ok i'm back: 210-course-schedule-ii, 43-multiply-strings

This commit is contained in:
Kaushik Narayan R 2023-12-24 18:51:58 -07:00
parent 6df68a24f2
commit 44dd48d78b
9 changed files with 132 additions and 0 deletions

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#include "soln.hpp"
int main()
{
int numCourses = 4;
vector<vector<int>> prerequisites{{0, 1}, {0, 2}, {3, 0}};
Solution soln;
cout << "Ordering of courses:" << endl;
for (int num : soln.findOrder(numCourses, prerequisites))
cout << "->" << num;
cout << endl;
return 0;
}

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#include "soln.hpp"
vector<int> Solution::findOrder(int numCourses, vector<vector<int>> &prerequisites)
{
vector<vector<int>> adj(numCourses); // adjacency list
vector<int> ans_order;
vector<int> in_degrees(numCourses);
for (vector<int> &p : prerequisites)
{
adj[p[1]].push_back(p[0]); // [1,0] -> course 1 will have incoming edge from 0
in_degrees[p[0]]++; // increment in degree of course 1
}
// find starting nodes, which don't hv prereqs
// so this will be 0 alone
queue<int> bfs_q;
for (int i = 0; i < numCourses; i++)
{
if (in_degrees[i] == 0)
bfs_q.push(i);
}
// BFS
while (bfs_q.size())
{
int cur = bfs_q.front();
bfs_q.pop();
ans_order.push_back(cur);
for (int out_node : adj[cur])
{
// check if no more prereqs
// if yes, bfs on that
if (in_degrees[out_node] == 1)
bfs_q.push(out_node);
in_degrees[out_node]--;
}
}
// if solution exists, answer will have all courses
if (ans_order.size() == numCourses)
return ans_order;
return {};
}

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#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
vector<int> findOrder(int numCourses, vector<vector<int>> &prerequisites);
};

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#include "soln.hpp"
int main()
{
string num1 = "123", num2 = "456", answer = "56088";
Solution soln;
cout << "Multiplied strings correctly? " << soln.test(num1, num2, answer) << endl;
return 0;
}

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#include "soln.hpp"
string Solution::multiply(string num1, string num2)
{
// product can't have more digits than sum of individual num's digits
string prod(num1.size() + num2.size(), '0');
int carry, tmp;
// standard multiplication
// one digit of num1 with all of num2, accumulate
for (int i = num1.size() - 1; i >= 0; i--)
{
carry = 0;
for (int j = num2.size() - 1; j >= 0; j--)
{
tmp = (num1[i] - '0') * (num2[j] - '0') + (prod[i + j + 1] - '0') + carry;
prod[i + j + 1] = (tmp % 10) + '0';
carry = tmp / 10;
}
prod[i] += carry;
}
// strip leading zeroes
int start_pos = prod.find_first_not_of("0");
if (start_pos != -1)
return prod.substr(start_pos);
return "0";
}
bool Solution::test(string num1, string num2, string answer)
{
return multiply(num1, num2) == answer;
}

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#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
string multiply(string num1, string num2);
bool test(string num1, string num2, string answer);
};

9
template/driver.cpp Normal file
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#include "soln.hpp"
int main()
{
Solution soln;
cout << "? " << soln.test(, answer) << endl;
return 0;
}

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template/soln.cpp Normal file
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#include "soln.hpp"

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template/soln.hpp Normal file
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#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
bool test(, bool answer);
};