grid47 Exploring patterns and algorithms
Sep 18, 2024
6 min read
Solution to LeetCode 494: Target Sum Problem
You are given an integer array nums and an integer target. You must build an expression by adding ‘+’ or ‘-’ before each element in nums and concatenate them to form an expression. Return the number of different expressions that result in the target value.
Problem
Approach
Steps
Complexity
Input: The input consists of an integer array nums and an integer target.
Example: nums = [1, 2, 3, 4, 5], target = 7
Constraints:
• 1 <= nums.length <= 20
• 0 <= nums[i] <= 1000
• 0 <= sum(nums[i]) <= 1000
• -1000 <= target <= 1000
Output: Return the number of distinct expressions that can be formed, which evaluate to the target value.
Example: Output: 3
Constraints:
• The result will be an integer representing the number of expressions.
Goal: The goal is to compute the number of expressions whose evaluated result matches the target value.
Steps:
• 1. Use dynamic programming or memoization to track the number of ways to achieve each sum with the given operations.
• 2. Use recursion to consider both adding and subtracting each number at every step.
• 3. Use a memoization table to store intermediate results for efficient computation.
Goal: The constraints for the problem are as follows.
Steps:
• 1 <= nums.length <= 20
• 0 <= nums[i] <= 1000
• 0 <= sum(nums[i]) <= 1000
• -1000 <= target <= 1000
Assumptions:
• The array will always have at least one element.
• Both operations '+' and '-' can be applied to any element in nums.
• Input: nums = [1, 2, 3, 4, 5], target = 7
• Explanation: There are 3 ways to form an expression that evaluates to 7: +1 +2 +3 -4 +5, +1 +2 -3 +4 +5, and +1 -2 +3 +4 +5.
• Input: nums = [10, 2, 3], target = 5
• Explanation: There is only 1 way to form an expression that evaluates to 5: +10 -2 -3.
Approach: The solution involves dynamic programming and memoization to recursively find the number of ways to form the target sum by applying the '+' and '-' operations to each element of nums.
Observations:
• The problem can be solved by exploring all possible combinations of '+' and '-' for each element.
• Dynamic programming with memoization should be used to efficiently calculate the number of ways to reach the target sum.
Steps:
• 1. Define a recursive function dp(target, idx) to explore both adding and subtracting nums[idx] at each step.
• 2. Use memoization to store the number of ways to reach the target sum for each subarray and target combination.
• 3. Return the final result after exploring all possible sums.
Empty Inputs:
• The problem guarantees that nums will always contain at least one element.
Large Inputs:
• The solution must be optimized to handle larger inputs efficiently, up to the constraint limits.
Special Values:
• If the target is 0, we need to count the number of ways to make the sum 0.
Constraints:
• Ensure that the algorithm handles both small and large arrays as specified in the constraints.
Declares a map `mp` which stores the number of ways to achieve a specific target using a given index, enabling memoization for efficient calculation.
2 : Variable Declaration
vector<int> nums;
Declares a vector `nums` to store the input numbers that will be used in the calculations.
3 : Function Definition
intdp(int target, int idx) {
Defines the `dp` function, which is a recursive function that calculates the number of ways to reach a specific target sum starting from a given index.
4 : Base Case
if(idx == nums.size()) return target ==0;
Base case: If all elements have been processed (i.e., `idx` equals the size of `nums`), return 1 if the target is 0 (meaning we’ve successfully formed the target sum), otherwise return 0.
5 : Memoization Check
if(mp.count(target))
Checks if the result for the current target has already been calculated for the given index using memoization.
6 : Memoization Lookup
if(mp[target].count(idx)) return mp[target][idx];
If the result for the current target and index is already stored, return the memoized value to avoid redundant calculations.
7 : Recursive Calls
int res = dp(target - nums[idx], idx +1);
Recursively calls `dp` by subtracting the current number (`nums[idx]`) from the target and moving to the next index.
8 : Recursive Calls
res += dp(target + nums[idx], idx +1);
Recursively calls `dp` again, but this time by adding the current number (`nums[idx]`) to the target and moving to the next index. Both results are added to account for both possibilities.
9 : Memoization
return mp[target][idx] = res;
Stores the computed result in the memoization table `mp` for future use, associating it with the current target and index.
10 : Function Definition
intfindTargetSumWays(vector<int>& nums, int target) {
Defines the `findTargetSumWays` function, which initializes the memoization structure and calls the recursive `dp` function to calculate the result.
11 : Input Assignment
this->nums = nums;
Assigns the input vector `nums` to the class-level `nums` variable, so it can be used in the `dp` function.
12 : Return Statement
returndp(target, 0);
Calls the `dp` function starting from index 0 and the target value, and returns the result, which is the number of ways to achieve the target sum.
Best Case: O(n * target)
Average Case: O(n * target)
Worst Case: O(n * target)
Description: The time complexity is O(n * target) because each subproblem is computed once and memoized.
Best Case: O(n * target)
Worst Case: O(n * target)
Description: The space complexity is O(n * target) due to the memoization table storing results for each target and index combination.
