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Weekly Contest 284 LeetCode Solution
Problem 1 – Find All K-Distant Indices in an Array Leetcode Solution
You are given a 0-indexed integer array nums and two integers key and k. A k-distant index is an index i of nums for which there exists at least one index j such that |i - j| <= k and nums[j] == key.
Return a list of all k-distant indices sorted in increasing order.
Example 1:
Input: nums = [3,4,9,1,3,9,5], key = 9, k = 1
Output: [1,2,3,4,5,6]
Explanation: Here, nums[2] == key and nums[5] == key.
- For index 0, |0 - 2| > k and |0 - 5| > k, so there is no j where |0 - j| <= k and nums[j] == key. Thus, 0 is not a k-distant index.
- For index 1, |1 - 2| <= k and nums[2] == key, so 1 is a k-distant index.
- For index 2, |2 - 2| <= k and nums[2] == key, so 2 is a k-distant index.
- For index 3, |3 - 2| <= k and nums[2] == key, so 3 is a k-distant index.
- For index 4, |4 - 5| <= k and nums[5] == key, so 4 is a k-distant index.
- For index 5, |5 - 5| <= k and nums[5] == key, so 5 is a k-distant index.
- For index 6, |6 - 5| <= k and nums[5] == key, so 6 is a k-distant index.
Thus, we return [1,2,3,4,5,6] which is sorted in increasing order.
Example 2:
Input: nums = [2,2,2,2,2], key = 2, k = 2
Output: [0,1,2,3,4]
Explanation: For all indices i in nums, there exists some index j such that |i - j| <= k and nums[j] == key, so every index is a k-distant index.
Hence, we return [0,1,2,3,4].
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 1000keyis an integer from the arraynums.1 <= k <= nums.length
Find All K-Distant Indices in an Array Leetcode Solution in C++
vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
vector<int> res;
for (int i = 0, j = 0; i < nums.size(); ++i)
if (nums[i] == key)
for (j = max(j, i - k); j <= i + k && j < nums.size(); ++j)
res.push_back(j);
return res;
}
Find All K-Distant Indices in an Array Leetcode Solution in Java
public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
List<Integer> idx = new ArrayList<>();
List<Integer> ans = new ArrayList<>();
for(int i = 0 ; i < nums.length; i++){
if(nums[i] == key){
idx.add(i);
}
}
int last = 0;
for(int ind : idx){
int i = Math.max(last,ind-k);
for(; i <= ind+k && i < nums.length; i++){
ans.add(i);
}
last = i;
}
return ans;
}
Find All K-Distant Indices in an Array Leetcode Solution in Python
class Solution:
def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
lis=deque([])
prev_popped=-1
for i in range(len(nums)):
if(nums[i]==key):
lis.append(i)
ans=[]
for i in range(len(nums)):
if(len(lis)>0 and lis[0]<i):
prev_popped = lis.popleft()
if(prev_popped!=-1 and (i-prev_popped) <=k):
ans.append(i)
elif(len(lis)>0 and (lis[0]-i)<=k):
ans.append(i)
return ans
Problem 2 – Count Artifacts That Can Be Extracted Leetcode Solution
There is an n x n 0-indexed grid with some artifacts buried in it. You are given the integer n and a 0-indexed 2D integer array artifacts describing the positions of the rectangular artifacts where artifacts[i] = [r1i, c1i, r2i, c2i] denotes that the ith artifact is buried in the subgrid where:
(r1i, c1i)is the coordinate of the top-left cell of theithartifact and(r2i, c2i)is the coordinate of the bottom-right cell of theithartifact.
You will excavate some cells of the grid and remove all the mud from them. If the cell has a part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact are uncovered, you can extract it.
Given a 0-indexed 2D integer array dig where dig[i] = [ri, ci] indicates that you will excavate the cell (ri, ci), return the number of artifacts that you can extract.
The test cases are generated such that:
- No two artifacts overlap.
- Each artifact only covers at most
4cells. - The entries of
digare unique.
Example 1:
Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1]]
Output: 1
Explanation:
The different colors represent different artifacts. Excavated cells are labeled with a 'D' in the grid.
