Combination Sum IV - Practice Coding Problems

Combination Sum IV - Problem

Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.

The test cases are generated so that the answer can fit in a 32-bit integer.

Note: Different sequences are counted as different combinations. For example, [1,2] and [2,1] are considered different combinations.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3], target = 4

› Output: 7

💡 Note: The possible combinations are: [1,1,1,1], [1,1,2], [1,2,1], [2,1,1], [1,3], [3,1], [2,2]. Total of 7 combinations.

Example 2 — Single Element

$ Input: nums = [9], target = 3

› Output: 0

💡 Note: Since 9 > 3, there's no way to form target 3 using only number 9. Answer is 0.

Example 3 — Perfect Match

$ Input: nums = [1,2], target = 3

› Output: 3

💡 Note: The possible combinations are: [1,1,1], [1,2], [2,1]. Total of 3 combinations.

Constraints

1 ≤ nums.length ≤ 200
1 ≤ nums[i] ≤ 1000
All elements in nums are unique
1 ≤ target ≤ 1000

Visualization

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Understanding the Visualization

Input

Array nums=[1,2,3] and target=4

Find Combinations

Count all sequences that sum to target

Output

Return total count of combinations

Key Takeaway

🎯 Key Insight: This is an unbounded knapsack where order matters - use DP to count all possible sequences

Asked in

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

The key insight is that this is an unbounded knapsack problem where order matters. We can use dynamic programming where dp[i] represents the number of ways to form amount i. For each amount, we try adding each number from nums. Time: O(target × nums.length), Space: O(target)

Common Approaches

Approach	Time	Space	Notes
✓ Recursive Brute Force	O(target^nums.length)	O(target)	Try every possible combination recursively without optimization
Bottom-Up Dynamic Programming	O(target × nums.length)	O(target)	Build solutions iteratively from 0 to target using a DP array
Recursion with Memoization	O(target × nums.length)	O(target)	Cache results of recursive calls to avoid redundant calculations

Recursive Brute Force — Algorithm Steps

Base case: if target is 0, return 1 (one way to make 0)
For each number in nums, recursively count combinations for (target - num)
Sum up all valid combinations from recursive calls

Visualization

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Step-by-Step Walkthrough

Start

Begin with target = 4

Branch

Try subtracting each number [1,2,3]

Recurse

Continue until target becomes 0 or negative

Code -

solution.c — C

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

int backtrack(int* nums, int numsSize, int remaining) {
    if (remaining == 0) return 1;
    if (remaining < 0) return 0;
    
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        count += backtrack(nums, numsSize, remaining - nums[i]);
    }
    return count;
}

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

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int target;
    scanf("%d", &target);
    
    printf("%d\n", solution(nums, numsSize, target));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(target^nums.length)

Each recursive call branches into nums.length calls, creating exponential growth

✓ Linear Growth

Space Complexity

O(target)

Recursion stack depth can go up to target in worst case

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Recursive Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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