You are given an integer n and an integer p representing an array arr of length n where all elements are set to 0's, except position p which is set to 1. You are also given an integer array banned containing restricted positions. Perform the following operation on arr:
Reverse a subarray with size k if the single 1 is not set to a position in banned.
Return an integer array answer with n results where the ith result isthe minimum number of operations needed to bring the single 1 to position i in arr, or -1 if it is impossible.
Example 1:
Input:n = 4, p = 0, banned = [1,2], k = 4
Output:[0,-1,-1,1]
Explanation:
Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
We can never place 1 on the banned positions, so the answer for positions 1 and 2 is -1.
Perform the operation of size 4 to reverse the whole array.
After a single operation 1 is at position 3 so the answer for position 3 is 1.
Example 2:
Input:n = 5, p = 0, banned = [2,4], k = 3
Output:[0,-1,-1,-1,-1]
Explanation:
Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0.
We cannot perform the operation on the subarray positions [0, 2] because position 2 is in banned.
Because 1 cannot be set at position 2, it is impossible to set 1 at other positions in more operations.
Example 3:
Input:n = 4, p = 2, banned = [0,1,3], k = 1
Output:[-1,-1,0,-1]
Explanation:
Perform operations of size 1 and 1 never changes its position.
Constraints:
1 <= n <= 105
0 <= p <= n - 1
0 <= banned.length <= n - 1
0 <= banned[i] <= n - 1
1 <= k <= n
banned[i] != p
all values in banned are unique
Solutions
Solution 1: Ordered Set + BFS
We notice that for any index \(i\) in the subarray interval \([l,..r]\), the flipped index \(j = l + r - i\).
If the subarray moves one position to the right, then \(j = l + 1 + r + 1 - i = l + r - i + 2\), that is, \(j\) will increase by \(2\).
Similarly, if the subarray moves one position to the left, then \(j = l - 1 + r - 1 - i = l + r - i - 2\), that is, \(j\) will decrease by \(2\).
Therefore, for a specific index \(i\), all its flipped indices form an arithmetic progression with common difference \(2\), that is, all the flipped indices have the same parity.
Next, we consider the range of values ββof the index \(i\) after flipping \(j\).
If the boundary is not considered, the range of values ββof \(j\) is \([i - k + 1, i + k - 1]\).
If the subarray is on the left, then \([l, r] = [0, k - 1]\), so the flipped index \(j\) of \(i\) is \(0 + k - 1 - i\), that is, \(j = k - i - 1\), so the left boundary \(mi = max(i - k + 1, k - i - 1)\).
If the subarray is on the right, then \([l, r] = [n - k, n - 1]\), so the flipped index \(j= n - k + n - 1 - i\) is \(j = n \times 2 - k - i - 1\), so the right boundary of \(j\) is \(mx = min(i + k - 1, n \times 2 - k - i - 1)\).
We use two ordered sets to store all the odd indices and even indices to be searched, here we need to exclude the indices in the array \(banned\) and the index \(p\).
Then we use BFS to search, each time searching all the flipped indices \(j\) of the current index \(i\), that is, \(j = mi, mi + 2, mi + 4, \dots, mx\), updating the answer of index \(j\) and adding index \(j\) to the search queue, and removing index \(j\) from the corresponding ordered set.
When the search is over, the answer to all indices can be obtained.
The time complexity is \(O(n \times \log n)\) and the space complexity is \(O(n)\). Where \(n\) is the given array length in the problem.
