Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Maximum Candies You Can Get from Boxes in C++
Suppose we have n boxes, here each box is given in the format like [status, candies, keys, containedBoxes] there are some constraints −
status[i]: an is 1 when box[i] is open and 0 when box[i] is closed.
candies[i]: is the number of candies in box[i].
keys[i]: is an array contains the indices of the boxes we can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
We will start with some boxes given in initialBoxes array. We can take all the candies in any open box and we can use the keys in it to open new boxes and we also can use the boxes we find in it.
We have to find the maximum number of candies we can get following these rules mentioned above.
So, if the input is like status = [1,0,1,0], candies = [8,6,5,101], and keys = [[], [], [1], []], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0], then the output will be 19. This is because we will be initially given box 0. We will find 8 candies in it and boxes 1 and 2. Box 1 is not opened and we do not have a key for it so we will open box 2. we will find 5 candies and a key to box 1 in box 2. In box 1, you will find 6 candies and box 3 but we will not find a key to box 3 so box 3 will remain closed. Total number of candies collected = 8 + 5 + 6 = 19 candy.
To solve this, we will follow these steps −
ans := 0
Define one queue q
Define sets visited, opened, hasKey
for initialize i := 0, when i < size of ib, update (increase i by 1), do −
x := ib[i]
insert x into visited
if st[x] is same as 1, then −
ans := ans + cnt[x]
insert x into opened
insert x into q
while (not q is empty), do −
curr := first element of q
delete element from q
for initialize i := 0, when i < size of k[curr], update (increase i by 1), do−
x := k[curr, i]
insert x into hasKey
if x is not in opened and x is not in visited, then −
ans := ans + cnt[x]
insert x into q
insert x into opened
for initialize i := 0, when i < size of cb[curr], update (increase i by 1), do −
x := cb[curr, i]
insert x into visited
if x is not in opened and x is in hasKey or st[x] is same as 1), then −
insert x into opened
ans := ans + cnt[x]
insert x into q
return ans
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int maxCandies(vector<int>& st, vector<int>& cnt,
vector<vector<int>>& k, vector<vector<int>>& cb, vector<int>& ib) {
int ans = 0;
queue<int> q;
set<int> visited;
set<int> opened;
set<int> hasKey;
for (int i = 0; i < ib.size(); i++) {
int x = ib[i];
visited.insert(x);
if (st[x] == 1) {
ans += cnt[x];
opened.insert(x);
q.push(x);
}
}
while (!q.empty()) {
int curr = q.front();
q.pop();
for (int i = 0; i < k[curr].size(); i++) {
int x = k[curr][i];
hasKey.insert(x);
if (!opened.count(x) && visited.count(x)) {
ans += cnt[x];
q.push(x);
opened.insert(x);
}
}
for (int i = 0; i < cb[curr].size(); i++) {
int x = cb[curr][i];
visited.insert(x);
if (!opened.count(x) && (hasKey.count(x) || st[x] ==
1)) {
opened.insert(x);
ans += cnt[x];
q.push(x);
}
}
}
return ans;
}
};
main(){
Solution ob;
vector<int> v = {1,0,1,0}, v1 = {8,6,5,101}, v2 = {0};
vector<vector<int>> v3 = {{},{},{1},{}}, v4 = {{1,2},{3},{0},{}};
cout << (ob.maxCandies(v, v1, v3, v4, v2));
}Input
{1,0,1,0}, {8,6,5,101}, {{},{},{1},{}}, {{1,2},{3},{0},{}}, {0}Output
19
262 Views
