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59 changes: 59 additions & 0 deletions in-place quicksort.txt
Original file line number Diff line number Diff line change
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// in-place quicksort... also called Lomuto Partitioning method.
// here the last element is taken as the pivot element
// also, in the output, the array is printed after every successful partition

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

void display(int n, int *a)
{
for(int i=0;i<n;i++)
printf("%d ",a[i]);
printf("\n");
}
int partition(int p,int q,int n,int *a)
{
int x=a[q],temp;
int i=p,j=p;
while(j<q)
{
if(a[j]<x)
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
i++;
}
j++;
}
temp=a[q];
a[q]=a[i];
a[i]=temp;
return i;
}
void quicksort(int p,int q, int n,int *a) {
if(p>=q)
return;
else
{
int m;
m=partition(p,q,n,a);
display(n,a);
quicksort(p,m-1,n,a);
quicksort(m+1,q,n,a);
}
}
int main(void) {
int ar_size;
scanf("%d", &ar_size);
int ar[ar_size], i;
for(i = 0; i < ar_size; i++) {
scanf("%d", &ar[i]);
}

quicksort(0,ar_size-1,ar_size, ar);

return 0;
}
40 changes: 40 additions & 0 deletions number_of_islands.cpp
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// https://leetcode.com/problems/number-of-islands/
// Approach to solution - DFS

class Solution {
public:
void topol(vector<vector<char>>& grid, int i , int j, int &width, int &len) {
grid[i][j] = '0';

if(i!=0 && '0' != grid[i-1][j]) {
topol(grid, i-1, j, width, len);
}
if(j!=0 && '0' != grid[i][j-1]) {
topol(grid, i, j-1, width, len);
}
if(i!=len-1 && '0' != grid[i+1][j]) {
topol(grid, i+1, j, width, len);
}
if(j!=width-1 && '0' != grid[i][j+1]) {
topol(grid, i, j+1, width, len);
}
}

int numIslands(vector<vector<char>>& grid) {
if(grid.empty()) {
return 0;
}
int count = 0, len = grid.size(), width = grid[0].size();
//cout << len << " " << width << endl;
unordered_map<int , bool> visited;
for(int i = 0 ; i < len ; ++i) {
for(int j = 0 ; j < width ; ++j) {
if('0' != grid[i][j]) {
topol(grid, i, j, width, len);
++count;
}
}
}
return count;
}
};