Adding solution of 1012, 1013 problem

Author: Garvit244

1000-1100q/1012.py

@@ -0,0 +1,41 @@
+'''Every non-negative integer N has a binary representation. For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on. Note that except for N = 0, there are no leading zeroes in any binary representation.
+
+The complement of a binary representation is the number in binary you get when changing every 1 to a 0 and 0 to a 1. For example, the complement of "101" in binary is "010" in binary.
+
+For a given number N in base-10, return the complement of it's binary representation as a base-10 integer.
+
+ 
+
+Example 1:
+
+Input: 5
+Output: 2
+Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
+Example 2:
+
+Input: 7
+Output: 0
+Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
+Example 3:
+
+Input: 10
+Output: 5
+Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
+ 
+
+Note:
+
+0 <= N < 10^9
+'''
+
+class Solution(object):
+ def bitwiseComplement(self, N):
+ """
+ :type N: int
+ :rtype: int
+ """
+ if N == 0:
+ return 1
+ import math
+ bits = (int)(math.floor(math.log(N) /math.log(2))) + 1
+ return ((1 << bits) - 1) ^ N

1013.py

@@ -0,0 +1,51 @@
+'''
+In a list of songs, the i-th song has a duration of time[i] seconds. 
+
+Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.
+
+ 
+
+Example 1:
+
+Input: [30,20,150,100,40]
+Output: 3
+Explanation: Three pairs have a total duration divisible by 60:
+(time[0] = 30, time[2] = 150): total duration 180
+(time[1] = 20, time[3] = 100): total duration 120
+(time[1] = 20, time[4] = 40): total duration 60
+Example 2:
+
+Input: [60,60,60]
+Output: 3
+Explanation: All three pairs have a total duration of 120, which is divisible by 60.
+
+Note:
+
+1 <= time.length <= 60000
+1 <= time[i] <= 500
+'''
+
+class Solution(object):
+ def numPairsDivisibleBy60(self, time):
+ """
+ :type time: List[int]
+ :rtype: int
+ """
+ count_arr = [0]*60
+ result = 0
+ for t in time:
+ remainder = t%60
+ complement = (60-remainder)%60
+ result += count_arr[complement]
+ count_arr[remainder] += 1
+ return result
+ 
+ '''
+Explanation: 
+Q1: why create array of size 60? 
+it is similar to the map which store the count. Why only 60 because 60 modulo of number cannot be more than 60
+Q2: why we need complement?
+to check the pair if it exisit with given value or not example: if remainder is 20 then we need to check if we have any number with remainder 40 or not.
+Q3: why 60 modulo complement?
+for handle cause when remainder is zero 
+ '''

README.md

@@ -7,8 +7,10 @@ Python solution of problems from [LeetCode](https://leetcode.com/).
 ##### [Problems 1000-1100](./1000-1100q/)
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
-|1002|[Find Common Characters](https://leetcode.com/problems/find-common-characters)|[Python](./1000-1100q/1002.py)|Easy|
 |1014|[Capacity To Ship Packages Within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/) | [Python](./1000-1100q/1014.py) | Medium|
+|1013|[Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)|[Python](./1000-1100q/1013.py)|Easy|
+|1012|[Complement of Base 10 Integer](https://leetcode.com/problems/complement-of-base-10-integer/) | [Python](./1000-1100q/1012.py)|Easy|
+|1002|[Find Common Characters](https://leetcode.com/problems/find-common-characters)|[Python](./1000-1100q/1002.py)|Easy|
 
 ##### [Problems 900-1000](./900-1000q/)
 | # | Title | Solution | Difficulty |