LeetCode, GeeksForGeeks Problem of the day solution
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class Solution {
public:
int constrainedSubsetSum(vector& nums, int k) {
deque dq;
int s=INT_MIN;
for(int i=0;i0)
dq.push_back(nums[i]);
if(i>=k && dq.front()==nums[i-k] && !dq.empty()) dq.pop_front();
}
return s;
}
};
class Solution
{
public:
long long sumOfDivisors(int N)
{
long long ans=0;
for(int i=1;i<=N;i++){
ans += i*(N/i);
}
return ans;
}
};
class NestedIterator {
public:
vector v;
int index;
void Recursion(vector& nums) {
int i=0;
while(i &nestedList) {
index=0;
Recursion(nestedList);
}
int next() {
if(hasNext()){
return v[index++];
}
return -1;
}
bool hasNext() {
if(index
class Solution {
public:
int isPossible(int N, int arr[]) {
long long sum = 0;
for(int i=0;i
class Solution {
public:
bool backspaceCompare(string s, string t) {
stack st1, st2;
for(int i=0;i0){
st1.pop();
}
else if(s[i]!='#'){
st1.push(s[i]);
}
}
for(int i=0;i0){
st2.pop();
}
else if(t[i]!='#'){
st2.push(t[i]);
}
}
return st1==st2;
}
};
class Solution
{
public:
//Function to find the level of node X.
int nodeLevel(int V, vector adj[], int X)
{
queue q;
vector vis(V, 0);
int level = 0;
q.push(0);
vis[0] = 1;
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
int adjnode = q.front();
q.pop();
if (adjnode == X) {
return level;
}
for (auto x : adj[adjnode]) {
if (!vis[x]) {
vis[x] = 1;
q.push(x);
}
}
}
level++;
}
return -1;
}
};
class Solution {
public:
int minimumTime(int n, vector>& relations, vector& time) {
vectoradj[n];
vectorindegree(n,0);
for(auto relation:relations){
int u = relation[0]-1;
int v = relation[1]-1;
adj[u].push_back(v);
indegree[v]++;
}
queueq;
vectormaxTime(n,0);
for(int i=0;i
class Solution {
public:
vector eventualSafeNodes(int V, vector adj[]) {
int indegree[V]={0};
vector adjR[V];
for(int i=0;i q;
for(int i=0;i ans;
while(!q.empty()){
int node = q.front();
q.pop();
ans.push_back(node);
for(auto it:adjR[node]){
indegree[it]--;
if(indegree[it]==0){
q.push(it);
}
}
}
sort(ans.begin(), ans.end());
return ans;
}
};
class Solution {
public:
int findRoot(int n, vector<int>& left, vector<int>& right) {
unordered_set<int> children;
for(int i = 0; i < left.size(); i++) {
children.insert(left[i]);
}
for(int i = 0; i < right.size(); i++) {
children.insert(right[i]);
}
for(int i = 0; i < n; i++) {
if(children.find(i) == children.end()) {
return i;
}
}
return -1;
}
bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
int root = findRoot(n, leftChild, rightChild);
if(root == -1) {
return false;
}
unordered_set<int> seen;
stack<int> st;
seen.insert(root);
st.push(root);
while(!st.empty()) {
int node = st.top();
st.pop();
int children[] = {leftChild[node], rightChild[node]};
for(int i : children) {
if(i != -1) {
if(seen.find(i) != seen.end()) {
return false;
}
st.push(i);
seen.insert(i);
}
}
}
return seen.size() == n;
}
};
class Solution{
public:
vector> transitiveClosure(int N, vector> graph)
{
for(int via = 0; via < N; via++)
for(int src = 0; src < N; src++)
for(int dest = 0; dest < N; dest++)
if(graph[src][via] && graph[via][dest] || src == dest)
graph[src][dest] = 1;
return graph;
}
};