functionminReverseOperations(n:number,p:number,banned:number[],k:number):number[]{constans:number[]=Array(n).fill(-1);constts=[newTreeSet<number>(),newTreeSet<number>()];for(leti=0;i<n;++i){ts[i%2].add(i);}ans[p]=0;ts[p%2].delete(p);for(constiofbanned){ts[i%2].delete(i);}ts[0].add(n);ts[1].add(n);letq=[p];while(q.length){constt:number[]=[];for(constiofq){constmi=Math.max(i-k+1,k-i-1);constmx=Math.min(i+k-1,n*2-k-i-1);consts=ts[mi%2];for(letj=s.ceil(mi)!;j<=mx;j=s.ceil(j)!){t.push(j);ans[j]=ans[i]+1;s.delete(j);}}q=t;}returnans;}typeCompare<T>=(lhs:T,rhs:T)=>number;classRBTreeNode<T=number>{data:T;count:number;left:RBTreeNode<T>|null;right:RBTreeNode<T>|null;parent:RBTreeNode<T>|null;color:number;constructor(data:T){this.data=data;this.left=this.right=this.parent=null;this.color=0;this.count=1;}sibling():RBTreeNode<T>|null{if(!this.parent)returnnull;// sibling null if no parentreturnthis.isOnLeft()?this.parent.right:this.parent.left;}isOnLeft():boolean{returnthis===this.parent!.left;}hasRedChild():boolean{return(Boolean(this.left&&this.left.color===0)||Boolean(this.right&&this.right.color===0));}}classRBTree<T>{root:RBTreeNode<T>|null;lt:(l:T,r:T)=>boolean;constructor(compare:Compare<T>=(l:T,r:T)=>(l<r?-1:l>r?1:0)){this.root=null;this.lt=(l:T,r:T)=>compare(l,r)<0;}rotateLeft(pt:RBTreeNode<T>):void{constright=pt.right!;pt.right=right.left;if(pt.right)pt.right.parent=pt;right.parent=pt.parent;if(!pt.parent)this.root=right;elseif(pt===pt.parent.left)pt.parent.left=right;elsept.parent.right=right;right.left=pt;pt.parent=right;}rotateRight(pt:RBTreeNode<T>):void{constleft=pt.left!;pt.left=left.right;if(pt.left)pt.left.parent=pt;left.parent=pt.parent;if(!pt.parent)this.root=left;elseif(pt===pt.parent.left)pt.parent.left=left;elsept.parent.right=left;left.right=pt;pt.parent=left;}swapColor(p1:RBTreeNode<T>,p2:RBTreeNode<T>):void{consttmp=p1.color;p1.color=p2.color;p2.color=tmp;}swapData(p1:RBTreeNode<T>,p2:RBTreeNode<T>):void{consttmp=p1.data;p1.data=p2.data;p2.data=tmp;}fixAfterInsert(pt:RBTreeNode<T>):void{letparent=null;letgrandParent=null;while(pt!==this.root&&pt.color!==1&&pt.parent?.color===0){parent=pt.parent;grandParent=pt.parent.parent;/* Case : A Parent of pt is left child of Grand-parent of pt */if(parent===grandParent?.left){constuncle=grandParent.right;/* Case : 1 The uncle of pt is also red Only Recoloring required */if(uncle&&uncle.color===0){grandParent.color=0;parent.color=1;uncle.color=1;pt=grandParent;}else{/* Case : 2 pt is right child of its parent Left-rotation required */if(pt===parent.right){this.rotateLeft(parent);pt=parent;parent=pt.parent;}/* Case : 3 pt is left child of its parent Right-rotation required */this.rotateRight(grandParent);this.swapColor(parent!,grandParent);pt=parent!;}}else{/* Case : B Parent of pt is right child of Grand-parent of pt */constuncle=grandParent!.left;/* Case : 1 The uncle of pt is also red Only Recoloring required */if(uncle!=null&&uncle.color===0){grandParent!.color=0;parent.color=1;uncle.color=1;pt=grandParent!;}else{/* Case : 2 pt is left child of its parent Right-rotation required */if(pt===parent.left){this.rotateRight(parent);pt=parent;parent=pt.parent;}/* Case : 3 pt is right child of its parent Left-rotation required */this.rotateLeft(grandParent!);this.swapColor(parent!,grandParent!);pt=parent!;}}}this.root!.color=1;}delete(val:T):boolean{constnode=this.find(val);if(!node)returnfalse;node.count--;if(!node.count)this.deleteNode(node);returntrue;}deleteAll(val:T):boolean{constnode=this.find(val);if(!node)returnfalse;this.deleteNode(node);returntrue;}deleteNode(v:RBTreeNode<T>):void{constu=BSTreplace(v);// True when u and v are both blackconstuvBlack=(u===null||u.color===1)&&v.color===1;constparent=v.parent!;if(!u){// u is null therefore v is leafif(v===this.root)this.root=null;// v