of size `subset + 1` with all elements set to 0, except `dp[0] = 1` (representing the base case where an empty subset has a sum of 0).
5
Step 5: Iterate through the `nums` array. For each number `num`, iterate backward through the `dp` array from `subset` down to `num`. Update `dp[s]` by adding `dp[s - num]` to it. This represents the number of ways to form the sum `s` either without including `num` (which is `dp[s]`) or by including `num` (which is `dp[s - num]`). Iterating backwards ensures that we are considering each number only once when building subsets.
6
Step 6: After iterating through all the numbers, `dp[subset]` will contain the number of subsets that sum up to the calculated `subset` value. Return `dp[subset]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * S)

💾

Space Complexity

O(S)

📝

Why?

The time complexity is determined by the nested loops. The outer loop iterates through each number in `nums` (n times), and the inner loop iterates through the possible subset sums up to `S` (S times). The space complexity is determined by the size of the `dp` array, which is of size `S+1` to store the number of ways to form sums from 0 to S.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Backtracking

A recursive approach to explore all possible combinations of '+' and '-' signs. For each number, we can either add it or subtract it from the current sum and recursively call the function for the next number. When we reach the end of the array, we check if the current sum equals the target and increment a counter if it does.

Time: O(2^n), where n is the length of the nums array. Each number has two possibilities (+ or -), leading to 2^n possible combinations.Space: O(n), where n is the length of the nums array. This is due to the recursion depth.

Trade-offs: Backtracking is simpler to implement but less efficient than dynamic programming, especially for larger input arrays. It is likely to result in Time Limit Exceeded (TLE) errors on LeetCode for larger test cases. DP is preferable for its improved time complexity.

✨

Key Insights

1Insight 1: The core idea is to transform the problem into a subset sum problem. Instead of finding combinations of '+/-' signs, we can divide the numbers into two groups: P (positive) and N (negative). The goal is to find the number of ways to choose numbers for the P group such that sum(P) - sum(N) = target.
2Insight 2: We can further simplify the equation: sum(P) - sum(N) = target. Since sum(P) + sum(N) = sum(nums) = total, adding the two equations gives us 2 * sum(P) = target + total, or sum(P) = (target + total) / 2. Therefore, the problem is equivalent to finding the number of subsets whose sum equals (target + total) / 2.
3Insight 3: Dynamic programming is suitable for solving the subset sum problem. We can use a 1D DP array to store the number of ways to achieve each possible sum from 0 up to (target + total) / 2. Iterate through the nums array, and for each number, update the DP array by considering the possibility of including that number in the subset.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `target` is very large (greater than the sum of all elements in `nums`). The code handles this by checking `Math.abs(target) > total` and returning 0 in this case, as no combination of signs can achieve the target sum.
Edge case 2: `(total + target)` is odd. The code handles this by checking `((total + target) & 1) == 1` (or `(total + target) % 2 != 0` in Python) and returning 0 in this case. This is because the required subset sum `(total + target) / 2` would not be an integer, and therefore no valid subset can produce the target sum.
Edge case 3: `nums` contains zero. The algorithm correctly handles zero values. When iterating through the nums, if `num` is 0, the inner loop still executes, adding `dp[s-0]` which is just `dp[s]` to `dp[s]`. This effectively accounts for the cases where the zero is either added or subtracted, doubling the count for `dp[s]` if it was non-zero before.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if `(total + target)` is even. If it's odd, there are no possible solutions because the subset sum will not be an integer.

✗Common mistake 2: Incorrect initialization of the DP array. `dp[0]` should be initialized to 1, representing the base case where an empty subset has a sum of 0.

✗Common mistake 3: Iterating the inner loop of the DP solution in the wrong order (forward instead of backward). Iterating forward would cause the same number to be used multiple times in the same subset, which is incorrect.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine you have a set of tasks, each with a certain cost. You want to find the number of ways to select tasks so that their total cost equals a specific budget. This is analogous to the target sum problem where costs are 'nums' and budget is 'target'.

🏦Real-world scenario 2: Financial planning. A financial advisor might need to determine different investment strategies (represented by the '+' or '-' signs indicating gains or losses) to achieve a particular financial goal (the 'target'). The algorithm helps to find the number of combinations of investment options that meet this goal.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Target Sum"?

For this medium-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?