From 9c45c6411034429d2c289ebdd66271b22c361795 Mon Sep 17 00:00:00 2001
From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:33:31 -0700
Subject: [PATCH 1/6] Create 698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
---
 ...98.Partition-to-K-Equal-Sum-Subsets_v2.cpp &#124; 24 +++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp b/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
new file mode 100644
index 000000000..66733f92c
--- /dev/null
+++ b/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
@@ -0,0 +1,24 @@
+class Solution {
+public:
+ bool canPartitionKSubsets(vector<int>& nums, int k) 
+ {
+ int n = nums.size();
+ int sum = accumulate(nums.begin(), nums.end(), 0);
+ if (sum%k!=0) return false;
+ int target = sum / k;
+ 
+ vector<int>dp(1<<n, -1); + dp[0] = 0; + for (int state=0; state<(1<<n); state++) + { + if (dp[state]==-1) continue; + for (int i=0; i<n; i++) + { + if (!((state>>i)&1) && (dp[state]+nums[i] <= target)) + dp[state + (1<<i)] = (dp[state]+nums[i]) % target; + } + } + + return dp[(1<<n)-1]==0; + } +}; From 78428ce345b60d8baf5816ba5f72f4fff2957cc3 Mon Sep 17 00:00:00 2001 From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:33:43 -0700
Subject: [PATCH 2/6] Rename 698.Partition-to-K-Equal-Sum-Subsets.cpp to
 698.Partition-to-K-Equal-Sum-Subsets_v1.cpp
---
 ...um-Subsets.cpp => 698.Partition-to-K-Equal-Sum-Subsets_v1.cpp} &#124; 0
 1 file changed, 0 insertions(+), 0 deletions(-)
 rename DFS/698.Partition-to-K-Equal-Sum-Subsets/{698.Partition-to-K-Equal-Sum-Subsets.cpp => 698.Partition-to-K-Equal-Sum-Subsets_v1.cpp} (100%)
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets.cpp b/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v1.cpp
similarity index 100%
rename from DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets.cpp
rename to DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v1.cpp
From 0a73a820e17a7e4618ca74774a104aaa68acf11a Mon Sep 17 00:00:00 2001
From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:53:37 -0700
Subject: [PATCH 3/6] Update Readme.md
---
 DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md &#124; 11 +++++++++++
 1 file changed, 11 insertions(+)
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
index c03b00350..70a464126 100644
--- a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
+++ b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
@@ -1,5 +1,6 @@
 ### 698.Partition-to-K-Equal-Sum-Subsets
 
+#### 解法1:搜索
 此题据说是NP-hard,没有什么特别高明的算法,就是老老实实地DFS,尝试将所有元素挨个尝试放入k个分类里,直至找到满足条件的分类。
 
 设计如下的递归函数:```DFS(nums, curPos, curGroup, curSum)```.
@@ -27,5 +28,15 @@ for (int i=curPos; i<n; i++) 与本题非常相似的题目有: 473. Matchsticks to Square, 416. Partition Equal Subset Sum, 996. Number of Squareful Arrays +#### 解法2:状态压缩DP +此题有一种非常巧妙的状压DP的算法。因为本题中nums的个数最多只有16个,那么任意subset的总数不超过2^16=65536,可以用n位的二进制数来表示。另外我们令 + +本题里的dp[state]的定义比较有意思:以state所代表的subset,如果它可以分割成若干个&quot;元素和是target的集合&quot;,再加上若干个总和不超过target的元素零头,那么dp[state]就表示为subset的元素总和对于target的取余(也就是所说的零头);否则dp[state]设为-1. + +举个例子,如果state代表的是{1,2,3,4},target是4,那么因为这个state可以分割成{4},{1+3},{2},所以dp[state]=2. 为什么这么定义dp呢?这是因为,如果当前dp[state]的值不是-1,那么我们只需要关心的今后是否能有新的元素可以补足目前的零头凑成另一个target。以这个例子来说,如果再来一个新元素是1加入目前的subset,新元素1和旧元素2放入待定区,那么可以就得到dp[state2]=3;同理,如果再来一个新元素是2加入目前的subset,新元素2和旧元素2恰好凑成一组target的集合,于是dp[state2]=0;但是如果再来一个新元素是3,已有的零头2加上3就超过了target,说明新来的3无法与目前的零头凑成一组target的集合,所以我们就放弃dp[state2]的更新。 + +所以我们本题状态转移的思想是,根据当前的state,考察可以加入哪些新元素(比如说i)得到新的state2,使得```dp[state]+nums[i]<=target```。满足的话,那么就可以更新dp[state2]为该组合的元素之和对于target的取余(零头)。 + +最终的答案就是考察state=(1<<n)-1时的dp值是否不为-1. (其实,如果dp值不为-1的话,它一定是0) [Leetcode Link](https://leetcode.com/problems/partition-to-k-equal-sum-subsets) From b16426bc63277df043a9681b0dcd03e0621398e2 Mon Sep 17 00:00:00 2001 From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:54:17 -0700
Subject: [PATCH 4/6] Update Readme.md
---
 DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md &#124; 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
index 70a464126..da33cb80d 100644
--- a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
+++ b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
@@ -37,6 +37,6 @@ for (int i=curPos; i<n; i++) 所以我们本题状态转移的思想是,根据当前的state,考察可以加入哪些新元素(比如说i)得到新的state2,使得```dp[state]+nums[i]<=target```。满足的话,那么就可以更新dp[state2]为该组合的元素之和对于target的取余(零头)。 -最终的答案就是考察state=(1<<n)-1时的dp值是否不为-1. (其实,如果dp值不为-1的话,它一定是0) +最终的答案就是考察state=(1<<n)-1时的dp值是否不为-1. (其实,如果dp值不为-1的话,它一定是0,因为我们预判了所有元素之和必然要被target整除) [Leetcode Link](https://leetcode.com/problems/partition-to-k-equal-sum-subsets) From 248b8a76cfa9c0ab07cb3f3f7300069feb81e788 Mon Sep 17 00:00:00 2001 From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:54:49 -0700
Subject: [PATCH 5/6] Update 698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
---
 .../698.Partition-to-K-Equal-Sum-Subsets_v2.cpp &#124; 3 ++-
 1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp b/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
index 66733f92c..b3c90ccd8 100644
--- a/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
+++ b/DFS/698.Partition-to-K-Equal-Sum-Subsets/698.Partition-to-K-Equal-Sum-Subsets_v2.cpp
@@ -19,6 +19,7 @@ class Solution {
 }
 }
 
- return dp[(1<<n)-1]==0; + return dp[(1<<n)-1] != -1; + } }; From 4b4e9e580213c4dd289f1ebb65f64025e8bfd542 Mon Sep 17 00:00:00 2001 From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 3 Jun 2022 23:55:12 -0700
Subject: [PATCH 6/6] Update Readme.md
---
 DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md &#124; 2 ++
 1 file changed, 2 insertions(+)
diff --git a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
index da33cb80d..6be5ed84e 100644
--- a/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
+++ b/DFS/698.Partition-to-K-Equal-Sum-Subsets/Readme.md
@@ -39,4 +39,6 @@ for (int i=curPos; i<n; i++)
 
 最终的答案就是考察state=(1<<n)-1时的dp值是否不为-1. (其实,如果dp值不为-1的话,它一定是0,因为我们预判了所有元素之和必然要被target整除)
 
+该解法的时间复杂度是```2^n*n```.
+
 [Leetcode Link](https://leetcode.com/problems/partition-to-k-equal-sum-subsets)
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