is root, making root nullelse{if(uvBlack){// u and v both black// v is leaf, fix double black at vthis.fixDoubleBlack(v);}else{// u or v is redif(v.sibling()){// sibling is not null, make it red"v.sibling()!.color=0;}}// delete v from the treeif(v.isOnLeft())parent.left=null;elseparent.right=null;}return;}if(!v.left||!v.right){// v has 1 childif(v===this.root){// v is root, assign the value of u to v, and delete uv.data=u.data;v.left=v.right=null;}else{// Detach v from tree and move u upif(v.isOnLeft())parent.left=u;elseparent.right=u;u.parent=parent;if(uvBlack)this.fixDoubleBlack(u);// u and v both black, fix double black at uelseu.color=1;// u or v red, color u black}return;}// v has 2 children, swap data with successor and recursethis.swapData(u,v);this.deleteNode(u);// find node that replaces a deleted node in BSTfunctionBSTreplace(x:RBTreeNode<T>):RBTreeNode<T>|null{// when node have 2 childrenif(x.left&&x.right)returnsuccessor(x.right);// when leafif(!x.left&&!x.right)returnnull;// when single childreturnx.left??x.right;}// find node that do not have a left child// in the subtree of the given nodefunctionsuccessor(x:RBTreeNode<T>):RBTreeNode<T>{lettemp=x;while(temp.left)temp=temp.left;returntemp;}}fixDoubleBlack(x:RBTreeNode<T>):void{if(x===this.root)return;// Reached rootconstsibling=x.sibling();constparent=x.parent!;if(!sibling){// No sibiling, double black pushed upthis.fixDoubleBlack(parent);}else{if(sibling.color===0){// Sibling redparent.color=0;sibling.color=1;if(sibling.isOnLeft())this.rotateRight(parent);// left caseelsethis.rotateLeft(parent);// right casethis.fixDoubleBlack(x);}else{// Sibling blackif(sibling.hasRedChild()){// at least 1 red childrenif(sibling.left&&sibling.left.color===0){if(sibling.isOnLeft()){// left leftsibling.left.color=sibling.color;sibling.color=parent.color;this.rotateRight(parent);}else{// right leftsibling.left.color=parent.color;this.rotateRight(sibling);this.rotateLeft(parent);}}else{if(sibling.isOnLeft()){// left rightsibling.right!.color=parent.color;this.rotateLeft(sibling);this.rotateRight(parent);}else{// right rightsibling.right!.color=sibling.color;sibling.color=parent.color;this.rotateLeft(parent);}}parent.color=1;}else{// 2 black childrensibling.color=0;if(parent.color===1)this.fixDoubleBlack(parent);elseparent.color=1;}}}}insert(data:T):boolean{// search for a position to insertletparent=this.root;while(parent){if(this.lt(data,parent.data)){if(!parent.left)break;elseparent=parent.left;}elseif(this.lt(parent.data,data)){if(!parent.right)break;elseparent=parent.right;}elsebreak;}// insert node into parentconstnode=newRBTreeNode(data);if(!parent)this.root=node;elseif(this.lt(node.data,parent.data))parent.left=node;elseif(this.lt(parent.data,node.data))parent.right=node;else{parent.count++;returnfalse;}node.parent=parent;this.fixAfterInsert(node);returntrue;}find(data:T):RBTreeNode<T>|null{letp=this.root;while(p){if(this.lt(data,p.data)){p=p.left;}elseif(this.lt(p.data,data)){p=p.right;}elsebreak;}returnp??null;}*inOrder(root:RBTreeNode<T>=this.root!):Generator<T,undefined,void>{if(!root)return;for(constvofthis.inOrder(root.left!))yieldv;yieldroot.data;for(constvofthis.inOrder(root.right!))yieldv;}*reverseInOrder(root:RBTreeNode<T>=this.root!):Generator<T,undefined,void>{if(!root)return;for(constvofthis.reverseInOrder(root.right!))yieldv;yieldroot.data;for(constvofthis.reverseInOrder(root.left!))yieldv;}}classTreeSet<T=number>{_size:number;tree:RBTree<T>;compare:Compare<T>;constructor(collection:T[]|Compare<T>=[],compare:Compare<T>=(l:T,r:T)=>(l<r?