2968. Apply Operations to Maximize Frequency Score

class Solution {
    private boolean ableToGetFrequency(int proposedLength, int numLength, long k, int[] nums, long[] prefixSum) {
        int L = 0, R = 0;

        while(R < numLength) {
            int subArrayLength = (R - L + 1);

            if(subArrayLength < proposedLength) {
                R += 1;
            } else {
                int medianIndex = L + (subArrayLength / 2);
                long leftHalfPrefixSum = prefixSum[medianIndex];
                if(L > 0) {
                    leftHalfPrefixSum -= prefixSum[L - 1];
                }

                long rightHalfPrefixSum = prefixSum[R];
                rightHalfPrefixSum -= prefixSum[medianIndex];

                long requiredOperations = ((long)nums[medianIndex] * (medianIndex - L + 1)) - leftHalfPrefixSum;
                requiredOperations += rightHalfPrefixSum - ((long)nums[medianIndex] * (R - medianIndex));

                if(requiredOperations <= k) {
                    return true;
                }

                L += 1;
                R += 1;
            }
        }

        return false;
    }

    public int maxFrequencyScore(int[] nums, long k) {
        int numLength = nums.length;

        Arrays.sort(nums);

        long[] prefixSum = new long[numLength];
        for(int i = 0; i < numLength; i++) {
            prefixSum[i] = nums[i];
            if(i > 0) prefixSum[i] += prefixSum[i - 1];
        }

        int result = 1;
        int L = 1, R = numLength;

        while (L <= R) {
            int mid = L + ((R - L) >> 1);

            if(ableToGetFrequency(mid, numLength, k, nums, prefixSum)) {
                result = mid;
                L = mid + 1;
            } else {
                R = mid - 1;
            }
        }

        return result;
    }
}