Today I have done Find N Unique Integers Sum up to Zero and leetcode's May LeetCoding Challenge with cpp.

Find N Unique Integers Sum up to Zero

Description

Given an integer n, return any array containing n unique integers such that they add up to 0.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

• 1 <= n <= 1000

Solution

class Solution {
public:
  vector<int> sumZero(int n) {
    vector<int> answer(n);
    for(int i = 0; i * 2 + 1 < n; ++i) {
      answer[i * 2] = i + 1;
      answer[i * 2 + 1] = -i - 1;
    }
    return move(answer);
  }
};

May LeetCoding Challenge 4

Description

Non-decreasing Array

Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element.

We define an array is non-decreasing if nums[i] <= nums[i + 1] holds for every i (0-based) such that (0 <= i <= n - 2).

Example 1:

Input: nums = [4,2,3]
Output: true
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

Example 2:

Input: nums = [4,2,1]
Output: false
Explanation: You can't get a non-decreasing array by modify at most one element.

Constraints:

• n == nums.length
• 1 <= n <= 104
• -10 <= nums[i] <= 105

Solution

class Solution {
public:
  bool checkPossibility(vector<int>& nums) {
    int len = nums.size();
    int addCount = 0;
    int minusCount = 0;
    vector<int> copies = nums;
    for(int i = 0; i < len - 1; ++i) {
      if(nums[i] > nums[i + 1]) {
        nums[i + 1] = nums[i];
        addCount += 1;
      }
    }
    for(int i = len - 1; i > 0; --i) {
      if(copies[i] < copies[i - 1]) {
        copies[i - 1] = copies[i];
        minusCount += 1;
      }
    }
    return addCount < 2 || minusCount < 2;
  }
};