[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.

 1 public boolean checkPairSum(int[] arr, int target) {
 2     if(arr == null || arr.length < 2) {
 3         return false;
 4     }
 5     int end = 0, start = 0;
 6     for(; end < arr.length - 1; end++) {
 7         if(arr[end] > arr[end + 1]) {
 8             break;
 9         }
10     }
11     start = (end + 1) % arr.length;
12     while(start != end) {
13         int sum = arr[start] + arr[end];
14         if(sum == target) {
15             return true;
16         }
17         else if(sum < target) {
18             start = (start + 1) % arr.length;
19         }
20         else {
21             end = (end - 1 + arr.length) % arr.length;
22         }
23     }
24     return false;
25 }