667. Beautiful Arrangement II
1. Description
Given two integers n and k, you need to construct a list which contains n different positive integers ranging from 1 to n and obeys the following requirement:
Suppose this list is [a1, a2, a3, … , an], then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, … , |an-1 - an|] has exactly k distinct integers.
If there are multiple answers, print any of them.
2. Example
Example 1:
Input: n = 3, k = 1
Output: [1, 2, 3]
Explanation: The [1, 2, 3] has three different positive integers ranging from 1 to 3, and the [1, 1] has exactly 1 distinct integer: 1.
Example 2:
Input: n = 3, k = 2
Output: [1, 3, 2]
Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
3. Note
- The n and k are in the range 1 <= k < n <= $10^4$.
4. Solutions
My Accepted Solution
Time complexity: O(n)
Space complexity: O(n)
// we could set the array as the rule: 1, n, 2, n-1, 3, n-2, ...
// then, every two nums have a different difference, so we could use this rule to create k - 1 pair numbers, then, we could set last numbers keep progressive increase or progressive decrease, so every two numbers have a difference as 1
class Solution {
public:
vector<int> constructArray(int n, int k) {
bool increase = true;
int left = 1, right = n, index = 0;
vector<int> beautifulArray(n);
for (int i = 1; i < k; ++i) {
if (increase) {
// put the '++' or '--' operation into other codes could decrease lines dramatically, but it will make codes less understandable, so I don't want to do that
beautifulArray[index] = left;
++left;
} else {
beautifulArray[index] = right;
--right;
}
++index;
increase = !increase;
}
if (increase) {
while (index < n) {
beautifulArray[index] = left;
++index;
++left;
}
} else {
while (index < n) {
beautifulArray[index] = right;
++index;
--right;
}
}
return beautifulArray;
}
};