There is 1 artifact that can be extracted, namely the red artifact.
The blue artifact has one part in cell (1,1) which remains uncovered, so we cannot extract it.
Thus, we return 1.
Example 2:
Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1],[1,1]]
Output: 2
Explanation: Both the red and blue artifacts have all parts uncovered (labeled with a 'D') and can be extracted, so we return 2.
Constraints:
1 <= n <= 10001 <= artifacts.length, dig.length <= min(n2, 105)artifacts[i].length == 4dig[i].length == 20 <= r1i, c1i, r2i, c2i, ri, ci <= n - 1r1i <= r2ic1i <= c2i- No two artifacts will overlap.
- The number of cells covered by an artifact is at most
4. - The entries of
digare unique.
Count Artifacts That Can Be Extracted Leetcode Solution in Java
class Solution {
public int digArtifacts(int n, int[][] artifacts, int[][] dig) {
HashSet<String> set = new HashSet<>();
for (int d[] : dig) set.add(d[0] + " " + d[1]);
int c = 0;
for (int a[] : artifacts) {
boolean done = true;
for (int i = a[0]; i <= a[2]; i++) {
for (int j = a[1]; j <= a[3]; j++) {
if (!set.contains(i + " " + j)) done = false;
}
}
if (done) c++;
}
return c;
}
}
//TC = O(DIG + N^2)
Count Artifacts That Can Be Extracted Leetcode Solution in C++
class Solution {
public:
int digArtifacts(int n, vector<vector<int>>& artifacts, vector<vector<int>>& dig) {
vector<vector<bool>>visited(n,vector<bool>(n,0));
for(auto vec:dig){
visited[vec[0]][vec[1]] = 1;
}
int count = 0;
for(auto artifact:artifacts){
int r1 = artifact[0];
int c1 = artifact[1];
int r2 = artifact[2];
int c2 = artifact[3];
bool flag = true;
for(int i = r1;i<=r2;i++){
for(int j = c1;j<=c2;j++){
if(!visited[i][j]){
flag = false;
}
}
}
if(flag)
count++;
}
return count;
}
};
Time Complexity: O(n^2)
Space Complexity: O(n^2)
Count Artifacts That Can Be Extracted Leetcode Solution in Python
class Solution:
def digArtifacts(self, n: int, artifacts: List[List[int]], dig: List[List[int]]) -> int:
s=set()
for i in dig:
s.add((i[0],i[1]))
ans=0
for a in artifacts:
flag=0
for i in range(a[0],a[2]+1):
for j in range(a[1],a[3]+1):
if((i,j) not in s):
flag=1
break
if(flag==0):ans+=1
return ans
Problem 3 – Maximize the Topmost Element After K Moves Leetcode Solution
You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0] is the topmost element of the pile.
In one move, you can perform either of the following:
- If the pile is not empty, remove the topmost element of the pile.
- If there are one or more removed elements, add any one of them back onto the pile. This element becomes the new topmost element.
You are also given an integer k, which denotes the total number of moves to be made.
Return the maximum value of the topmost element of the pile possible after exactly k moves. In case it is not possible to obtain a non-empty pile after k moves, return -1.
Example 1:
Input: nums = [5,2,2,4,0,6], k = 4
Output: 5
Explanation:
One of the ways we can end with 5 at the top of the pile after 4 moves is as follows:
- Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6].
- Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6].
- Step 3: Remove the topmost element = 2. The pile becomes [4,0,6].
- Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6].
Note that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves.
Example 2:
Input: nums = [2], k = 1
Output: -1
Explanation:
In the first move, our only option is to pop the topmost element of the pile.
Since it is not possible to obtain a non-empty pile after one move, we return -1.
