LeetCode, GeeksForGeeks Problem of the day solution
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class Solution
{
public:
/*You are required to complete below method */
void inorder(Node *root,int &sum)
{
if(root==NULL)
return ;
if(root->left==NULL && root->right==NULL)
{ sum=sum+root->data;
return ;}
inorder(root->left,sum);
inorder(root->right,sum);
}
public:
int sumOfLeafNodes(Node *root ){
int sum=0;
inorder(root,sum);
return sum;
}
};
class Solution {
public:
int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
priority_queue<int, vector<int>, greater<int>> pq;
for(int i = 1; i < heights.size(); i++) {
int diff = heights[i] - heights[i-1];
if(diff > 0) pq.push(diff);
if(pq.size() > ladders) {
if(pq.top() <= bricks) bricks -= pq.top(), pq.pop();
else return i-1;
}
}
return heights.size()-1;
}
};
class Solution{
public:
bool isMaxHeap(int arr[], int n)
{
for(int i=0;i<=(n-2)/2;i++) {
if(arr[2*i+1] > arr[i]) {
return false;
}
else if(2*i+2<n && arr[2*i+2] > arr[i]) {
return false;
}
}
return true;
}
};
class Solution {
public:
int findLeastNumOfUniqueInts(vector& arr, int k) {
unordered_mapmp;
for(auto it:arr) mp[it]++;
vector>dp;
for(auto it:mp){
dp.push_back(it);
}
sort(begin(dp),end(dp),[](auto a,auto b){
return a.second=it.second){
k-=it.second;
count++;
}else{
break;
}
}
return size(dp)-count;
}
};
class Solution
{
public:
void helper(Node *root,Node * &newNode,Node * &r){
if(!root) return ;
helper(root->left,newNode,r);
if(!newNode) newNode = root;
if(r){
r->right = root;
r->left = NULL;
}
r = root;
helper(root->right,newNode,r);
}
public:
Node *flattenBST(Node *root)
{
Node * newNode = NULL,*r = NULL;
helper(root,newNode,r);
r->left = NULL;
return newNode;
}
};
class Solution {
public:
long long largestPerimeter(vector& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
long long ans = 0, cur = 0;
for(int i=0; i nums[i] && i >= 2){
ans = max(cur, ans);
ans += nums[i];
}
cur += nums[i];
}
if(ans == 0) return -1;
return ans ;
}
};
class Solution {
public:
int isPossible(vector>paths){
for(int i=0;i
class Solution {
public:
long long largestPerimeter(vector& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
long long ans = 0, cur = 0;
for(int i=0; i nums[i] && i >= 2){
ans = max(cur, ans);
ans += nums[i];
}
cur += nums[i];
}
if(ans == 0) return -1;
return ans ;
}
};
class Solution {
public:
vector rearrangeArray(vector& nums) {
vector positive;
vector negative;
vector ans;
int n = nums.size();
for(int i = 0; i < n; i++){
if(nums[i] < 0){
negative.push_back(nums[i]);
}
else{
positive.push_back(nums[i]);
}
}
for(int i = 0; i < n/2; i++){
ans.push_back(positive[i]);
ans.push_back(negative[i]);
}
return ans;
}
};
class Solution {
public:
void dfs(int node , int parent , vector&low ,vector&tin , vector&vis , vector adj[] , int &timer, vector> & bridges ){
vis[node] = 1;
low[node] = timer;
tin[node] = timer;
timer++;
for(auto nbr : adj[node]){
if(nbr == parent) continue;
else if(!vis[nbr]){
dfs(nbr,node , low , tin , vis , adj , timer , bridges);
low[node] = min(low[node] , low[nbr]);
if(low[nbr] > tin[node]){
if(nbr> node) bridges.push_back({node,nbr});
else bridges.push_back({nbr,node});
}
}
else{
low[node] = min(low[node] , low[nbr]);
}
}
}
vector>criticalConnections(int v, vector adj[]){
// Code here
vector> bridges;
vector low(v,0);
vector tin(v,0);
vector vis(v,0);
int parent = -1;
int timer = 0;
dfs(0 , parent , low , tin , vis , adj , timer , bridges);
sort(bridges.begin() , bridges.end());
return bridges;
}
};
