add complement of base 10 integer soln
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0476_number-complement/README.md
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0476_number-complement/README.md
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The **complement** of an integer is the integer you get when you flip all the `0`'s to `1`'s and all the `1`'s to `0`'s in its binary representation.
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* For example, The integer `5` is `"101"` in binary and its **complement** is `"010"` which is the integer `2`.
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Given an integer `num`, return _its complement_.
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**Example 1:**
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Input: num = 5
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Output: 2
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Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
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**Example 2:**
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Input: num = 1
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Output: 0
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Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
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**Constraints:**
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* `1 <= num < 231`
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**Note:** This question is the same as 1009: [https://leetcode.com/problems/complement-of-base-10-integer/](https://leetcode.com/problems/complement-of-base-10-integer/)
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0476_number-complement/python3/linear.py
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0476_number-complement/python3/linear.py
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# Time: O(N) where N is number of bits
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class Solution:
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def findComplement(self, n: int) -> int:
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if n == 0:
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return 1
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cn, bit = n, 1
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while n > 0:
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# XOR 1 can be used to flip bit.
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# 0 XOR 1 = 1
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# 1 XOR 1 = 0
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cn = cn ^ bit
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bit = bit << 1
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n = n >> 1
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return cn
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0476_number-complement/python3/solution.py
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0476_number-complement/python3/solution.py
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class Solution:
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def findComplement(self, n: int) -> int:
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if n == 0:
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return 1
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mask = n
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# Create mask of length matching num of bits
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# in n (taken from Hacker's Delight, Figure 3-1)
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mask |= mask >> 1 # \
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mask |= mask >> 2 # \
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mask |= mask >> 4 # |> Totalling 31 bits
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mask |= mask >> 8 # /
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mask |= mask >> 16 # /
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# XOR 1 will flip the bit
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return n ^ mask
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1054_complement-of-base-10-integer/README.md
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1054_complement-of-base-10-integer/README.md
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The **complement** of an integer is the integer you get when you flip all the `0`'s to `1`'s and all the `1`'s to `0`'s in its binary representation.
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* For example, The integer `5` is `"101"` in binary and its **complement** is `"010"` which is the integer `2`.
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Given an integer `n`, return _its complement_.
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**Example 1:**
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Input: n = 5
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Output: 2
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Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
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**Example 2:**
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Input: n = 7
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Output: 0
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Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
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**Example 3:**
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Input: n = 10
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Output: 5
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Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
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**Constraints:**
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* `0 <= n < 109`
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**Note:** This question is the same as 476: [https://leetcode.com/problems/number-complement/](https://leetcode.com/problems/number-complement/)
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1054_complement-of-base-10-integer/python3/linear.py
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1054_complement-of-base-10-integer/python3/linear.py
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# Time: O(N) where N is number of bits
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class Solution:
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def bitwiseComplement(self, n: int) -> int:
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if n == 0:
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return 1
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cn, bit = n, 1
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while n > 0:
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# XOR 1 can be used to flip bit.
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# 0 XOR 1 = 1
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# 1 XOR 1 = 0
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cn = cn ^ bit
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bit = bit << 1
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n = n >> 1
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return cn
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1054_complement-of-base-10-integer/python3/solution.py
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1054_complement-of-base-10-integer/python3/solution.py
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class Solution:
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def bitwiseComplement(self, n: int) -> int:
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if n == 0:
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return 1
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mask = n
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# Create mask of length matching num of bits
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# in n (taken from Hacker's Delight, Figure 3-1)
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mask |= mask >> 1 # \
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mask |= mask >> 2 # \
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mask |= mask >> 4 # |> Totalling 31 bits
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mask |= mask >> 8 # /
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mask |= mask >> 16 # /
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# XOR 1 will flip the bit
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return n ^ mask
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