You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.
The second greater integer of nums[i] is nums[j] such that:
j > i
nums[j] > nums[i]
There exists exactly one index k such that nums[k] > nums[i] and i < k < j.
If there is no such nums[j], the second greater integer is considered to be -1.
For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3,Β and that of 3 and 4 is -1.
Return an integer array answer, where answer[i] is the second greater integer of nums[i].
Β
Example 1:
Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].
Example 2:
Input: nums = [3,3]
Output: [-1,-1]
Explanation:
We return [-1,-1] since neither integer has any integer greater than it.
Β
Constraints:
1 <= nums.length <= 105
0 <= nums[i] <= 109
Solutions
Solution 1: Sorting + Ordered Set
We can convert the elements in the array into pairs \((x, i)\), where \(x\) is the value of the element and \(i\) is the index of the element. Then sort by the value of the elements in descending order.
Next, we traverse the sorted array, maintaining an ordered set that stores the indices of the elements. When we traverse to the element \((x, i)\), the indices of all elements greater than \(x\) are already in the ordered set. We only need to find the index \(j\) of the next element after \(i\) in the ordered set, then the element corresponding to \(j\) is the second largest element of \(x\). Then, we add \(i\) to the ordered set. Continue to traverse the next element.
The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array.
functionsecondGreaterElement(nums:number[]):number[]{constn=nums.length;constarr:number[][]=[];for(leti=0;i<n;++i){arr.push([nums[i],i]);}arr.sort((a,b)=>(a[0]==b[0]?a[1]-b[1]:b[0]-a[0]));constans=Array(n).fill(-1);constts=newTreeSet<number>();for(const[_,i]ofarr){letj=ts.higher(i);if(j!==undefined){j=ts.higher(j);if(j!==undefined){ans[i]=nums[j];}}ts.add(i);}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;}}
Solution 2: Double Stacks
We maintain two decreasing monotonic stacks:
stackOne: stores elements that have not yet encountered any greater element to their right.
stackTwo: stores elements that have encountered exactly one greater element to their right.
Algorithm:
As we iterate through the array nums, for current element \(nums[k]\), have these steps:
While stackTwo is not empty and its top element is less than \(nums[k]\), these top elements have found their second next greater element. Pop them and record \(nums[k]\) as the answer for their respective indices.
While stackOne is not empty and its top element is less than \(nums[k]\), these top elements have found their first next greater element. Pop them to move them to a temporary list/vector called transporter.
Pop all elements from the back of transporter and push them onto stackTwo. These elements will naturally maintain the decreasing order in stackTwo.
Push current element \(nums[k]\) into stackOne for future comparisons.
Our time complexity is \(O(n)\), where \(n\) is the length of the array nums.
This is because each element is pushed and popped across two stacks for at most 4 times in total.
Space complexity is \(O(n)\), as all elements must be stored across two stacks.
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classSolution:defsecondGreaterElement(self,nums:list[int])->list[int]:second_next_greater=[-1]*len(nums)stack_1:list[tuple[int,int]]=[]# Decreasing monotonic stacks: (num, idx).stack_2:list[tuple[int,int]]=[]# Transport tuples from stack 1 to stack 2.transporter:list[tuple[int,int]]=[]foridx,numinenumerate(nums):whilestack_2andstack_2[-1][0]<num:_,past_idx=stack_2.pop(-1)second_next_greater[past_idx]=numwhilestack_1andstack_1[-1][0]<num:transporter.append(stack_1.pop(-1))whiletransporter:stack_2.append(transporter.pop(-1))# Ensure decreasing monotonicity.stack_1.append((num,idx))returnsecond_next_greater