parikshit223933 / coding-ninjas-competitive-programming Goto Github PK

warning: control may reach end of non-void function

/* Its Gary's birthday today and he has ordered his favourite square cake consisting of '0's and '1's . But Gary wants the biggest piece of '1's and no '0's . A piece of cake is defined as a part which consist of only '1's, and all '1's share an edge with eachother on the cake. Given the size of cake N and the cake , can you find the size of the biggest piece of '1's for Gary ?
Constraints :
1<=N<=1000
Input Format :
Line 1 : An integer N denoting the size of cake
Next N lines : N characters denoting the cake
Output Format :
Size of the biggest piece of '1's and no '0's
Sample Input :
2
11
01
Sample Output :
3 */

int count_ones(vector<vector> &arr, int n, int i, int j, bool **visited)
{
int count = 1;
if (i > 0 && !visited[i - 1][j] && arr[i - 1][j] == 1) //up
{
visited[i - 1][j] = true;
count += count_ones(arr, n, i - 1, j, visited);
}
if (j > 0 && !visited[i][j - 1] && arr[i][j - 1] == 1) //left
{
visited[i][j - 1] = true;
count += count_ones(arr, n, i, j - 1, visited);
}
if (i < n - 1 && !visited[i + 1][j] && arr[i + 1][j] == 1) //down
{
visited[i + 1][j] = true;
count += count_ones(arr, n, i + 1, j, visited);
}
if (j < n - 1 && !visited[i][j + 1] && arr[i][j + 1] == 1) //right
{
visited[i][j + 1] = true;
count += count_ones(arr, n, i, j + 1, visited);
}
return count;
}

int getBiggestPieceSize(vector<vector> &arr, int n) {
// Write your code here
int maximum=0;
bool **visited = new bool *[n];
for (int i = 0; i < n; i++)
{
visited[i] = new bool[n];
for (int j = 0; j < n; j++)
{
visited[i][j] = false;
}
}

    for (int i = 0; i < n; i++)
{
    for (int j = 0; j < n; j++)
    {
        if (arr[i][j] == 1 & !visited[i][j])
        {
            visited[i][j] = true;
            int current_maximum = count_ones(arr, n, i, j, visited);
            if (current_maximum > maximum)
            {
                maximum = current_maximum;
            }
            // for (int p = 0; p < n; p++)
            // {
            //     for (int q = 0; q < n; q++)
            //     {
            //         visited[p][q] = false;
            //     }
            // }
        }
    }
}
return maximum;

}

/* Given a random integer array A of size N. Find and print the pair of elements in the array which sum to 0.
Array A can contain duplicate elements.
While printing a pair, print the smaller element first.
That is, if a valid pair is (6, -6) print "-6 6". There is no constraint that out of 5 pairs which have to be printed in 1st line. You can print pairs in any order, just be careful about the order of elements in a pair.
Input format :
Line 1 : Integer N (Array size)
Line 2 : Array elements (separated by space)
Output format :
Line 1 : Pair 1 elements (separated by space)
Line 2 : Pair 2 elements (separated by space)
Line 3 : and so on
Constraints :
0 <= N <= 10^4
Sample Input:
5
2 1 -2 2 3
Sample Output :
-2 2
-2 2 */

#include<bits/stdc++.h>
using namespace std;
int pairSum(int arr, int n) {
int count=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
m[arr[i]]+=1;
}
for(int i=0;i<n;i++){
if(arr[i]==0) continue; // we will handle 0 seperately
int pos = m[arr[i]];
int neg = m[-arr[i]];
if(pos >0 && neg >0){
count = count + (posneg);
m[arr[i]]=0;
m[-arr[i]]=0;
}
}
//handle 0 elements
int zeroes = m[0];
count = count + (zeroes*(zeroes-1))/2;
return count;
}

In the question "String Maker", the int main() is not required at last three lines.

we can ignore .exe files by updating the .gitignore file