[GeeksForGeeks] Find if there is a pair with a given sum in a sorted and rotated array.

Given a sorted array, this array is rotated by some unknown times. Find if there is a pair in this array 

that sums to a given value.

Solution 1. O(n * log n) runtime, sort this array then use two pointers to check if there is a pair. 

Solution 2. O(n) runtime, rotate this array back to its sorted order then use two pointers to check if there is a pair. 

How to rotate: first find the split point where the natural order is broke; then apply the 3-step rotation algorithm.

Solution 3. optimal:  O(n) runtime, two pointers without modifying the given array.

public boolean checkPairSum(int[] arr, int target) {
    if(arr == null || arr.length < 2) {
        return false;
    }
    int end = 0, start = 0;
    for(; end < arr.length - 1; end++) {
        if(arr[end] > arr[end + 1]) {
            break;
        }
    }
    start = (end + 1) % arr.length;
    while(start != end) {
        int sum = arr[start] + arr[end];
        if(sum == target) {
            return true;
        }
        else if(sum < target) {
            start = (start + 1) % arr.length;
        }
        else {
            end = (end - 1 + arr.length) % arr.length;
        }
    }
    return false;
}