.

Author: hqztrue

0301-0400/312. Burst Balloons.txt

@@ -0,0 +1,2 @@
+DP, let f[i][j] denote the maximum coins we can collect by busting [i,j]. O(n^3).
+

0701-0800/743. Network Delay Time.txt

@@ -0,0 +1,2 @@
+compute eccentricity, using shortest path in weighted directed graphs. O(m+n log log n) \cite{thorup2004integer}.
+

0701-0800/752. Open the Lock.txt

@@ -0,0 +1,2 @@
+bfs. O(10^4).
+

0701-0800/753. Cracking the Safe.txt

@@ -0,0 +1,2 @@
+Eulerian path, there always exist a solution. O(output).
+

0901-1000/933. Number of Recent Calls.txt

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

0901-1000/934. Shortest Bridge.txt

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

0901-1000/935. Knight Dialer.txt

@@ -0,0 +1,2 @@
+DP, use matrix and fast exponentiation. O(log n).
+

0901-1000/936. Stamping The Sequence.txt

@@ -0,0 +1,4 @@
+note that we are not required to optimize the number of moves.
+use greedy, remove the stamps in reverse order. whenever we find a match, we can remove that substring and replace it by '?'s.
+for a fast implementation, we first find the leftmost full match, it will split the target into two parts. for the left part, the next match must be a prefix of stamp, and we can repeatedly do this until we remove the left part. the right part is of the form ?...?t, and we repeatedly remove a prefix of t that matches with a suffix of stamp. we then find the next full match, which again split t into two parts. the left part is of the form ?...?r?...?. we then repeatedly remove a suffix of r, until r matches a substring of stamp. use string hashing to compare two strings. O(n).
+

0901-1000/937. Reorder Data in Log Files.txt

@@ -0,0 +1,2 @@
+string sorting. O(sort(n,L)).
+

0901-1000/945. Minimum Increment to Make Array Unique 48ms.cpp

@@ -0,0 +1,26 @@
+const int N=40005;
+int c[N];
+class Solution {
+public:
+	int minIncrementForUnique(vector<int>& a) {
+		int n=a.size(),ans=0,m=0,s=0;
+		if (!n)return 0;
+		for (int *i=&a[0],*end=i+n;i!=end;++c[*i],++i)
+			if (*i>m)m=*i;
+		for (int *i=c,*end=c+m;i<=end;++i){
+			s+=*i;
+			if (s>=1)--s,ans+=s;
+		}
+		ans+=s*(s-1)/2;
+		for (int i=0;i<n;++i)--c[a[i]];
+		return ans;
+	}
+};
+
+//IO
+int _IO=[](){
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+

0901-1000/945. Minimum Increment to Make Array Unique.pdf

@@ -0,0 +1,2 @@
+note that the values are unique. simulation. O(n).
+

0901-1000/946. Validate Stack Sequences.txt

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

0901-1000/981. Time Based Key-Value Store.txt

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

0901-1000/982. Triples with Bitwise AND Equal To Zero.pdf

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

0901-1000/983. Minimum Cost For Tickets.txt

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

0901-1000/984. String Without AAA or BBB.txt

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

0901-1000/985. Sum of Even Numbers After Queries.txt

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

0901-1000/986. Interval List Intersections.txt

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

0901-1000/987. Vertical Order Traversal of a Binary Tree.txt

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

0901-1000/989. Add to Array-Form of Integer.txt

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

0901-1000/993. Cousins in Binary Tree.txt

@@ -0,0 +1,2 @@
+greedily flip the leftmost 1. O(n).
+

0901-1000/994. Rotting Oranges.txt

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

0901-1000/995. Minimum Number of K Consecutive Bit Flips.txt

@@ -1,2 +1,3 @@
 1. suffix tree. O(n).
 2. minimum representation. O(n).
+

0901-1000/997. Find the Town Judge.txt

@@ -8,11 +8,12 @@ void dfs(int x){
 		if (k<=t)break;
 		if (!v[k])dfs(k);
 	}
-	if (x>t)
-	for (int j=1;j<=p1;++j){
-		int k=x*p[j];
-		if (k>n)break;
-		if (!v[k])dfs(k);
+	if (x>t){
+		for (int j=1;j<=p1;++j){
+			int k=x*p[j];
+			if (k>n)break;
+			if (!v[k])dfs(k);
+		}
 	}
 }
 class Solution {

