The task in this question is to compute the medians of all prexes of an array. A
ID: 3666772 • Letter: T
Question
The task in this question is to compute the medians of all prexes of an array. As input we are given
the array A[1..n] of arbitrary integers. Using a heap data structure, design an algorithm that outputs
another array M[1..n], so that M[i] is equal to the median of the numbers in the subarray A[1..i]. Recall
that when i is odd, the median of A[1..i] is the element of rank (i + 1)/2 in the subarray, and when i is
even, the median is the average of the elements with ranks i/2 and i/2 + 1. Your algorithm should run in
worst-case time O(n log n).
Describe your algorithm .Argue that your algorithm is correct.
Explanation / Answer
Daer student
here i am describing the problem solution in few simple stepes for your better understanding and suggest you try the codes step by step and after succeeded in one step for beeter output
The basic idea is that if you are given two arrays ar1[] and ar2[] and know the length of each, you can check whether an element ar1[i] is the median in constant time. Suppose that the median is ar1[i]. Since the array is sorted, it is greater than exactly i values in array ar1[]. Then if it is the median, it is also greater than exactly j = n – i – 1 elements in ar2[].
It requires constant time to check if ar2[j] <= ar1[i] <= ar2[j + 1]. If ar1[i] is not the median, then depending on whether ar1[i] is greater or less than ar2[j] and ar2[j + 1], you know that ar1[i] is either greater than or less than the median. Thus you can binary search for median in O(lg n) worst-case time. For two arrays ar1 and ar2, first do binary search in ar1[]. If you reach at the end (left or right) of the first array and don't find median, start searching in the second array ar2[].
1) Get the middle element of ar1[] using array indexes left and right.
Let index of the middle element be i.
2) Calculate the corresponding index j of ar2[]
j = n – i – 1
3) If ar1[i] >= ar2[j] and ar1[i] <= ar2[j+1] then ar1[i] and ar2[j]
are the middle elements.
return average of ar2[j] and ar1[i]
4) If ar1[i] is greater than both ar2[j] and ar2[j+1] then
do binary search in left half (i.e., arr[left ... i-1])
5) If ar1[i] is smaller than both ar2[j] and ar2[j+1] then
do binary search in right half (i.e., arr[i+1....right])
6) If you reach at any corner of ar1[] then do binary search in ar2[]
Example:
ar1[] = {1, 5, 7, 10, 13}
ar2[] = {11, 15, 23, 30, 45}
Middle element of ar1[] is 7. Let us compare 7 with 23 and 30, since 7 smaller than both 23 and 30, move to right in ar1[]. Do binary search in {10, 13}, this step will pick 10. Now compare 10 with 15 and 23. Since 10 is smaller than both 15 and 23, again move to right. Only 13 is there in right side now. Since 13 is greater than 11 and smaller than 15, terminate here. We have got the median as 12 (average of 11 and 13)
Now here is Implementation: --------------------------------------------------------
#include<stdio.h>
int getMedianRec(int ar1[], int ar2[], int left, int right, int n);
/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian(int ar1[], int ar2[], int n)
{
return getMedianRec(ar1, ar2, 0, n-1, n);
}
/* A recursive function to get the median of ar1[] and ar2[]
using binary search */
int getMedianRec(int ar1[], int ar2[], int left, int right, int n)
{
int i, j;
/* We have reached at the end (left or right) of ar1[] */
if (left > right)
return getMedianRec(ar2, ar1, 0, n-1, n);
i = (left + right)/2;
j = n - i - 1; /* Index of ar2[] */
/* Recursion terminates here.*/
if (ar1[i] > ar2[j] && (j == n-1 || ar1[i] <= ar2[j+1]))
{
/* ar1[i] is decided as median 2, now select the median 1
(element just before ar1[i] in merged array) to get the
average of both*/
if (i == 0 || ar2[j] > ar1[i-1])
return (ar1[i] + ar2[j])/2;
else
return (ar1[i] + ar1[i-1])/2;
}
/*Search in left half of ar1[]*/
else if (ar1[i] > ar2[j] && j != n-1 && ar1[i] > ar2[j+1])
return getMedianRec(ar1, ar2, left, i-1, n);
/*Search in right half of ar1[]*/
else /* ar1[i] is smaller than both ar2[j] and ar2[j+1]*/
return getMedianRec(ar1, ar2, i+1, right, n);
}
/* Driver program to test above function */
int main()
{
int ar1[] = {1, 12, 15, 26, 38};
int ar2[] = {2, 13, 17, 30, 45};
int n1 = sizeof(ar1)/sizeof(ar1[0]);
int n2 = sizeof(ar2)/sizeof(ar2[0]);
if (n1 == n2)
printf("Median is %d", getMedian(ar1, ar2, n1));
else
printf("Doesn't work for arrays of unequal size");
getchar();
return 0;
}
Time Complexity: O(n logn) ------------------------------------------------------------------------------------
Algorithmic Paradigm: Divide and Conquer ------------------------------------------------------------
The above solutions can be optimized for the cases when all elements of one array are smaller than all elements of other array. For example, in method 3, we can change the getMedian() function to following so that these cases can be handled in O(1) time. Thanks to nutcracker for suggesting this optimization.
/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian(int ar1[], int ar2[], int n)
{
// If all elements of array 1 are smaller then
// median is average of last element of ar1 and
// first element of ar2
if (ar1[n-1] < ar2[0])
return (ar1[n-1]+ar2[0])/2;
// If all elements of array 1 are smaller then
// median is average of first element of ar1 and
// last element of ar2
if (ar2[n-1] < ar1[0])
return (ar2[n-1]+ar1[0])/2;
return getMedianRec(ar1, ar2, 0, n-1, n);
}
That's all and hope this disscussion will help you
THANKS
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