LeetCode in C

494. Target Sum

Medium

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

Explanation: There are 5 ways to assign symbols to make the sum of nums be target 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

Solution

#include <stdio.h>
#include <stdlib.h>

int findTargetSumWays(int* nums, int numsSize, int s) {
    int sum = 0;
    s = abs(s);
    for (int i = 0; i < numsSize; i++) {
        sum += nums[i];
    }
    
    // Invalid target case, return 0
    if (s > sum || (sum + s) % 2 != 0) {
        return 0;
    }
    
    int target = (sum + s) / 2;
    
    // Initialize the dp array
    int** dp = (int**)malloc((target + 1) * sizeof(int*));
    for (int i = 0; i <= target; i++) {
        dp[i] = (int*)calloc(numsSize + 1, sizeof(int));
    }
    
    dp[0][0] = 1;
    
    // Initialize the base case
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == 0) {
            dp[0][i + 1] = dp[0][i] * 2;
        } else {
            dp[0][i + 1] = dp[0][i];
        }
    }
    
    // Fill the dp table
    for (int i = 1; i <= target; i++) {
        for (int j = 0; j < numsSize; j++) {
            dp[i][j + 1] = dp[i][j];
            if (nums[j] <= i) {
                dp[i][j + 1] += dp[i - nums[j]][j];
            }
        }
    }
    
    int result = dp[target][numsSize];
    
    // Free the dynamically allocated memory
    for (int i = 0; i <= target; i++) {
        free(dp[i]);
    }
    free(dp);
    
    return result;
}

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