2022-05-26

70. Climbing Stairs

Description

You are climbing a staircase. It takes n steps to reach the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Example 1:

Input: n = 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: n = 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

Constraints:

• 1 <= n <= 45

Solution

Approach #0

func climbStairs(n int) int {
    p, q, r := 0, 0, 1
    for i := 0; i < n; i++ {
        p = q
        q = r
        r = p + q
    }
    return r
}

198. House Robber

Description

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

Constraints:

• 1 <= nums.length <= 100

• 0 <= nums[i] <= 400

Solution

Approach #0

func rob(nums []int) int {
    if len(nums) == 0 {
        return 0
    }
    r0, r1 := 0, nums[0]
    for i := 1; i < len(nums); i++ {
        r0, r1 = r1, max(r1, r0+nums[i])
    }
    return r1
}

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

120. Triangle

Description

Given a triangle array, return the minimum path sum from top to bottom.

For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.

Example 1:

Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11
Explanation: The triangle looks like:
   2
  3 4
 6 5 7
4 1 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).

Example 2:

Input: triangle = [[-10]]
Output: -10

Constraints:

• 1 <= triangle.length <= 200

• triangle[0].length == 1

• triangle[i].length == triangle[i - 1].length + 1

• -104 <= triangle[i][j] <= 104

Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

Solution

Approach #0: Better than the official solution

func minimumTotal(triangle [][]int) int {
    if len(triangle) == 1 {
        return triangle[0][0]
    }
    for i := 1; i < len(triangle); i++ {
        size := len(triangle[i])
        triangle[i][0] = triangle[i-1][0] + triangle[i][0]
        for j := 1; j < size-1; j++ {
            triangle[i][j] = min(triangle[i-1][j], triangle[i-1][j-1]) + triangle[i][j]
        }
        triangle[i][size-1] = triangle[i-1][size-2] + triangle[i][size-1]
    }
    r := triangle[len(triangle)-1][0]
    for _, item := range triangle[len(triangle)-1] {
        r = min(r, item)
    }
    return r
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

699. Falling Squares

Description

There are several squares being dropped onto the X-axis of a 2D plane.

You are given a 2D integer array positions where positions[i] = [lefti, sideLengthi] represents the ith square with a side length of sideLengthi that is dropped with its left edge aligned with X-coordinate lefti.

Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands on the top side of another square or on the X-axis. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved.

After each square is dropped, you must record the height of the current tallest stack of squares.

Return an integer array ans where ans[i] represents the height described above after dropping the ith square.

Example 1:

Input: positions = [[1,2],[2,3],[6,1]]
Output: [2,5,5]
Explanation:
After the first drop, the tallest stack is square 1 with a height of 2.
After the second drop, the tallest stack is squares 1 and 2 with a height of 5.
After the third drop, the tallest stack is still squares 1 and 2 with a height of 5.
Thus, we return an answer of [2, 5, 5].

Example 2:

Input: positions = [[100,100],[200,100]]
Output: [100,100]
Explanation:
After the first drop, the tallest stack is square 1 with a height of 100.
After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.
Thus, we return an answer of [100, 100].
Note that square 2 only brushes the right side of square 1, which does not count as landing on it.

Constraints:

• 1 <= positions.length <= 1000

• 1 <= lefti <= 10^8

• 1 <= sideLengthi <= 10^6

Solution

Approach #0

func fallingSquares(positions [][]int) []int {
    n := len(positions)
    heights := make([]int, n)
    for i, p := range positions {
        left1, right1 := p[0], p[1]+p[0]-1
        heights[i] = p[1]
        for j, q := range positions[:i] {
            left2, right2 := q[0], q[1]+q[0]-1
            if left1 <= right2 && right1 >= left2 {
                heights[i] = max(heights[i], heights[j]+p[1])
            }
        }
    }
    for i := 1; i < n; i++ {
        heights[i] = max(heights[i], heights[i-1])
    }
    return heights
}

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

27. Remove Element

Description

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

• 0 <= nums.length <= 100

• 0 <= nums[i] <= 50

• 0 <= val <= 100

Solution

Approach #0

func removeElement(nums []int, val int) int {
    i := 0
    for j := 0; j < len(nums); j++ {
        if nums[j] != val {
            nums[i], nums[j] = nums[j], nums[i]
            i++
        }
    }
    return i
}

Approach #1

func removeElement(nums []int, val int) int {
    i, j := 0, len(nums)
    for i < j {
        if nums[i] != val {
            i++
        } else {
            nums[i] = nums[j-1]
            j--
        }
    }
    return i
}