.

Author: hqztrue

0201-0300/300. Longest Increasing Subsequence.pdf

@@ -1,2 +1,2 @@
-fast exponentiation. Let M=1337, b>>M, we can shrink b to O(M) by Euler's theorem.
+fast exponentiation. Let M=1337, if b is much larger than M, we can shrink b to O(M) by Euler's theorem.
 

0301-0400/372. Super Pow.txt

@@ -127,8 +127,8 @@ class Solution {
 		for (int i=1;i<n;++i)s[i]+=s[i-1];
 		for (int i=n-2;i;--i)M[mp(s[n-1]-s[i],a[i])]=i;
 		auto minmax=minmax_element(a.begin(),a.end());
-        int minv=*minmax.first,maxv=*minmax.second;
-        int mins=(s[n-1]-3*maxv)/4,maxs=(s[n-1]-3*minv)/4;
+		int minv=*minmax.first,maxv=*minmax.second;
+		int mins=(s[n-1]-3*maxv)/4,maxs=(s[n-1]-3*minv)/4;
 		for (int i=1;i<n-5;++i){
 			int s1=s[i-1];
 			if (s1<mins||s1>maxs)continue;

0501-0600/548. Split Array with Equal Sum 8ms.cpp

@@ -13,9 +13,9 @@ int interpolation_search(int a[],int n,int x){
 }
 class Solution {
 public:
-    int search(vector<int>& nums, int t) {
-        int n=nums.size(),p=interpolation_search(&nums[0]-1,n,t);
-        return p<=n&&nums[p-1]==t?p-1:-1;
-    }
+	int search(vector<int>& nums, int t) {
+		int n=nums.size(),p=interpolation_search(&nums[0]-1,n,t);
+		return p<=n&&nums[p-1]==t?p-1:-1;
+	}
 };
 

0701-0800/704. Binary Search.cpp

@@ -10,7 +10,7 @@ size_t get_hash(vector<int> &v){
 		for (int j=0;j<u.size();++j)res=((long long)res*M+u[j])%P;
 	}*/
 	sort(v.begin(),v.end());
-    for (int j=0;j<v.size();++j)res=((long long)res*M+v[j])%P;
+	for (int j=0;j<v.size();++j)res=((long long)res*M+v[j])%P;
 	return ((long long)res*M+n)%P;
 }
 struct point{

0701-0800/711. Number of Distinct Islands II 40ms.cpp

@@ -0,0 +1,2 @@
+predecessor search. O(n*pred(n)).
+

0701-0800/729. My Calendar I.txt

@@ -0,0 +1,2 @@
+balanced tree. O(n log n).
+

0701-0800/731. My Calendar II.txt

@@ -0,0 +1,2 @@
+O(nm).
+

0701-0800/733. Flood Fill.txt

@@ -0,0 +1,2 @@
+stack. O(n).
+

0701-0800/735. Asteroid Collision.txt

@@ -0,0 +1,2 @@
+stack. O(n).
+

0701-0800/736. Parse Lisp Expression.txt

@@ -0,0 +1,2 @@
+O(log n).
+

0701-0800/738. Monotone Increasing Digits.txt

@@ -0,0 +1,2 @@
+partition the numbers into consecutive intervals, then run DP. O(n).
+

0701-0800/740. Delete and Earn.txt

@@ -0,0 +1,2 @@
+binary search. O(log n) after reading the input.
+

0701-0800/744. Find Smallest Letter Greater Than Target.txt

@@ -0,0 +1,2 @@
+DP. O(n).
+

0701-0800/746. Min Cost Climbing Stairs.txt

@@ -0,0 +1,2 @@
+O(n).
+

0701-0800/747. Largest Number At Least Twice of Others.txt

@@ -0,0 +1,2 @@
+O(L).
+

0701-0800/748. Shortest Completing Word.txt

@@ -0,0 +1,2 @@
+First repeatedly go towards the target until we go pass it. Then we need at most 2 more moves, and flip at most 2 previous moves, depending whether the difference is even or odd. O(1).
+

