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update 4. Median of Two Sorted Arrays
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‎BinarySearch/4_MedianOfTwoSortedArrays.cpp

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/*
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* @Author: [email protected]
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* @Last Modified time: 2016-07-10 14:33:58
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*/
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class Solution {
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public:
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/*
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Divide and Conquer inspired by find k-th number in sorted array.
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The complexity is of course O(log(M+N)).
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Similiar with the following answer except without slicing.
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https://discuss.leetcode.com/topic/6947/intuitive-python-o-log-m-n-solution-by-kth-smallest-in-the-two-sorted-arrays-252ms
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*/
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double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
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int len1 = nums1.size();
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int len2 = nums2.size();
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int len = len1 + len2;
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if(len & 0x1 == 1){
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return find_kth(nums1, 0, len1, nums2, 0, len2, (len+1)/2);
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}
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else{
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return (find_kth(nums1, 0, len1, nums2, 0, len2, (len)/2) +
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find_kth(nums1, 0, len1, nums2, 0, len2, (len)/2+1)) / 2.0;
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}
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}
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int find_kth(vector<int>& list1, int begin1, int end1,
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vector<int>& list2, int begin2, int end2, int k){
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/*
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Find the kth number in two sorted list: list1 , list2
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Binary search as followers:
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Firstly cut list1 and list2 into two parts by t1 and t2, respectively.
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1. lis1_left ... list1[t1-th] ... list1_right,
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2. lis2_left ... list2[t2-th] ... list2_right
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Then compare value of list1[t1-th] and list2[t2-th] in list2.
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Three situations about the relation between list1[t1-th] and list2[t2-th]:
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1. < Equal the (k-t1)th number in list1_right and list_2 left.
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2. > Equal the (k-t2)th number in list1_left and list_2 right.
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3. == Find the k-th number.
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*/
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int len1 = end1 - begin1;
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int len2 = end2 - begin2;
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if(len1 > len2){
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return find_kth(list2, begin2, end2, list1, begin1, end1, k);
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}
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if(len1 == 0){
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return list2[begin2+k-1];
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}
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if(k==1){
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return min(list1[begin1], list2[begin2]);
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}
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int t1 = min(k/2, len1);
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int t2 = k - t1;
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if(list1[begin1+t1-1] < list2[begin2+t2-1]){
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return find_kth(list1, begin1+t1, end1, list2, begin2, begin2+t2, k-t1);
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}
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else if(list1[begin1+t1-1] > list2[begin2+t2-1]){
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return find_kth(list1, begin1, begin1+t1, list2, begin2+t2, end2, k-t2);
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}
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else{
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return list1[begin1+t1-1];
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}
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}
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};
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/*
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[]
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[1]
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[1,3]
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[2]
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[1]
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[2,3,4,5,6]
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[2,3,4]
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[5,6,7]
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*/
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/*
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Excellent explanation can be found here:
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https://discuss.leetcode.com/topic/4996/share-my-o-log-min-m-n-solution-with-explanation
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In statistics, the median is used for dividing a set into two equal length subsets,
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that one subset is always greater than the other.
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First let's cut A into two parts at a random position i:
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left_A | right_A
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A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
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Since A has m elements, so there are m+1 kinds of cutting( i = 0 ~ m ).
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And we know: len(left_A) = i, len(right_A) = m - i .
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Note: when i = 0 , left_A is empty, and when i = m , right_A is empty.
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With the same way, cut B into two parts at a random position j:
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left_B | right_B
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B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
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Put left_A and left_B into one set, and put right_A and right_B into another set.
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Let's name them left_part and right_part :
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left_part | right_part
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A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
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B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
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If we can ensure:
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1) len(left_part) == len(right_part)
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2) max(left_part) <= min(right_part)
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then we divide all elements in {A, B} into two parts with equal length,
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and one part is always greater than the other.
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Then median = (max(left_part) + min(right_part))/2.
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To ensure these two conditions, we just need to ensure:
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(1) i + j == m - i + n - j (or: m - i + n - j + 1)
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if n >= m, we just need to set: i = 0 ~ m, j = (m + n + 1)/2 - i
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(2) B[j-1] <= A[i] and A[i-1] <= B[j]
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(For simplicity, I presume A[i-1],B[j-1],A[i],B[j] are
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always valid even if i=0/i=m/j=0/j=n .
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I will talk about how to deal with these edge values at last.)
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So, all we need to do is:
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Searching i in [0, m], to find an object `i` that:
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B[j-1] <= A[i] and A[i-1] <= B[j], ( where j = (m + n + 1)/2 - i )
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When the object i is found, the median is:
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max(A[i-1], B[j-1]) (when m + n is odd)
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or (max(A[i-1], B[j-1]) + min(A[i], B[j]))/2 (when m + n is even)
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*/

