Description
You are given a 1-indexed array of integers nums of length n.
We define a function greaterCount such that greaterCount(arr, val) returns the number of elements in arr that are strictly greater than val.
You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the ith operation:
- If
greaterCount(arr1, nums[i]) > greaterCount(arr2, nums[i]), appendnums[i]toarr1. - If
greaterCount(arr1, nums[i]) < greaterCount(arr2, nums[i]), appendnums[i]toarr2. - If
greaterCount(arr1, nums[i]) == greaterCount(arr2, nums[i]), appendnums[i]to the array with a lesser number of elements. - If there is still a tie, append
nums[i]toarr1.
The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].
Return the integer array result.
Example 1:
Input: nums = [2,1,3,3] Output: [2,3,1,3] Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1]. In the 3rd operation, the number of elements greater than 3 is zero in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 3 is zero in both arrays. As the length of arr2 is lesser, hence, append nums[4] to arr2. After 4 operations, arr1 = [2,3] and arr2 = [1,3]. Hence, the array result formed by concatenation is [2,3,1,3].
Example 2:
Input: nums = [5,14,3,1,2] Output: [5,3,1,2,14] Explanation: After the first 2 operations, arr1 = [5] and arr2 = [14]. In the 3rd operation, the number of elements greater than 3 is one in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 1 is greater in arr1 than arr2 (2 > 1). Hence, append nums[4] to arr1. In the 5th operation, the number of elements greater than 2 is greater in arr1 than arr2 (2 > 1). Hence, append nums[5] to arr1. After 5 operations, arr1 = [5,3,1,2] and arr2 = [14]. Hence, the array result formed by concatenation is [5,3,1,2,14].
Example 3:
Input: nums = [3,3,3,3] Output: [3,3,3,3] Explanation: At the end of 4 operations, arr1 = [3,3] and arr2 = [3,3]. Hence, the array result formed by concatenation is [3,3,3,3].
Constraints:
3 <= n <= 1051 <= nums[i] <= 109
Code
Time Complexity: , Space Complexity:
這題重點在於如何使用 upper_bound 以及 insert。
upper_bound(arr.begin(), arr.end(), val); 會回傳嚴格大於 val 的 位置，如果沒有找到符合的，會回傳 arr.end()。
left.insert(left.end() - x , nums[i]); 會在 left.end() - x 這個位置上插入 nums[i]。
ll.insert(ll.end(), rr.begin(), rr.end()); 會在 ll 的尾端插入整個 rr vector。
class Solution {
public:
int greaterCount(vector<int>& arr, int val) {
return arr.end() - upper_bound(arr.begin(), arr.end(), val);
}
vector<int> resultArray(vector<int>& nums) {
int n = nums.size();
vector<int> left, right, ll, rr;
left.push_back(nums[0]);
right.push_back(nums[1]);
ll.push_back(nums[0]);
rr.push_back(nums[1]);
int l = 0, r = 0;
for (int i = 2; i < n; i++) {
int x = greaterCount(left, nums[i]);
int y = greaterCount(right, nums[i]);
if (x > y) {
left.insert(left.end() - x , nums[i]);
ll.push_back(nums[i]);
l++;
} else if (x < y) {
right.insert(right.end() - y, nums[i]);
rr.push_back(nums[i]);
r++;
} else {
if (l <= r) {
left.insert(left.end() - x, nums[i]);
ll.push_back(nums[i]);
l++;
} else {
right.insert(right.end() - y, nums[i]);
rr.push_back(nums[i]);
r++;
}
}
}
ll.insert(ll.end(), rr.begin(), rr.end());
return ll;
}
};