Tuesday, August 8th

 Value in a pivoted array

Given a sorted array without duplicates that may have been rotated, determine if the array contains a target value


As hinted by the required runtime complexity, this problem can be solved with binary search.  First binary search will be used to identify the pivot, then will run binary search on the two arrays divided on the pivot

        int n = nums.size();
int l = 0, r = n - 1;
        int mid;
while (l <= r) {
mid = (l + r) / 2;
if (nums[mid] > nums[n - 1]) {
l = mid + 1;
} else {
r = mid - 1;
}
}

Then perform binary search over the two arrays split by the pivot.  Binary search will be performed twice on the arrays, so we define a separate binary search function to call

        int binarySearch(vector<int> nums, int l, int r, int target) {
int mid;
while (l <= r) {
mid = (l+r) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > target) {
r = mid - 1;
} else {
l = mid + 1;
}
}
        return -1;

The binary search function will be called twice, once for the array [0, left of pivot] and [pivot, end of array].  If the first call returns -1, then the target isn't in the first array.

        int ans = binarySearch(nums, 0, l - 1, target);
if (ans != -1) return ans;
return binarySearch(nums, l, n - 1, target);

Solved 195/197 test cases with some devolved spaghetti code.  

Thank you to the Editorial for firming up binary search

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