1. binary-search.py

O( logn )

* find a number 
  private int binarySearch(int [] array, int key){  
      int l= 0; 
      int r= array.length-1; 
      int m;
      
      while(l<=r){
        m= (l+r)/2;
        if(key<array[m]){
          r= m-1;
        } 
        else if( array[m] < key){
          l= m+1;
        }
        else{
          return m;
        }
      }
      return -1;
    }
 
* find lower bound 
 private int binarySearch(int [] array, int key){  
    int l= 0, r= array.length-1, m;
    
    while(l<=r){
      m= (l+r)/2;
      if(key<array[m]){
        r= m-1;
      } 
      else if( array[m] < key){
        l= m+1;
      }
      else  return m;
    }
    return -1;
  }

nums= [0,1,2,3,3,4,5]
val= 3
print binarySearch(nums, val)

lower bound: 3
upper bound: 4
  
* insert a val
 private static int binarySearch(int [] array, int key){  
      int l= 0; 
      int r= array.length-1; 
      int m;
      
      while(l<=r){
        m= (l+r)/2;

        System.out.println(l+"/"+m+"/"+r);

        if(key<array[m]){
          r= m-1;
        } 
        else if( array[m] < key){
          l= m+1;
        }
        else{
          l= m;
          break;
        }
      }
      return l;
    }

nums= [0,1,2,3,3,4,5]
val= 3
print binarySearch(nums, val)
2

val= 2.5
print binarySearch(nums, val)
l,m,r
0 1 2
2 2 2
3 2 2
3

k close.py

* 658. Find K Closest Elements  
https://leetcode.com/problems/find-k-closest-elements/  
Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.
Example 1:
Input: [1,2,3,4,5], k=4, x=3
Output: [1,2,3,4]
Example 2:
Input: [1,2,3,4,5], k=4, x=-1
Output: [1,2,3,4]  
  
class Solution {
    public int findMin(int[] nums) {
        if(nums.length==1) return nums[0];
        
        int l=0, r=nums.length-1;
        while(l<=r) {
            if(nums[l]<nums[r]) return nums[l];
            
        	int m= (l+r)/2;
        	if( nums[m] > nums[m+1]) return  nums[m+1];
        	if( nums[m-1] > nums[m]) return  nums[m];
        	
        	if(nums[l] < nums[m]) l= m+1;
        	else r=m-1;
        
        }
    	
    	return -1;
    }
}

other.py

69. Sqrt(x)
https://leetcode.com/problems/sqrtx/
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.
class Solution(object):
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        l = 0
        r = x/2 +1

        while(l <= r):
            m = (r + l)/2
            
            if x < m*m: r = m - 1
            elif m*m < x: l = m + 1
            else:    return m

        return l - 1

peak.py

852. Peak Index in a Mountain Array
https://leetcode.com/problems/peak-index-in-a-mountain-array/
Let's call an array A a mountain if the following properties hold:
A.length >= 3
There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].
Example 1:
Input: [0,1,0]
Output: 1
Example 2:
Input: [0,2,1,0]
Output: 1  
  
class Solution(object):
    def peakIndexInMountainArray(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        l, r= 0, len(A)-1
        while l < r :
            m= (l+r)/2
            if  A[m-1] < A[m] > A[m+1]: return m
            elif A[m-1] < A[m] < A[m+1]: l=m
            #A[m-1] > A[m] > A[m+1] or A[m-1] > A[m] < A[m+1]
            else: r= m
              
              
162. Find Peak Element 
https://leetcode.com/problems/find-peak-element/
A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞.
Example 1:
Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.
Example 2:
Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5 
Explanation: Your function can return either index number 1 where the peak element is 2, 
             or index number 5 where the peak element is 6.
  
class Solution(object):
    def findPeakElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums= [float("-inf")]+ nums + [float("-inf") ]
        
        l, r= 0, len(nums)-1
        while l < r :
            m= (l+r)/2
            if nums[m-1] < nums[m] > nums[m+1]: return m-1
            elif  nums[m-1] < nums[m] < nums[m+1]: l= m
            #nums[m-1] > nums[m] > nums[m+1] or nums[m-1] > nums[m] < nums[m+1]
            else: r= m  
            
            
            
            

Two Sorted Arrays.py

4. Median of Two Sorted Arrays
https://leetcode.com/problems/median-of-two-sorted-arrays/
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5

class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        def findKth(a, b, k):
            if not a:   return b[k]
            if not b:   return a[k]
            
            ia, ib = len(a)/2 , len(b)/2
            ma, mb = a[ia], b[ib]

            # when k is bigger than the sum of a and b's median indices 
            if ia + ib < k:
                # if a's median is bigger than b's, b's first half doesn't include k
                if ma > mb: return findKth(a, b[ib + 1:], k - ib - 1)
                else:   return findKth(a[ia + 1:], b, k - ia - 1)
            # when k is smaller than the sum of a and b's indices
            else:
                # if a's median is bigger than b's, a's second half doesn't include k
                if ma > mb: return findKth(a[:ia], b, k)
                else:   return findKth(a, b[:ib], k)                
        
        l = len(nums1) + len(nums2)
        if l % 2 == 1:  return findKth(nums1, nums2, l/2)
        else: return ( findKth(nums1, nums2, l/2) + findKth(nums1, nums2, l/2-1) ) / 2.