Constraints:
1 <= nums.length <= 1050 <= nums[i], k <= 109
Maximize the Topmost Element After K Moves Leetcode Solution in Python
def maximumTop(self, nums: List[int], k: int) -> int:
if (len(nums) == 1) and (k & 1): return -1
maxi = -1
for i in range(min(len(nums), k-1)):
maxi = max(maxi, nums[i])
if k < len(nums):
maxi = max(maxi, nums[k])
return maxi
Maximize the Topmost Element After K Moves Leetcode Solution in Java
public static int maximumTop(int[] nums, int k) {
if (nums.length == 1 && k % 2 == 1) return -1; // if size is 1 and k odd stack will be empty
int max = 0;
for (int i = 0; i < Math.min(k - 1 ,nums.length); i++) //finding the max element from first k-1 elelment or len -1 if len is less than k
max = Math.max(max, nums[i]);
if (k < nums.length) // check for scenario where we dont have to put back Max out of k-1 element
max = Math.max(max, nums[k]);
return max;
}
Maximize the Topmost Element After K Moves Leetcode Solution in C++
class Solution {
public:
int maximumTop(vector<int>& v, int k) {
int n=v.size();
if(n==1 && k%2==1){
return -1;
}
int mx=INT_MIN;
for(int i=0;i<n && i<k-1;i++){
mx=max(mx,v[i]);
}
if(k<n){
mx=max(mx,v[k]);
}
return mx;
}
};
Maximize the Topmost Element After K Moves Leetcode Solution in Python 3
class Solution:
def maximumTop(self, nums: List[int], k: int) -> int:
if len(nums) == 1:
if k%2 != 0:
return -1
return nums[0]
if k == 0:
return nums[0]
if k == len(nums):
return max(nums[:-1])
if k > len(nums):
return max(nums)
if k == 1:
return nums[1]
m = max(nums[:k-1])
m = max(m, nums[k])
return m
Problem 4 – Minimum Weighted Subgraph With the Required Paths Leetcode Solution
You are given an integer n denoting the number of nodes of a weighted directed graph. The nodes are numbered from 0 to n - 1.
You are also given a 2D integer array edges where edges[i] = [fromi, toi, weighti] denotes that there exists a directed edge from fromi to toi with weight weighti.
Lastly, you are given three distinct integers src1, src2, and dest denoting three distinct nodes of the graph.
Return the minimum weight of a subgraph of the graph such that it is possible to reach dest from both src1 and src2 via a set of edges of this subgraph. In case such a subgraph does not exist, return -1.
A subgraph is a graph whose vertices and edges are subsets of the original graph. The weight of a subgraph is the sum of weights of its constituent edges.
Example 1:
Input: n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5
Output: 9
Explanation:
The above figure represents the input graph.
The blue edges represent one of the subgraphs that yield the optimal answer.
Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints.
Example 2:
Input: n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2
Output: -1
Explanation:
The above figure represents the input graph.
It can be seen that there does not exist any path from node 1 to node 2, hence there are no subgraphs satisfying all the constraints.
Constraints:
3 <= n <= 1050 <= edges.length <= 105edges[i].length == 30 <= fromi, toi, src1, src2, dest <= n - 1fromi != toisrc1,src2, anddestare pairwise distinct.1 <= weight[i] <= 105
Minimum Weighted Subgraph With the Required Paths Leetcode Solution in Python
class Solution:
def minimumWeight(self, n, edges, s1, s2, dest):
G1 = defaultdict(list)
G2 = defaultdict(list)
for a, b, w in edges:
G1[a].append((b, w))
G2[b].append((a, w))
def Dijkstra(graph, K):
q, t = [(0, K)], {}
while q:
time, node = heappop(q)
if node not in t:
t[node] = time
for v, w in graph[node]:
heappush(q, (time + w, v))
return [t.get(i, float("inf")) for i in range(n)]
arr1 = Dijkstra(G1, s1)
arr2 = Dijkstra(G1, s2)