0701-0800/754. Reach a Number.txt

@@ -0,0 +1,37 @@
+const int CH=255,N=5005,W=(1<<16)-1;
+int _a[26][N],*a[CH],a1[CH],b[N],l[N];
+class Solution {
+public:
+	int numMatchingSubseq(const string &s, vector<string>& w) {
+		int n=s.size(),m=w.size(),ans=0;
+		for (int i=0;i<26;++i)a['a'+i]=_a[i],a1['a'+i]=0;
+		for (int i=0;i<m;++i){
+			l[i]=w[i].size();
+			char c=w[i][0];
+			a[c][a1[c]++]=i<<16;
+		}
+		for (int i=0;i<n;++i){
+			int k=a1[s[i]];
+			memcpy(b,a[s[i]],sizeof(int)*k);
+			a1[s[i]]=0;
+			for (int j=0;j<k;++j){
+				int x=b[j]>>16,y=(b[j]&W)+1;
+				if (y==l[x])++ans;
+				else {
+					char c=w[x][y];
+					a[c][a1[c]++]=b[j]+1;
+				}
+			}
+		}
+		return ans;
+	}
+};
+
+//IO
+int _IO=[](){
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+
+

0701-0800/792. Number of Matching Subsequences 80ms.cpp

@@ -0,0 +1,2 @@
+O(1).
+

0701-0800/792. Number of Matching Subsequences.pdf

@@ -0,0 +1,2 @@
+O(n).
+

0701-0800/794. Valid Tic-Tac-Toe State.txt

@@ -0,0 +1,2 @@
+dfs. O(n+m+output).
+

0701-0800/795. Number of Subarrays with Bounded Maximum.txt

@@ -0,0 +1,2 @@
+count the number of x-y pairs and y-x pairs. O(n).
+

0701-0800/797. All Paths From Source to Target.txt

@@ -0,0 +1,2 @@
+find connected component. O(nm).
+

0901-1000/996. Number of Squareful Arrays.pdf

@@ -0,0 +1,2 @@
+trie. O(L).
+

1201-1300/1247. Minimum Swaps to Make Strings Equal.txt

@@ -0,0 +1,3 @@
+Let n denote the number of steps and let m denote the array length.
+DP. O(n*min(n,m)).
+

1201-1300/1267. Count Servers that Communicate.txt

@@ -0,0 +1,76 @@
+const int N=50005,L=16,M=12;
+int ch[N],sta[N],d[N],h[N],sz[N],nxt[N],mark[N],stk[N],chain[N*2],c1[N],id[N],msk[N],
+	g[N],kth[1<<M][M],_fa[N/M][L],*fa[N],*p,D,len,fa1,id1,_id1;
+void pre(){
+	for (int i=1;i<(1<<M);++i){
+		kth[i][0]=__builtin_ctz(i);
+		for (int j=1;j<M;++j)kth[i][j]=kth[i-(i&-i)][j-1];
+	}
+}
+class TreeAncestor {
+public:
+	void dfs1(int x){
+		id[id1++]=x; g[x]=_id1;
+		for (int i=sta[x];i<sta[x+1];++i){
+			int y=ch[i]; msk[y]=msk[x]|(1<<id1-_id1); dfs1(y);
+		}
+	}
+	void dfs(int x){
+		d[x]=D; stk[D++]=x; sz[x]=1; h[x]=0; nxt[x]=mark[x]=-1; bool flag=1;
+		for (int i=sta[x];i<sta[x+1];++i){
+			int y=ch[i]; dfs(y);
+			sz[x]+=sz[y];
+			if (sz[y]>M)flag=0,mark[x]=mark[y];
+			if (h[y]>h[x])h[x]=h[y],nxt[x]=y;
+		}
+		--D; ++h[x];
+		if (sz[x]<=M)flag=0;
+		if (flag){
+			mark[x]=x; fa[x]=_fa[fa1++];
+			for (int i=0;i<L&&(1<<i)<=D;++i)fa[x][i]=stk[D-(1<<i)];
+		}
+		for (int i=sta[x];i<sta[x+1];++i){
+			int y=ch[i];
+			if (mark[x]!=-1&&sz[y]<=M)_id1=id1,msk[y]=1,dfs1(y);
+			if (y!=nxt[x]){
+				len+=h[y];
+				for (int z=y;z!=-1;z=nxt[z])chain[c1[z]=--len]=z;
+				len+=h[y];
+				for (int end=len+h[y],z=x;z!=-1&&len<end;z=p[z])chain[len++]=z;
+			}
+		}
+	}
+	TreeAncestor(int n, vector<int>& _p) {
+		p=&_p[0];
+		if (!kth[2][1])pre();
+		memset(sta,0,sizeof(int)*n);
+		for (int i=1;i<n;++i)++sta[p[i]];
+		for (int i=0,s=0,t;i<n;++i)t=sta[i],sta[i]=s,s+=t;
+		for (int i=1;i<n;++i)ch[sta[p[i]]++]=i;
+		for (int i=n;i;--i)sta[i]=sta[i-1]; sta[0]=0;
+		D=len=fa1=id1=0; dfs(0); len+=h[0];
+		for (int z=0;z!=-1;z=nxt[z])chain[c1[z]=--len]=z; len+=h[0];
+		if (sz[0]<=M)_id1=id1,msk[0]=1,dfs1(0);
+	}
+	int getKthAncestor(int x, int k) {
+		k=d[x]-k; if (k<0)return -1;
+		if (mark[x]==-1){
+			int y=id[g[x]];
+			if (k<d[y])x=p[y];
+			else return id[g[x]+kth[msk[x]][k-d[y]]];
+		}
+		x=mark[x];
+		if (d[x]==k)return x;
+		int l=31-__builtin_clz(d[x]-k); x=fa[x][l];
+		return chain[c1[x]+d[x]-k];
+	}
+};
+
+//IO
+int _IO=[](){
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+
+

