Medium
Minimum Removals to Balance Array — C++
Full explanation · Time O(nlogn) · Space O(1)
// Time: O(nlogn)
// Space: O(1)
// sort, two pointers
class Solution {
public:
int minRemoval(vector<int>& nums, int k) {
sort(begin(nums), end(nums));
int left = 0;
for (int right = 0; right < size(nums); ++right) {
if (static_cast<int64_t>(nums[left]) * k < nums[right]) {
++left;
}
}
return left;
}
};
// Time: O(nlogn)
// Space: O(1)
// sort, two pointers
class Solution2 {
public:
int minRemoval(vector<int>& nums, int k) {
sort(begin(nums), end(nums));
int result = 0;
for (int right = 0, left = 0; right < size(nums); ++right) {
while (static_cast<int64_t>(nums[left]) * k < nums[right]) {
++left;
}
result = max(result, right - left + 1);
}
return size(nums) - result;
}
};