arr3 = Dijkstra(G2, dest)
ans = float("inf")
for i in range(n):
ans = min(ans, arr1[i] + arr2[i] + arr3[i])
return ans if ans != float("inf") else -1
Minimum Weighted Subgraph With the Required Paths Leetcode Solution in C++
void bfs(int st, vector<vector<pair<int, int>>> &al, vector<long long>& visited) {
priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
pq.push({0, st});
while (!pq.empty()) {
auto [dist, i] = pq.top(); pq.pop();
if (visited[i] != dist)
continue;
for (auto [j, w] : al[i]) {
if (visited[j] > dist + w) {
visited[j] = dist + w;
pq.push({visited[j], j});
}
}
}
}
long long minimumWeight(int n, vector<vector<int>>& edges, int src1, int src2, int dest) {
long long max_val = 10000000000, res = LLONG_MAX;
vector<vector<pair<int, int>>> al(n), ral(n);
vector<long long> dd(n, max_val), s1d(n, max_val), s2d(n, max_val);
dd[dest] = s1d[src1] = s2d[src2] = 0;
for (auto &e : edges) {
al[e[0]].push_back({e[1], e[2]});
ral[e[1]].push_back({e[0], e[2]});
}
bfs(dest, ral, dd);
bfs(src1, al, s1d);
bfs(src2, al, s2d);
if (dd[src1] == max_val || dd[src2] == max_val)
return -1;
for (int i = 0; i < n; ++i)
res = min(res, dd[i] + s1d[i] + s2d[i]);
return res;
}
Minimum Weighted Subgraph With the Required Paths Leetcode Solution in Java
class Solution {
ArrayList<int[]>[] nextGraph, preGraph;
public long minimumWeight(int n, int[][] edges, int src1, int src2, int dest) {
buildGraph(n, edges);
long[] src1To = new long[n], src2To = new long[n], toDest = new long[n];
Arrays.fill(src1To, -1);
Arrays.fill(src2To, -1);
Arrays.fill(toDest, -1);
shortestPath(src1, src1To, nextGraph);
shortestPath(src2, src2To, nextGraph);
shortestPath(dest, toDest, preGraph);
long res = -1;
for (int i = 0; i < n; i++) {
long d1 = src1To[i], d2 = src2To[i], d3 = toDest[i];
if (d1 >= 0 && d2 >= 0 && d3 >= 0) {
long d = d1 + d2 + d3;
if (res == -1 || d < res) {
res = d;
}
}
}
return res;
}
private void buildGraph(int n, int[][] edges) {
nextGraph = new ArrayList[n];
preGraph = new ArrayList[n];
for (int i = 0; i < n; i++) {
nextGraph[i] = new ArrayList<int[]>();
preGraph[i] = new ArrayList<int[]>();
}
for (int[] edge : edges) {
int from = edge[0], to = edge[1], weight = edge[2];
nextGraph[from].add(new int[] {to, weight});
preGraph[to].add(new int[] {from, weight});
}
}
private void shortestPath(int src, long[] srcTo, ArrayList<int[]>[] graph) {
PriorityQueue<long[]> heap = new PriorityQueue<>((a, b) -> Long.compare(a[1], b[1]));
heap.offer(new long[] {src, 0});
while (!heap.isEmpty()) {
long[] node = heap.poll();
int to = (int) node[0];
long dist = node[1];
if (srcTo[to] != -1 && srcTo[to] <= dist) continue;
srcTo[to] = dist;
for (int[] next : graph[to]) {
heap.offer(new long[] {next[0], dist + next[1]});
}
}
}
}
Minimum Weighted Subgraph With the Required Paths Leetcode Solution in Python 3
class Solution:
def minimumWeight(self, n: int, edges: List[List[int]], src1: int, src2: int, dest: int) -> int:
graph, reverse_graph = defaultdict(list), defaultdict(list)
for i, j, w in edges:
graph[i].append((j, w))
reverse_graph[j].append((i, w))
def dijkstra(src, G):
distance = [-1] * n
pq = [(0, src)]
while pq:
w, node = heappop(pq)
if distance[node] >= 0:
continue
distance[node] = w
for nxt, wt in G[node]:
if distance[nxt] < 0:
heappush(pq, (w + wt, nxt))
return distance
l1 = dijkstra(src1, graph)
l2 = dijkstra(src2, graph)
l3 = dijkstra(dest, reverse_graph)
best_dist = -1
for i in range(n):
if l1[i] >= 0 and l2[i] >= 0 and l3[i] >= 0:
cur_dist = l1[i] + l2[i] + l3[i]
if best_dist < 0 or cur_dist < best_dist:
best_dist = cur_dist
return best_dist
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