# 2022-05-27

Given an integer

`n`

, return *true*

*if it is a power of two. Otherwise, return**false*

.An integer

`n`

is a power of two, if there exists an integer `x`

such that `n == 2x`

.**Example 1:**

Input: n = 1

Output: true

Explanation: 20 = 1

**Example 2:**

Input: n = 16

Output: true

Explanation: 24 = 16

**Example 3:**

Input: n = 3

Output: false

**Constraints:**

`-2^31 <= n <= 2^31 - 1`

**Follow up:**Could you solve it without loops/recursion?

func isPowerOfTwo(n int) bool {

return n > 0 && n&(n-1) == 0

}

func isPowerOfTwo(n int) bool {

big := 1 << 30

return n > 0 && big%n == 0

}

Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

**Note:**

- Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
- In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in
**Example 3**, the input represents the signed integer.`-3`

.

**Example 1:**

Input: n = 00000000000000000000000000001011

Output: 3

Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

**Example 2:**

Input: n = 00000000000000000000000010000000

Output: 1

Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

**Example 3:**

Input: n = 11111111111111111111111111111101

Output: 31

Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

**Constraints:**

- The input must be a
**binary string**of length`32`

.

**Follow up:**If this function is called many times, how would you optimize it?

func hammingWeight(num uint32) int {

count := 0

for num > 0 {

num = num & (num - 1)

count++

}

return count

}

func hammingWeight(num uint32) int {

count := 0

for i := 0; i < 32; i++ {

if 1<<i&num > 0 {

count++

}

}

return count

}

Given the coordinates of two

**rectilinear**rectangles in a 2D plane, return*the total area covered by the two rectangles*.The first rectangle is defined by its

**bottom-left**corner`(ax1, ay1)`

and its **top-right**corner`(ax2, ay2)`

.The second rectangle is defined by its

**bottom-left**corner`(bx1, by1)`

and its **top-right**corner`(bx2, by2)`

.**Example 1:**

****

Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2

Output: 45

**Example 2:**

Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2

Output: 16

**Constraints:**

`-10^4 <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 10^4`

func computeArea(ax1 int, ay1 int, ax2 int, ay2 int, bx1 int, by1 int, bx2 int, by2 int) int {

area1 := (ax2 - ax1) * (ay2 - ay1)

area2 := (bx2 - bx1) * (by2 - by1)

overlapWidth := min(ax2, bx2) - max(ax1, bx1)

overlapHeight := min(ay2, by2) - max(ay1, by1)

overlapArea := max(overlapWidth, 0) * max(overlapHeight, 0)

return area1 + area2 - overlapArea

}

func max(a, b int) int {

if a > b {

return a

}

return b

}

func min(a, b int) int {

if a < b {

return a

}

return b

}

Last modified 9mo ago