This approach uses recursion and backtracking to generate all possible combinations by exploring each candidate. If a candidate is chosen, we explore further with the remaining target reduced by the candidate's value. We use backtracking to remove a candidate and try the next one. This way, we explore all possible combinations using depth-first search.
Time Complexity: O(2^T), where T is the target, as it potentially explores every combination of the candidates.
Space Complexity: O(T) for the recursion call stack and the path being explored.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5void findCombination(int* candidates, int candidatesSize, int target, int** result, int* returnColSize, int* returnSize, int* currentCombo, int currentComboSize, int startIndex) {
6 if (target < 0) return;
7 if (target == 0) {
8 result[*returnSize] = (int*)malloc(sizeof(int) * currentComboSize);
9 memcpy(result[*returnSize], currentCombo, sizeof(int) * currentComboSize);
10 returnColSize[*returnSize] = currentComboSize;
11 (*returnSize)++;
12 return;
13 }
14
15 for (int i = startIndex; i < candidatesSize; i++) {
16 currentCombo[currentComboSize] = candidates[i];
17 findCombination(candidates, candidatesSize, target - candidates[i], result, returnColSize, returnSize, currentCombo, currentComboSize + 1, i);
18 }
19}
20
21int** combinationSum(int* candidates, int candidatesSize, int target, int** columnSizes, int* returnSize) {
22 int** result = (int**)malloc(150 * sizeof(int*));
23 *columnSizes = (int*)malloc(150 * sizeof(int));
24 *returnSize = 0;
25 int* currentCombo = (int*)malloc(target * sizeof(int));
26 findCombination(candidates, candidatesSize, target, result, *columnSizes, returnSize, currentCombo, 0, 0);
27 free(currentCombo);
28 return result;
29}
30
31int main() {
32 int candidates[] = {2, 3, 6, 7};
33 int candidatesSize = 4;
34 int target = 7;
35 int* columnSizes;
36 int returnSize;
37 int** result = combinationSum(candidates, candidatesSize, target, &columnSizes, &returnSize);
38 for (int i = 0; i < returnSize; i++) {
39 printf("[");
40 for (int j = 0; j < columnSizes[i]; j++) {
41 printf("%d%s", result[i][j], j == columnSizes[i] - 1 ? "" : ", ");
42 }
43 printf("]\n");
44 free(result[i]);
45 }
46 free(result);
47 free(columnSizes);
48 return 0;
49}
50This C solution uses recursion to explore combinations and backtracks when a target less than zero is reached. If the target is zero, a valid combination is found and added to the result. The helper function findCombination iterates through candidates allowing reuse of the current candidate.
We can use a dynamic programming (DP) approach to solve the Combination Sum problem. We maintain a DP array where each index represents the number of ways to form that particular sum using the candidates. The approach involves iterating through candidates and updating the DP array for each possible sum that can be formed with that candidate.
Time Complexity: Roughly O(T * N), for target T and candidates N.
Space Complexity: O(T) for the dp array itself in context of storage.
1import java.util.ArrayList;
2import java.util.List;
3
4public class CombinationSumDP {
5 public static List<List<Integer>> combinationSum(int[] candidates, int target) {
6 List<List<Integer>>[] dp = new List[target + 1];
7 dp[0] = new ArrayList<>();
8 dp[0].add(new ArrayList<>());
9
10 for (int i = 1; i <= target; i++) {
11 dp[i] = new ArrayList<>();
12 for (int candidate : candidates) {
13 if (i >= candidate) {
14 for (List<Integer> combination : dp[i - candidate]) {
15 List<Integer> newComb = new ArrayList<>(combination);
16 newComb.add(candidate);
17 dp[i].add(newComb);
18 }
19 }
20 }
21 }
22
23 return dp[target];
24 }
25
26 public static void main(String[] args) {
27 int[] candidates = {2, 3, 6, 7};
28 int target = 7;
29 List<List<Integer>> results = combinationSum(candidates, target);
30 results.forEach(System.out::println);
31 }
32}
33In the Java solution, a DP array holds lists where combinations for each target sum are stored, built step-by-step through each candidate addition and classified into dp indices by their sum value.