1201-1300/1268. Search Suggestions System.txt

@@ -0,0 +1,2 @@
+O(nm).
+

1201-1300/1269. Number of Ways to Stay in the Same Place After Some Steps.txt

@@ -0,0 +1,2 @@
+monotone queue. O(n).
+

1401-1500/1483. Kth Ancestor of a Tree Node 308ms.cpp

@@ -6,13 +6,13 @@ class Solution {
 		int n=a.size(),ma=0,mi=inf;
 		int vmin[W+1],vmax[W+1],can[W+1];
 		for (int i=0;i<=W;++i)vmin[i]=inf,vmax[i]=0,can[i]=1;
-        for (int i=0;i<n;++i)b[i]=a[i]&1?a[i]*2:a[i];
+		for (int i=0;i<n;++i)b[i]=a[i]&1?a[i]*2:a[i];
 		for (int i=0;i<n;++i)a[i]>>=__builtin_ctz(a[i]),ma=max(ma,a[i]);
 		int c=__builtin_clz(ma);
 		for (int i=0;i<n;++i){
 			a[i]<<=__builtin_clz(a[i])-c;
 			if (a[i]>ma)a[i]>>=1;
-            a[i]=min(a[i],b[i]);
+			a[i]=min(a[i],b[i]);
 			mi=min(mi,a[i]);
 		}
 		int ans=ma-mi;

1401-1500/1483. Kth Ancestor of a Tree Node.pdf

@@ -0,0 +1,2 @@
+O(n).
+

1401-1500/1483. Kth Ancestor of a Tree Node.txt

@@ -0,0 +1,41 @@
+namespace Hash{
+	typedef unsigned int uint;
+	const uint S=17,S1=32-S,M=1996090921;
+	struct node{
+		int x,y,t;
+	}h[(1<<S)+1005];
+	int T=1;
+	inline void insert(int x,int y){
+		node *p=h+((uint)x*M>>S1);
+		for (;p->t==T;++p)
+			if (p->x==x){p->y+=y; return;}
+		p->t=T; p->x=x; p->y=y;
+	}
+	inline bool find(int x){
+		for (node *p=h+((uint)x*M>>S1);p->t==T;++p)
+			if (p->x==x&&p->y)return 1;
+		return 0;
+	}
+} using namespace Hash;
+
+class Solution {
+public:
+	int maxOperations(vector<int>& a, int k) {
+		int n=a.size(),ans=0; ++T;
+		for (int i=0;i<n;++i)
+			if (a[i]<k){
+				if (find(k-a[i]))insert(k-a[i],-1),++ans;
+				else insert(a[i],1);
+			}
+		return ans;
+	}
+};
+
+//IO
+int _IO=[](){
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+
+

1601-1700/1672. Richest Customer Wealth.txt

@@ -0,0 +1,5 @@
+Each subset can be viewed as an interval, we can prove that intervals cannot partially intersect.
+1. subset DP, let f[i][j] denote the minimum sum of incompatibilities when we distribute subset i, at most one subset is not full and starting at j. We add the subsets in the order of increasing right endpoint, so the last subset ends at the maximum element in i. O(2^n*n^2).
+
+Remark. is this problem NP-hard? If we don't require the subsets be of equal size, then the problem can be solved by DP.
+

