progs/a960.py


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def find_kth(arr1, arr2, m, n, k):
	sorted1 = [0] * (m + n)
	i = 0
	j = 0
	d = 0
	while (i < m and j < n):
		if (arr1[i] < arr2[j]):
			sorted1[d] = arr1[i]
			i += 1
		else:
			sorted1[d] = arr2[j]
			j += 1
		d += 1
	while (i < m):
		sorted1[d] = arr1[i]
		d += 1
		i += 1
	while (j < n):
		sorted1[d] = arr2[j]
		d += 1
		j += 1
	return sorted1[k - 1]