Today is Sunday, I gonna review the tasks I've done this week, and finish today's leetcode's March LeetCoding Challenge with cpp.

LeetCode Review

Reverse Subarray To Maximize Array Value

image array [..., a, b, ... , c, d, ...] and we reverse [b, ..., c], array value is only change at pair(a, b) and pair(c, d), which turns into pair(a, c) and pair(b, d). So we want to maximize this change.

see this image

class Solution {
public:
  int maxValueAfterReverse(vector<int>& nums) {
    int minMax = INT_MAX;
    int maxMin = INT_MIN;
    int len = nums.size();
    int value = 0;
    for(int i = 1; i < len; ++i) value += abs(nums[i] - nums[i - 1]);
    
    for(int i = 1; i < len; ++i) {
      minMax = min(minMax, max(nums[i - 1], nums[i]));
      maxMin = max(maxMin, min(nums[i - 1], nums[i]));
    }
    
    int change = 2 * (maxMin - minMax);
    for(int i = 1; i < len; ++i) {
      change = max(change, abs(nums[0] - nums[i]) - abs(nums[i] - nums[i - 1]));
    }
    for(int i = 0; i < len - 1; ++i) {
      change = max(change, abs(nums[len - 1] - nums[i]) - abs(nums[i] - nums[i + 1]));
    }
    return value + change;
  }
};

March LeetCoding Challenge 21

Description

Reordered Power of 2

Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2.

Example 1:

Input: 1
Output: true

Example 2:

Input: 10
Output: false

Example 3:

Input: 16
Output: true

Example 4:

Input: 24
Output: false

Example 5:

Input: 46
Output: true

Note:

1. 1 <= N <= 10^9

Solution

// sorted powers of 2
const set<string> st = {"011237","0122579","012356789","0124","0134449","0145678","01466788","0248","0368888","0469","1","112234778","11266777","122446","125","128","1289","13468","16","2","224588","23","23334455","234455668","23678","256","35566","4","46","8"};
class Solution {
public:
  bool reorderedPowerOf2(int N) {
    string n = to_string(N);
    sort(n.begin(), n.end());
    return st.count(n);
  }
};