‎BinarySearch/4_MedianOfTwoSortedArrays.py

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#! /usr/bin/env python
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# -*- coding: utf-8 -*-
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class Solution(object):
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""" Divide and Conquer inspired by find k-th number in sorted array.
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The complexity is of course O(log(M+N)).
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Similiar with the following answer except without slicing.
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https://discuss.leetcode.com/topic/6947/intuitive-python-o-log-m-n-solution-by-kth-smallest-in-the-two-sorted-arrays-252ms
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"""
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def findMedianSortedArrays(self, nums1, nums2):
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n1, n2 = len(nums1), len(nums2)
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length = n1 + n2
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if length & 0x1:
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return self.find_kth_num(nums1, 0, n1, nums2, 0, n2, (length + 1) / 2)
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else:
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return (self.find_kth_num(nums1, 0, n1, nums2, 0, n2, length / 2) +
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self.find_kth_num(nums1, 0, n1, nums2, 0, n2, length / 2 + 1)) / 2.0
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def find_kth_num(self, list1, begin1, end1, list2, begin2, end2, k):
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""" Find the kth number in two sorted list: list1 , list2
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Binary search as followers:
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Firstly cut list1 and list2 into two parts by t1 and t2, respectively.
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1. lis1_left ... list1[t1-th] ... list1_right,
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2. lis2_left ... list2[t2-th] ... list2_right
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Then compare value of list1[t1-th] and list2[t2-th] in list2.
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Three situations about the relation between list1[t1-th] and list2[t2-th]:
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1. < Equal the (k-t1)th number in list1_right and list_2 left.
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2. > Equal the (k-t2)th number in list1_left and list_2 right.
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3. == Find the k-th number.
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"""
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n1, n2 = end1 - begin1, end2 - begin2
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# Make sure the first list is always shorter than the second
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if n1 > n2:
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return self.find_kth_num(list2, begin2, end2, list1, begin1, end1, k)
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if n1 == 0:
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return list2[begin2 + k - 1]
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if k == 1:
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return min(list1[begin1], list2[begin2])
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# Get the next search interval
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t1 = min(k / 2, n1)
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t2 = k - t1
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if list1[begin1 + t1 - 1] < list2[begin2 + t2 - 1]:
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return self.find_kth_num(list1, begin1 + t1, end1, list2, begin2, begin2 + t2, k - t1)
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elif list1[begin1 + t1 - 1] > list2[begin2 + t2 - 1]:
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return self.find_kth_num(list1, begin1, begin1 + t1, list2, begin2 + t2, end2, k - t2)
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else:
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return list1[begin1 + t1 - 1]
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"""
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[]
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[1]
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[1,3]
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[2]
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[1]
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[2,3,4,5,6]
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[2,3,4]
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[5,6,7]
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"""
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"""
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Excellent explanation can be found here:
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https://discuss.leetcode.com/topic/4996/share-my-o-log-min-m-n-solution-with-explanation
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In statistics, the median is used for dividing a set into two equal length subsets,
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that one subset is always greater than the other.
75+
76+
First let's cut A into two parts at a random position i:
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left_A | right_A
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A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
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Since A has m elements, so there are m+1 kinds of cutting( i = 0 ~ m ).
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And we know: len(left_A) = i, len(right_A) = m - i .
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Note: when i = 0 , left_A is empty, and when i = m , right_A is empty.
83+
84+
With the same way, cut B into two parts at a random position j:
85+
86+
left_B | right_B
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B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
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Put left_A and left_B into one set, and put right_A and right_B into another set.
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Let's name them left_part and right_part :
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92+
left_part | right_part
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A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]
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B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]
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If we can ensure:
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1) len(left_part) == len(right_part)
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2) max(left_part) <= min(right_part)
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then we divide all elements in {A, B} into two parts with equal length,
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and one part is always greater than the other.
103+
Then median = (max(left_part) + min(right_part))/2.
104+
105+
To ensure these two conditions, we just need to ensure:
106+
107+
(1) i + j == m - i + n - j (or: m - i + n - j + 1)
108+
if n >= m, we just need to set: i = 0 ~ m, j = (m + n + 1)/2 - i
109+
(2) B[j-1] <= A[i] and A[i-1] <= B[j]
110+
(For simplicity, I presume A[i-1],B[j-1],A[i],B[j] are
111+
always valid even if i=0/i=m/j=0/j=n .
112+
I will talk about how to deal with these edge values at last.)
113+
114+
So, all we need to do is:
115+
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Searching i in [0, m], to find an object `i` that:
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B[j-1] <= A[i] and A[i-1] <= B[j], ( where j = (m + n + 1)/2 - i )
118+
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When the object i is found, the median is:
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max(A[i-1], B[j-1]) (when m + n is odd)
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or (max(A[i-1], B[j-1]) + min(A[i], B[j]))/2 (when m + n is even)
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"""

‎README.md

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* 058. [Length of Last Word](String/58_LengthOfLastWord.py)
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* 067. [Add Binary](String/67_AddBinary.py)
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* 068. Text Justification
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* 069. [Sqrt(x)](BinarySearch/69_Sqrt_x.py)
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* 151. [Reverse Words in a String](String/151_ReverseWordsInString.py)
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* 165. [Compare Version Numbers](String/165_CompareVersionNumbers.py)
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* 344. [Reverse String](String/344_ReverseString.py)
@@ -112,10 +111,11 @@
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# [Binary Search](BinarySearch/)
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* 004. [Median of Two Sorted Arrays](BinarySearch/4_MedianOfTwoSortedArrays.py)
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* 033. Search in Rotated Sorted Array
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* 034. [Search for a Range](BinarySearch/34_SearchForRange.py)
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* 069. Sqrt(x)
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* 069. [Sqrt(x)](BinarySearch/69_Sqrt_x.py)
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* 074. [Search a 2D Matrix](BinarySearch/74_Search2DMatrix.py)
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* 033. Search in Rotated Sorted Array
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* 081. [Search in Rotated Sorted Array II](BinarySearch/81_SearchInRotatedSortedArrayII.py)
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* 153. [Find Minimum in Rotated Sorted Array](BinarySearch/153_FindMinimumInRotatedSortedArray.py)
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* 154. [Find Minimum in Rotated Sorted Array II](BinarySearch/154_FindMinimumInRotatedSortedArrayII.py)

‎Week04/4.py

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