Target Sum

Target Sum - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,1,1,1,1], target = 3

› Output: 5

💡 Note: There are 5 ways to assign symbols: -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 — Simple Case

$ Input: nums = [1], target = 1

› Output: 1

💡 Note: Only one way: +1 = 1

Example 3 — Impossible Case

$ Input: nums = [1,2], target = 4

› Output: 0

💡 Note: Impossible to reach 4 with numbers [1,2]. Maximum possible is +1+2=3, minimum is -1-2=-3

Constraints

1 ≤ nums.length ≤ 20
0 ≤ nums[i] ≤ 1000
0 ≤ sum(nums[i]) ≤ 1000
-1000 ≤ target ≤ 1000

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8 M Microsoft 6

The key insight is transforming the problem from assigning +/- signs to finding subset sums. Instead of trying 2^n combinations, we realize that if subset P gets + signs and subset N gets - signs, then sum(P) - sum(N) = target and sum(P) + sum(N) = total. This gives us sum(P) = (total + target)/2. The optimal approach uses dynamic programming to count subsets with this exact sum. Time: O(n × S), Space: O(S) where S is the target subset sum.

Common Approaches

✓ Brute Force Recursion

⏱️ Time: O(2ⁿ) Space: O(n)

For each number in the array, we have two choices: add it or subtract it. We recursively explore all 2^n possibilities and count how many evaluate to the target sum.

Memoization (Top-Down DP)

⏱️ Time: O(n × S) Space: O(n × S)

Same recursive approach as brute force, but we cache results for (index, currentSum) pairs to avoid redundant calculations. This dramatically reduces time complexity.

Subset Sum Transformation

⏱️ Time: O(n × S) Space: O(S)

Key insight: If we assign + to subset P and - to subset N, then sum(P) - sum(N) = target and sum(P) + sum(N) = total. Solving gives sum(P) = (total + target)/2. Count subsets that sum to this value.

Brute Force Recursion — Algorithm Steps

Start with first number and current sum of 0
For each number, try both adding and subtracting
When we reach the end, check if sum equals target
Count all valid combinations

Visualization

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Step-by-Step Walkthrough

Start

Begin with index 0, sum 0

Branch

For each number, try both + and -

Count

Count paths that reach target

Code -

solution.c — C

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

int backtrack(int* nums, int numsSize, int index, int currentSum, int target) {
    if (index == numsSize) {
        return currentSum == target ? 1 : 0;
    }
    
    // Try adding the current number
    int count = backtrack(nums, numsSize, index + 1, currentSum + nums[index], target);
    // Try subtracting the current number
    count += backtrack(nums, numsSize, index + 1, currentSum - nums[index], target);
    
    return count;
}

int solution(int* nums, int numsSize, int target) {
    return backtrack(nums, numsSize, 0, 0, target);
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int target;
    scanf("%d", &target);
    
    int result = solution(nums, numsSize, target);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ)

For each of n numbers, we make 2 recursive calls (+ or -)

✓ Linear Growth

Space Complexity

O(n)

Recursion depth goes up to n levels deep

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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