2856. Minimum Array Length After Pair Removals

solution:

First take a look at a few cases:

• 1,2,3,4 -> 0
• 1,1,1,1,1,2 -> 4
• 1,1,1,1,2 -> 3

Which makes us thinking about something to do with the most frequent number in the array.

class Solution {
    public int minLengthAfterRemovals(List<Integer> nums) {
        int n = nums.size();
        // if a number appears more than a half times, 
        // it must appear at the middle of the sequence
        // we can use bin-search to get the first occurance and last, to calculate the count
        int maxFreqNum = nums.get(n / 2);

        int maxFreqCnt = findLastIdx(nums, maxFreqNum) - findFirstIdx(nums, maxFreqNum) + 1;

        if (maxFreqCnt <= n / 2) {
            return n % 2 == 0 ? 0 : 1;
        } else {
            return 2 * maxFreqCnt - n; // n - 2(n - maxFreCnt)
        }
    }

    public int findFirstIdx(List<Integer> nums, int target) {
        int left = 0, right = nums.size() - 1;
        int res = -1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums.get(mid) == target) {
                res = mid;
                right = mid - 1;
            } else if (nums.get(mid) > target) {
                right = mid -1;
            } else {
                left = mid + 1;
            }
        }
        return res;
    }

    public int findLastIdx(List<Integer> nums, int target) {
        int left = 0, right = nums.size() - 1;
        int res = -1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if (nums.get(mid) == target) {
                res = mid;
                left = mid + 1;
            } else if (nums.get(mid) > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return res;
    }
}