1601-1700/1673. Find the Most Competitive Subsequence.txt

@@ -0,0 +1,3 @@
+1. DP, maintain the best and second best solution. O(n^2).
+2. similar to LCS, O(n^2/polylog n).
+

1601-1700/1675. Minimize Deviation in Array 196ms.cpp

@@ -0,0 +1,2 @@
+O(n).
+

1601-1700/1678. Goal Parser Interpretation.txt

@@ -0,0 +1,2 @@
+prefix sum. O(n).
+

1601-1700/1679. Max Number of K-Sum Pairs 212ms.cpp

@@ -0,0 +1,2 @@
+They will greedily take the stones according to a[i]+b[i] in decreasing order. O(sort(n)).
+

1601-1700/1679. Max Number of K-Sum Pairs.pdf

@@ -0,0 +1,2 @@
+the answer is n-1. O(1).
+

1601-1700/1679. Max Number of K-Sum Pairs.txt

@@ -0,0 +1,2 @@
+we can prove that the answer is the maximum digit. O(n).
+

1601-1700/1681. Minimum Incompatibility.txt

@@ -0,0 +1,2 @@
+DP. O(n^2).
+

1601-1700/1682. Longest Palindromic Subsequence II.txt

@@ -0,0 +1,2 @@
+sorting according to max(w,l,h), then compute 2D LIS. O(n log^2 n).
+

1601-1700/1684. Count the Number of Consistent Strings.txt

@@ -0,0 +1,2 @@
+The problem can be solved by a sliding window approach. We can prove that in the optimal solution, we always erase a "maximal" interval which contains unique elements. by "maximal" we mean, fix the left endpoint of the interval, we keep moving the right endpoint to the right, until the interval no longer contain unique elements. Then we can repeatedly move the left endpoint one step to the right, until the interval contains unique elements again. To check whether the current interval contain unique elements, we can use a hash table to maintain the elements within the interval. The running time is O(n).
+

1601-1700/1685. Sum of Absolute Differences in a Sorted Array.txt

@@ -0,0 +1,4 @@
+The problem can be solved by DP. Let's start with a simple solution.
+Let f[i] denote the largest score we can get when we end at position i. The transition is f[i]=max_{i-k<=j<=i-1} f[j]+nums[i]. The final output is f[n-1]. The running time is O(n^2).
+Now we further improve the running time. Use a monotone queue to maintain max_{i-k<=j<=i-1} f[j]. The running time is O(n).
+

1601-1700/1686. Stone Game VI.txt

@@ -1197,9 +1197,34 @@ @inproceedings{coppersmith2002almost
   organization={Springer}
 }
 
+@inproceedings{demaine2009cartesian,
+  title={On cartesian trees and range minimum queries},
+  author={Demaine, Erik D and Landau, Gad M and Weimann, Oren},
+  booktitle={International Colloquium on Automata, Languages, and Programming},
+  pages={341--353},
+  year={2009},
+  organization={Springer}
+}
 
+@inproceedings{chan2014succinct,
+  title={Succinct indices for path minimum, with applications to path reporting},
+  author={Chan, Timothy M and He, Meng and Munro, J Ian and Zhou, Gelin},
+  booktitle={European Symposium on Algorithms},
+  pages={247--259},
+  year={2014},
+  organization={Springer}
+}
 
-
+@article{bjorklund2014determinant,
+  title={Determinant sums for undirected hamiltonicity},
+  author={Bjorklund, Andreas},
+  journal={SIAM Journal on Computing},
+  volume={43},
+  number={1},
+  pages={280--299},
+  year={2014},
+  publisher={SIAM}
+}
 
 
 

1601-1700/1688. Count of Matches in Tournament.txt

@@ -203,13 +203,13 @@ namespace Hash{
 		int x,y,t;
 	}h[(1<<S)+1005];
 	int T=1;
-	void insert(int x,int y){
+	inline void insert(int x,int y){
 		node *p=h+((uint)x*M>>S1);
 		for (;p->t==T;++p)
 			if (p->x==x){p->y=y; return;}
 		p->t=T; p->x=x; p->y=y;
 	}
-	int* find(int x){
+	inline int* find(int x){
 		for (node *p=h+((uint)x*M>>S1);p->t==T;++p)
 			if (p->x==x)return &p->y;
 		return 0;