

def find_kth(self, A, i, j, k):
    p = self.pivot(A, i, self)
    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/VarMisuse@28/31 `j`->`self`