2019-02-23

.461. Hamming Distance
.944. Delete Columns to Make Sorted

note:
bin(x^y)

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461. Hamming Distance

1) Description

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 2³¹.

Example:

Input:x = 1, y = 4
Output:2
Explanation:
1 (0 0 0 1)
4 (0 1 0 0)
↑ ↑
The above arrows point to positions where the corresponding bits are different.

2) Solution

class Solution:
    def hammingDistance(self, x: 'int', y: 'int') -> 'int': 
        A = list(bin(x).replace('0b','')) 
        B = list(bin(y).replace('0b',''))
        res=0
        for i in range(max(0,len(A),len(B))):
            if i > len(A) -1:
                A.insert(0,"0")  
            if i > len(B) -1:
                B.insert(0,"0")
            if A[~i] != B[~i]:
                res = res+1
        return res

944. Delete Columns to Make Sorted

1) Description

We are given an array A of N lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have an array A = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"], and the remaining columns of A are ["b","v"], ["e","y"], and ["f","z"]. (Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]].)

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.

Return the minimum possible value of D.length.
Example 1:

Input: ["cba","daf","ghi"]
Output: 1
Explanation:
After choosing D = {1}, each column ["c","d","g"] and ["a","f","i"] are in non-decreasing sorted order.
If we chose D = {}, then a column ["b","a","h"] would not be in non-decreasing sorted order.

Example 2:

Input: ["a","b"]
Output: 0
Explanation: D = {}

2) Solution

class Solution:
    def minDeletionSize(self, A: List[str]) -> int:
        res=0
        for i in A:
            i=list(i)
        for k in range(len(A[-1])):
            for j in range(len(A)-1):
                if ord(A[j][k]) > ord(A[j+1][k]):
                    res = res+1
                    break
        return res

other 's

class Solution(object):
    def minDeletionSize(self, A):
        ans = 0
        for col in zip(*A):
            if any(col[i] > col[i+1] for i in xrange(len(col) - 1)):
                ans += 1
        return ans