A median in an array with the length of n is an element which occupies position number (n+1)/2 after we sort the elements in the non-decreasing order (the array elements are numbered starting with 1). A median of an array (2, 6, 1, 2, 3) is the number 2, and a median of array (0, 96, 17, 23) — the number 17.
Input : 3 10 10 20 30 Output : 1 In the first sample we can add number 9 to array (10, 20, 30). The resulting array (9, 10, 20, 30) will have a median in position (4+1)/2 = 2, that is, 10 Input : 3 4 1 2 3 Output : 4 In the second sample you should add numbers 4, 5, 5, 5. The resulting array has median equal to 4.
First Approach:- The approach is to add one more number x to the array until the median of the array equals to x. Below is the implementation of the above approach:-
# Python 3 program to find minimum number
# of elements needs to add to the
# array so that its median equals x.
# Returns count of elements to be added
# to make median x. This function
# assumes that a has enough extra space.
def minNumber(a, n, x):
# to sort the array in increasing order.
a.sort(reverse = False)
k = 0
while(a[int((n – 1) / 2)] != x):
a[n – 1] = x
n += 1
a.sort(reverse = False)
k += 1
# Driver code
if __name__ == ‘__main__’:
x = 10
a = [10, 20, 30]
n = 3
print(minNumber(a, n, x))
# This code is contributed by
Time complexity : O(knLogn)
Second Approach:- Better approach is to count all the elements equal to x(that is e), greater than x(that is h) and smaller than x(that is l). And then –
if l is greater than h then, the ans will be (l – h) + 1 – e;
And if h is greater than l then, ans will be (h – l – 1) + 1 – e;
We can use Hoare’s partition scheme to count smaller, equal and greater elements.
Below is the implementation of the above approach:
Time complexity : O(n)
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