2022-06-16

384. Shuffle an Array

Description

Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.

Implement the Solution class:

• Solution(int[] nums) Initializes the object with the integer array nums.

• int[] reset() Resets the array to its original configuration and returns it.

• int[] shuffle() Returns a random shuffling of the array.

Example 1:

Input
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
Output
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.shuffle();    // Shuffle the array [1,2,3] and return its result.
                       // Any permutation of [1,2,3] must be equally likely to be returned.
                       // Example: return [3, 1, 2]
solution.reset();      // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
solution.shuffle();    // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]

Constraints:

• 1 <= nums.length <= 50

• -10^6 <= nums[i] <= 10^6

• All the elements of nums are unique.

• At most 10^4 calls in total will be made to reset and shuffle.

Solution

Approach #0

type Solution struct {
    nums, original []int
}

func Constructor(nums []int) Solution {
    return Solution{nums, append([]int(nil), nums...)}
}

func (this *Solution) Reset() []int {
    copy(this.nums, this.original)
    return this.nums
}

func (this *Solution) Shuffle() []int {
    shuffle := make([]int, len(this.nums))
    for i := range shuffle {
        j := rand.Intn(len(this.nums))
        shuffle[i] = this.nums[j]
        this.nums = append(this.nums[:j], this.nums[j+1:]...)
    }
    this.nums = shuffle
    return this.nums
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(nums);
 * param_1 := obj.Reset();
 * param_2 := obj.Shuffle();
 */

Approach #1

type Solution struct {
    nums, original []int
}

func Constructor(nums []int) Solution {
    return Solution{nums, append([]int(nil), nums...)}
}

func (this *Solution) Reset() []int {
    copy(this.nums, this.original)
    return this.nums
}

func (this *Solution) Shuffle() []int {
    n := len(this.nums)
    for i := range this.nums {
        j := i + rand.Intn(n-i)
        this.nums[i], this.nums[j] = this.nums[j], this.nums[i]
    }
    return this.nums
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(nums);
 * param_1 := obj.Reset();
 * param_2 := obj.Shuffle();
 */

202. Happy Number

Description

Write an algorithm to determine if a number n is happy.

A happy number is a number defined by the following process:

• Starting with any positive integer, replace the number by the sum of the squares of its digits.

• Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.

• Those numbers for which this process ends in 1 are happy.

Return true if n is a happy number, and false if not.

Example 1:

Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

Example 2:

Input: n = 2
Output: false

Constraints:

• 1 <= n <= 2^31 - 1

Solution

Approach #0: Fast ans slow pointer

func isHappy(n int) bool {
    slow, fast := n, step(n)
    for fast != 1 && fast != slow {
        slow = step(slow)
        fast = step(step(fast))
    }
    return fast == 1
}

func step(n int) int {
    var sum int
    for n > 0 {
        sum += (n % 10) * (n % 10)
        n /= 10
    }
    return sum
}

149. Max Points on a Line

Description

Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line.

Example 1:

Input: points = [[1,1],[2,2],[3,3]]
Output: 3

Example 2:

Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4

Constraints:

• 1 <= points.length <= 300

• points[i].length == 2

• -104 <= xi, yi <= 104

• All the points are unique.

Solution

Approach #0

func maxPoints(points [][]int) (ans int) {
    n := len(points)
    if n <= 2 {
        return n
    }
    for i, p := range points {
        if ans >= n-i || ans > n/2 {
            break
        }
        cnt := make(map[int]int)
        for _, q := range points[i+1:] {
            x, y := p[0]-q[0], p[1]-q[1]
            if x == 0 {
                y = 1
            }
            if y == 0 {
                x = 1
            }
            if y < 0 {
                x, y = -x, -y
            }
            g := gcd(abs(x), abs(y))
            x /= g
            y /= g
            cnt[y+20001*x]++
        }
        for _, c := range cnt {
            ans = max(ans, c+1)
        }
    }
    return
}

func gcd(a, b int) int {
    for a > 0 {
        a, b = b%a, a
    }
    return b
}

func abs(a int) int {
    if a < 0 {
        return -a
    }
    return a
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

532. K-diff Pairs in an Array

Description

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

• 0 <= i, j < nums.length

• i != j

• nums[i] - nums[j] == k

Notice that |val| denotes the absolute value of val.

Example 1:

Input: nums = [3,1,4,1,5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input: nums = [1,2,3,4,5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Constraints:

• 1 <= nums.length <= 10^4

• -10^7 <= nums[i] <= 10^7

• 0 <= k <= 10^7

Solution

Approach #0

func findPairs(nums []int, k int) (ans int) {
    sort.Ints(nums)
    j, n := 0, len(nums)
    for i, num := range nums {
        if i == 0 || nums[i-1] != num {
            for j < n && (nums[j] < k+num || j <= i) {
                j++
            }
            if j < n && nums[j] == k+num {
                ans++
            }
        }
    }
    return
}