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Huahua's Tech Road

花花酱 LeetCode 1323. Maximum 69 Number

Given a positive integer num consisting only of digits 6 and 9.

Return the maximum number you can get by changing at most one digit (6 becomes 9, and 9 becomes 6).

Example 1:

Input: num = 9669
Output: 9969
Explanation: 
Changing the first digit results in 6669.
Changing the second digit results in 9969.
Changing the third digit results in 9699.
Changing the fourth digit results in 9666. 
The maximum number is 9969.

Example 2:

Input: num = 9996
Output: 9999
Explanation: Changing the last digit 6 to 9 results in the maximum number.

Example 3:

Input: num = 9999
Output: 9999
Explanation: It is better not to apply any change.

Constraints:

  • 1 <= num <= 10^4
  • num‘s digits are 6 or 9.

Solution: Greedy

Replace the highest 6 to 9, if no 6, return the original number.

Time complexity: O(1)
Space complexity: O(1)

C++

花花酱 LeetCode 1325. Delete Leaves With a Given Value

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value targetif it’s parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can’t).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). 
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

  • 1 <= target <= 1000
  • Each tree has at most 3000 nodes.
  • Each node’s value is between [1, 1000].

Solution: Recursion

Post-order traversal

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 1326. Minimum Number of Taps to Open to Water a Garden

There is a one-dimensional garden on the x-axis. The garden starts at the point 0 and ends at the point n. (i.e The length of the garden is n).

There are n + 1 taps located at points [0, 1, ..., n] in the garden.

Given an integer n and an integer array ranges of length n + 1 where ranges[i] (0-indexed) means the i-th tap can water the area [i - ranges[i], i + ranges[i]] if it was open.

Return the minimum number of taps that should be open to water the whole garden, If the garden cannot be watered return -1.

Example 1:

Input: n = 5, ranges = [3,4,1,1,0,0]
Output: 1
Explanation: The tap at point 0 can cover the interval [-3,3]
The tap at point 1 can cover the interval [-3,5]
The tap at point 2 can cover the interval [1,3]
The tap at point 3 can cover the interval [2,4]
The tap at point 4 can cover the interval [4,4]
The tap at point 5 can cover the interval [5,5]
Opening Only the second tap will water the whole garden [0,5]

Example 2:

Input: n = 3, ranges = [0,0,0,0]
Output: -1
Explanation: Even if you activate all the four taps you cannot water the whole garden.

Example 3:

Input: n = 7, ranges = [1,2,1,0,2,1,0,1]
Output: 3

Example 4:

Input: n = 8, ranges = [4,0,0,0,0,0,0,0,4]
Output: 2

Example 5:

Input: n = 8, ranges = [4,0,0,0,4,0,0,0,4]
Output: 1

Constraints:

  • 1 <= n <= 10^4
  • ranges.length == n + 1
  • 0 <= ranges[i] <= 100

Solution 1: Greedy

Reduce to 1024. Video Stitching

Time complexity: O(nlogn)
Space complexity: O(n)

C++

Solution 2: Greedy

Reduce to 45. Jump Game II

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 42. Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.


The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

Solution 1: Brute Force

r[i] = min(max(h[0:i+1]), max(h[i:n]))
ans = sum(r[i])

Time complexity: O(n^2)
Space complexity: O(1)

C++

Solution 2: DP

l[i] := max(h[0:i+1])
r[i] := max(h[i:n])
ans = sum(min(l[i], r[i]) – h[i])

Time complexity: O(n)
Space complexity: O(n)

C++

Solution 3: Two Pointers

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 1318. Minimum Flips to Make a OR b Equal to c

Given 3 positives numbers ab and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation).
Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.

Example 1:

Input: a = 2, b = 6, c = 5
Output: 3
Explanation: After flips a = 1 , b = 4 , c = 5 such that (a OR b == c)

Example 2:

Input: a = 4, b = 2, c = 7
Output: 1

Example 3:

Input: a = 1, b = 2, c = 3
Output: 0

Constraints:

  • 1 <= a <= 10^9
  • 1 <= b <= 10^9
  • 1 <= c <= 10^9

Solution: Bit operation

If the bit of c is 1, a / b at least has one 1. cost = 1 – ((a | b) & 1)
If the bit of c is 0, a / b must be 0, cost = (a & 1) + (b & 1)

Time complexity: O(32)
Space complexity: O(1)

C++