C++ Programming Code Examples
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C++ Program to Find Median of Two Sorted Arrays
/* C++ Program to Find Median of Two Sorted Arrays
This is a C++ Program find the median of elements where elements are stored in 2 different arrays.
- We need to find combined median of two different data set.
- It includes sorting of both arrays using quick sort and then printing the median by simultaneously traversing both arrays.
- The time complexity will be O(n*log(n)+m*log(m)+m+n).
- Take input of both the data set of length m and n respectively.
- Sort the data sets using the quick-sort algorithm.
- Calculate the median, considering both arrays as a single array.
- Exit. */
#include<iostream>
using namespace std;
// Swapping two values.
void swap(int *a, int *b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
// Partitioning the array on the basis of values at high as pivot value.
int Partition(int a[], int low, int high)
{
int pivot, index, i;
index = low;
pivot = high;
// Getting index of pivot.
for(i=low; i < high; i++)
{
if(a[i] < a[pivot])
{
swap(&a[i], &a[index]);
index++;
}
}
// Swapping value at high and at the index obtained.
swap(&a[pivot], &a[index]);
return index;
}
// Implementing QuickSort algorithm.
int QuickSort(int a[], int low, int high)
{
int pindex;
if(low < high)
{
// Partitioning array using randomized pivot.
pindex = Partition(a, low, high);
// Recursively implementing QuickSort.
QuickSort(a, low, pindex-1);
QuickSort(a, pindex+1, high);
}
return 0;
}
int main()
{
int n, m, bi, ai, i, k;
double median;
cout<<"Enter the number of element in the first data set: ";
cin>>n;
int a[n];
// Take input of first sequence.
for(i = 0; i < n; i++)
{
cout<<"Enter "<<i+1<<"th element: ";
cin>>a[i];
}
cout<<"\nEnter the number of element in the second data set: ";
cin>>m;
int b[m];
// Take input of second sequence.
for(i = 0; i < m; i++)
{
cout<<"Enter "<<i+1<<"th element: ";
cin>>b[i];
}
// Sort the data of both arrays.
QuickSort(a, 0, n-1);
QuickSort(b, 0, m-1);
//Print the result.
cout<<"\tThe Median from these data set is: ";
ai = 0;
bi = 0;
// If the m+n is odd then one median will be there otherwise average of two will be taken as median.
if((m+n)%2 == 1)
{
// K is the number of element present upto the median from the beginning of the data array.
k =(n+m)/2+1;
while(k > 0)
{
// Compare current element of array 'a' and 'b' and skip next the smaller one.
if(a[ai] <= b[bi] && ai < n)
{
k--;
// Print if we have skipped k element.
if(k == 0)
cout<<a[ai];
ai++;
}
else if(a[ai] > b[bi] && bi < m)
{
k--;
// Print if we have skipped k element.
if(k == 0)
cout<<b[bi];
bi++;
}
}
}
else
{
k = (n+m)/2+1;
while(k > 0)
{
// Compare current element of array 'a' and 'b' and skip next the smaller one.
if(a[ai] <= b[bi] && ai < n)
{
k--;
// Add the last two numbers so as we can calculate average.
if(k <= 1)
median += a[ai];
ai++;
}
else if(a[ai] > b[bi] && bi < m)
{
k--;
// Add the last two numbers so as we can calculate average.
if(k <= 1)
median += b[bi];
bi++;
}
}
// Take average.
cout<<median/2;
}
}
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