Table of Contents
Combination Sum IV – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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.
Example 1:
Input: nums = [1,2,3], target = 4 Output: 7 Explanation: The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations.
Example 2:
Input: nums = [9], target = 3 Output: 0
Constraints:
1 <= nums.length <= 2001 <= nums[i] <= 1000- All the elements of
numsare unique. 1 <= target <= 1000
C++ Combination Sum IV LeetCode Solution
class Solution {
public:
/**
* @param A: an integer array.
* @param k: a positive integer (k <= length(A))
* @param target: a integer
* @return an integer
*/
int kSum(vector<int> A, int k, int target) {
// wirte your code here
const int n = A.size();
/** dp[i][j][target] : # of ways to start from vector[0..i-1], choose j elements to sum to target **/
vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(k + 1, vector<int>(target + 1, 0)));
for(int i = 1; i <= n; i++) {
if(A[i-1] <= target) {
for(int j = i; j <= n; j++) {
dp[j][1][A[i-1]] =1;
}
}
}
/** for position i, we can choose it or not **/
for(int i = 1; i <= n; i++) {
for(int j = min(i, k); j > 1; j--) {
for(int p = 1; p <= target; p++) {
dp[i][j][p] = dp[i - 1][j][p];
if(p - A[i - 1] >= 0) {
dp[i][j][p] += dp[i - 1][j - 1][p - A[i - 1]];
}
}
}
}
return dp[n][k][target];
}
};
Java Combination Sum IV LeetCode Solution
private int[] dp;
public int combinationSum4(int[] nums, int target) {
dp = new int[target + 1];
Arrays.fill(dp, -1);
dp[0] = 1;
return helper(nums, target);
}
private int helper(int[] nums, int target) {
if (dp[target] != -1) {
return dp[target];
}
int res = 0;
for (int i = 0; i < nums.length; i++) {
if (target >= nums[i]) {
res += helper(nums, target - nums[i]);
}
}
dp[target] = res;
return res;
}
Python 3 Combination Sum IV LeetCode Solution
class Solution(object):
def combinationSum4WithLength(self, nums, target, length, memo=collections.defaultdict(int)):
if length <= 0: return 0
if length == 1: return 1 * (target in nums)
if (target, length) not in memo:
for num in nums:
memo[target, length] += self.combinationSum4(nums, target - num, length - 1)
return memo[target, length]
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