1101-1200/1163. Last Substring in Lexicographical Order.txt

@@ -0,0 +1,31 @@
+const int N=40005;
+int c[N],del[N];
+class Solution {
+public:
+	int eatenApples(vector<int>& a, vector<int>& d) {
+		int n=a.size(),m=0,s=0,s1=0,t,ans=0;
+		for (int i=n-1;i>=0;--i)
+			if (i+d[i]>m)m=i+d[i];
+		memcpy(c,&a[0],sizeof(int)*n);
+		memset(c+n,0,sizeof(int)*(m+1-n));
+		memset(del,0,sizeof(int)*(m+1));
+		for (int i=0;i<n;++i)del[i+d[i]]+=a[i];
+		for (int i=0;i<=m;++i){
+			s+=c[i]; t=del[i];
+			if (t){
+				if (s1>=t)s1-=t;
+				else s-=t-s1,s1=0;
+			}
+			if (s)++s1,--s,++ans;
+		}
+		return ans;
+	}
+};
+
+//IO
+int _IO=[](){
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+

1601-1700/1627. Graph Connectivity With Threshold 284ms.cpp

@@ -0,0 +1,63 @@
+typedef unsigned int uint;
+const int N=1000005,u=3,U=1<<u,L=30/u;
+struct node{
+	node* c[U];
+	int min; uint w;
+}p[N],*root=p;
+int g[U][U][1<<U],p1;
+void init(){
+	for (int i=0;i<(1<<U);++i)
+		for (int j=0;j<U;++j){
+			int ma=-1,t=-1;
+			for (int k=0;k<U;++k){
+				if ((i&(1<<k))&&(j^k)>ma)ma=j^k,t=k;
+				g[j][k][i]=t;
+			}
+		}
+}
+class Solution {
+public:
+	vector<int> maximizeXor(vector<int>& a, vector<vector<int>>& q) {
+		int n=a.size(),m=q.size(),mi=1<<30; vector<int> ans(m);
+		root->w=0; p1=0;
+		for (int i=0;i<n;++i){
+			node *x=root; int y=a[i];
+			if (y<mi)mi=y;
+			for (int j=L-1;j>=0;--j){
+				int t=(y>>j*u)&(U-1);
+				if (x->w&(1<<t)){
+					x=x->c[t];
+					int &z=x->min; if (y<z)z=y;
+				}
+				else {
+					x->w|=1<<t;
+					x=x->c[t]=p+(++p1);
+					x->w=0; x->min=y;
+				}
+			}
+		}
+		for (int i=0;i<m;++i){
+			node *x=root; int y=q[i][0],b=q[i][1],j;
+			if (b<mi){ans[i]=-1; continue;}
+			for (j=L-1;j>=0;--j){
+				int t=(y>>j*u)&(U-1);
+				int r=(b>>j*u)&(U-1),h=g[t][r][x->w];
+				if (h<r){x=x->c[h]; break;}
+				else if (x->c[h]->min>b){x=x->c[g[t][r][x->w-(1<<r)]]; break;}
+				x=x->c[h];
+			}
+			for (--j;j>=0;--j)x=x->c[g[(y>>j*u)&(U-1)][U-1][x->w]];
+			ans[i]=x->min^y;
+		}
+		return ans;
+	}
+};
+
+//IO
+int _IO=[](){
+	init();
+	ios::sync_with_stdio(0);
+	cin.tie(0); //cout.tie(0);
+	return 0;
+}();
+

1701-1800/1705. Maximum Number of Eaten Apples 88ms.cpp

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+the integers are distinct. find the maximum of the first n-k+1 numbers. O(n).
+

1701-1800/1705. Maximum Number of Eaten Apples.pdf

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

1701-1800/1705. Maximum Number of Eaten Apples.txt

@@ -0,0 +1,2 @@
+Reduce to two sum. O(n log U).
+

1701-1800/1707. Maximum XOR With an Element From Array 424ms.cpp

@@ -0,0 +1,2 @@
+enumerate the first split, use monotone pointers to maintain the range of the second split. O(n).
+

1701-1800/1707. Maximum XOR With an Element From Array.pdf

@@ -0,0 +1,2 @@
+LCS. the numbers in target are distinct, so we can reduce to LIS. O(n log log n) \cite{fredman1975computing}.
+

1701-1800/1707. Maximum XOR With an Element From Array.txt

@@ -1237,4 +1237,16 @@ @article{asathulla2019faster
   publisher={ACM New York, NY, USA}
 }
 
+@article{thorup2004integer,
+  title={Integer priority queues with decrease key in constant time and the single source shortest paths problem},
+  author={Thorup, Mikkel},
+  journal={Journal of Computer and System Sciences},
+  volume={69},
+  number={3},
+  pages={330--353},
+  year={2004},
+  publisher={Elsevier}
+}
+
+