Merging Two Sorted Arrays

Approach:

Use two pointers starting from the end of both arrays and fill the result array from the end to avoid overwriting.

Implementation:

class ArrayMerger {
    public void combineSortedArrays(int[] primary, int size1, int[] secondary, int size2) {
        int position = primary.length - 1;
        int ptr1 = size1 - 1, ptr2 = size2 - 1;
        
        while (ptr1 >= 0 && ptr2 >= 0) {
            if (primary[ptr1] > secondary[ptr2]) {
                primary[position--] = primary[ptr1--];
            } else {
                primary[position--] = secondary[ptr2--];
            }
        }
        
        while (ptr1 >= 0) primary[position--] = primary[ptr1--];
        while (ptr2 >= 0) primary[position--] = secondary[ptr2--];
    }
}

Removing Specific Elements

Approach:

Miantain two pointers where one traverses the array while the other builds the result by skipping unwanted values.

Implementation:

class ElementRemover {
    public int filterElements(int[] data, int target) {
        int resultIndex = 0;
        for (int current : data) {
            if (current != target) {
                data[resultIndex++] = current;
            }
        }
        return resultIndex;
    }
}

Eliminating Duplicates from Sorted Array

Approach:

Use two pointers to track unique elements, comparing each element with its predecessor.

Implementation:

class DuplicateHandler {
    public int removeDuplicates(int[] sortedData) {
        if (sortedData.length == 0) return 0;
        int uniqueCount = 1;
        for (int i = 1; i < sortedData.length; i++) {
            if (sortedData[i] != sortedData[i-1]) {
                sortedData[uniqueCount++] = sortedData[i];
            }
        }
        return uniqueCount;
    }
}

Handling Duplicates with Limited Occurrences

Approach:

Allow up to two duplicates by comparing elements with the one two positions back in the result array.

Implementation:

class LimitedDuplicateHandler {
    public int removeExcessDuplicates(int[] nums) {
        if (nums.length <= 2) return nums.length;
        int validIndex = 2;
        for (int j = 2; j < nums.length; j++) {
            if (nums[j] != nums[validIndex-2]) {
                nums[validIndex++] = nums[j];
            }
        }
        return validIndex;
    }
}

Finding the Dominant Element

Approach:

The majority element will always be at the middle position in a sorted array.

Implementation:

class MajorityFinder {
    public int findDominantElement(int[] elements) {
        Arrays.sort(elements);
        return elements[elements.length / 2];
    }
}

Rotating Array Elements

Approach:

Reverse the entire array, then reverse the first k elements and the remaining elements separately.

Implementation:

class ArrayRotator {
    public void circularShift(int[] arr, int rotations) {
        rotations %= arr.length;
        reverseSegment(arr, 0, arr.length - 1);
        reverseSegment(arr, 0, rotations - 1);
        reverseSegment(arr, rotations, arr.length - 1);
    }
    
    private void reverseSegment(int[] array, int start, int end) {
        while (start < end) {
            int temp = array[start];
            array[start++] = array[end];
            array[end--] = temp;
        }
    }
}