-1:l>r?1:0),){if(typeofcollection==='function'){compare=collection;collection=[];}this._size=0;this.compare=compare;this.tree=newRBTree(compare);for(constvalofcollection)this.add(val);}size():number{returnthis._size;}has(val:T):boolean{return!!this.tree.find(val);}add(val:T):boolean{constsuccessful=this.tree.insert(val);this._size+=successful?1:0;returnsuccessful;}delete(val:T):boolean{constdeleted=this.tree.deleteAll(val);this._size-=deleted?1:0;returndeleted;}ceil(val:T):T|undefined{letp=this.tree.root;lethigher=null;while(p){if(this.compare(p.data,val)>=0){higher=p;p=p.left;}else{p=p.right;}}returnhigher?.data;}floor(val:T):T|undefined{letp=this.tree.root;letlower=null;while(p){if(this.compare(val,p.data)>=0){lower=p;p=p.right;}else{p=p.left;}}returnlower?.data;}higher(val:T):T|undefined{letp=this.tree.root;lethigher=null;while(p){if(this.compare(val,p.data)<0){higher=p;p=p.left;}else{p=p.right;}}returnhigher?.data;}lower(val:T):T|undefined{letp=this.tree.root;letlower=null;while(p){if(this.compare(p.data,val)<0){lower=p;p=p.right;}else{p=p.left;}}returnlower?.data;}first():T|undefined{returnthis.tree.inOrder().next().value;}last():T|undefined{returnthis.tree.reverseInOrder().next().value;}shift():T|undefined{constfirst=this.first();if(first===undefined)returnundefined;this.delete(first);returnfirst;}pop():T|undefined{constlast=this.last();if(last===undefined)returnundefined;this.delete(last);returnlast;}*[Symbol.iterator]():Generator<T,void,void>{for(constvalofthis.values())yieldval;}*keys():Generator<T,void,void>{for(constvalofthis.values())yieldval;}*values():Generator<T,undefined,void>{for(constvalofthis.tree.inOrder())yieldval;returnundefined;}/** * Return a generator for reverse order traversing the set */*rvalues():Generator<T,undefined,void>{for(constvalofthis.tree.reverseInOrder())yieldval;returnundefined;}}classTreeMultiSet<T=number>{_size:number;tree:RBTree<T>;compare:Compare<T>;constructor(collection:T[]|Compare<T>=[],compare:Compare<T>=(l:T,r:T)=>(l<r?-1:l>r?1:0),){if(typeofcollection==='function'){compare=collection;collection=[];}this._size=0;this.compare=compare;this.tree=newRBTree(compare);for(constvalofcollection)this.add(val);}size():number{returnthis._size;}has(val:T):boolean{return!!this.tree.find(val);}add(val:T):boolean{constsuccessful=this.tree.insert(val);this._size++;returnsuccessful;}delete(val:T):boolean{constsuccessful=this.tree.delete(val);if(!successful)returnfalse;this._size--;returntrue;}count(val:T):number{constnode=this.tree.find(val);returnnode?node.count:0;}ceil(val:T):T|undefined{letp=this.tree.root;lethigher=null;while(p){if(this.compare(p.data,val)>=0){higher=p;p=p.left;}else{p=p.right;}}returnhigher?.data;}floor(val:T):T|undefined{letp=this.tree.root;letlower=null;while(p){if(this.compare(val,p.data)>=0){lower=p;p=p.right;}else{p=p.left;}}returnlower?.data;}higher(val:T):T|undefined{letp=this.tree.root;lethigher=null;while(p){if(this.compare(val,p.data)<0){higher=p;p=p.left;}else{p=p.right;}}returnhigher?.data;}lower(val:T):T|undefined{letp=this.tree.root;letlower=null;while(p){if(this.compare(p.data,val)<0){lower=p;p=p.right;}else{p=p.left;}}returnlower?.data;}first():T|undefined{returnthis.tree.inOrder().next().value;}last():T|undefined{returnthis.tree.reverseInOrder().next().value;}shift():T|undefined{constfirst=this.first();if(first===undefined)returnundefined;this.delete(first);returnfirst;}pop():T|undefined{constlast=this.last();if(last===undefined)returnundefined;this.delete(last);returnlast;}*[Symbol.iterator]():Generator<T,void,void>{yield*this.values();}*keys():Generator<T,void,void>{for(constvalofthis.values())yieldval;}*values():Generator<T,undefined,void>{for(constvalofthis.tree.inOrder()){letcount=this.count(val);while(count--)yieldval;}returnundefined;}/** * Return a generator for reverse order traversing the multi-set */*rvalues():Generator<T,undefined,void>{for(constvalofthis.tree.reverseInOrder()){letcount=this.count(val);while(count--)yieldval;}returnundefined;}}