Smallest Subarray with a given sum (easy) Longest Substring with K Distinct Characters (medium) Fruits into Baskets (medium) No-repeat Substring (hard) Longest Substring with Same Letters after Replacement (hard) Longest Subarray with Ones after Replacement (hard) 2. Pattern: two points, 双指针...
classSolution {publicbooleancanPartitionKSubsets(int[] nums,intk) {intsum =sum(nums);//check if possible to have K equal sum subsetsif(sum % k != 0) {returnfalse; }intsubSum = sum /k; Arrays.sort(nums);intbeginIndex = nums.length - 1;//check if the largest num is greater than...
class Solution: def subsetsWithDup(self, nums: List[int]) -> List[List[int]]: def bt(nums, path, start): res.append(copy.deepcopy(path)) # 每一个path都加入res for i in range(start,len(nums)): # 遍历选择列表 if i == start or nums[i] != nums[i-1]: # 剪枝条件 path.appe...
classSolution {publicbooleancanPartitionKSubsets(int[] nums,intk) {if(nums ==null|| nums.length == 0) {returnfalse; }intsum = 0;for(inti = 0; i < nums.length; i++) { sum+=nums[i]; }if(sum % k != 0) {returnfalse; }inttarget = sum /k;boolean[] isUsed =newboolean[num...
class Solution { vector<vector<int>> res; vector<int> path; bool used[11] = {}; int n; public: vector<vector<int>> subsets(vector<int>& nums) { n = nums.size(); dfs(nums, 0); return res; } void dfs(vector<int>& nums, int pos) { res.push_back(path); for (int i = ...
Can you solve this real interview question? Partition to K Equal Sum Subsets - Given an integer array nums and an integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: num
public:boolcanPartitionKSubsets(vector<int>& nums,intk) {intsum = accumulate(nums.begin(), nums.end(),0);if(sum % k !=0)returnfalse; vector<bool> visited(nums.size(),false);returnhelper(nums, k, sum / k,0,0, visited);
输入:n=4,k=2输出:[[2,4],[3,4],[2,3],[1,2],[1,3],[1,4],] 示例2: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 输入:n=1,k=1输出:[[1]] 💡 思路: 这道题是回溯算法的经典题目。 我们来看一下这道题的抽象树形结构: ...
partition-to-k-equal-sum-subsets partitioning-into-minimum-number-of-deci-binary-numbers pascals-triangle-ii pascals-triangle path-sum-ii path-sum-iii path-sum path-with-minimum-effort paths-with-sum-lcci peak-index-in-a-mountain-array peeking-iterator percentage-of-letter-in-st...
class Solution: def subsetsWithDup(self, nums: List[int]) -> List[List[int]]: if nums == []: return [[]] def backtrack(idx): # 不需要等一条路径回溯结束再保存值,每一步都要保存值。 res.append(path[:]) for i in range(idx, len(nums)): if i > idx and nums[i] == nums[...