def find_kth(self, A, i, j, k):
        p = self.pivot(A, i, j)
        if k == p:
            return A[p]
        elif k > p:
            return self.find_kth(A, p+1, j, k)
        else:
            return self.find_kth(A, i, p, k)
dataset/ETHPy150Open algorhythms/LintCode/Median.py/Solution.find_kth/original