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/**

* Given an array of integers sorted in ascending order, find the starting and

* ending position of a given target value.

*

* Your algorithm's runtime complexity must be in the order of O(log n).

*

* If the target is not found in the array, return [-1, -1].

*

* For example,

* Given [5, 7, 7, 8, 8, 10] and target value 8,

* return [3, 4].

*/

public class SearchForARange34 {

public int[] searchRange(int[] nums, int target) {

int start = 0;

int end = nums.length - 1;

while (start <= end) {

if (nums[start] < target) {

start++;

} else if (nums[end] > target) {

end--;

} else {

break;

}

}

if (start > nums.length - 1 || end < 0 || start > end) {

return new int[]{-1, -1};

}

return new int[]{start, end};

}

public int[] searchRange2(int[] nums, int target) {

int start = 0;

int end = nums.length - 1;

int mid = (end - start) >> 1 + start;

while (start <= end) {

if (nums[mid] < target) {

start = mid + 1;

} else if (nums[mid] > target) {

end = mid - 1;

} else {

break;

}

mid = (end - start) / 2 + start;

}

while (start <= end) {

if (nums[start] < target) {

start++;

} else if (nums[end] > target) {

end--;

} else {

break;

}

}

if (start > nums.length - 1 || end < 0 || start > end) {

return new int[]{-1, -1};

}

return new int[]{start, end};

}

/**

* https://discuss.leetcode.com/topic/6327/a-very-simple-java-solution-with-only-one-binary-search-algorithm

*/

public int[] searchRange3(int[] A, int target) {

int start = Solution.firstGreaterEqual(A, target);

if (start == A.length || A[start] != target) {

return new int[]{-1, -1};

}

return new int[]{start, Solution.firstGreaterEqual(A, target + 1) - 1};

}

//find the first number that is greater than or equal to target.

//could return A.length if target is greater than A[A.length-1].

//actually this is the same as lower_bound in C++ STL.

private static int firstGreaterEqual(int[] A, int target) {

int low = 0, high = A.length;

while (low < high) {

int mid = low + ((high - low) >> 1);

//low <= mid < high

if (A[mid] < target) {

low = mid + 1;

} else {

//should not be mid-1 when A[mid]==target.

//could be mid even if A[mid]>target because mid<high.

high = mid;

}

}

return low;

}

/**

* https://discuss.leetcode.com/topic/10692/simple-and-strict-o-logn-solution-in-java-using-recursion

*/

public int[] searchRange4(int[] A, int target) {

int[] range = {A.length, -1};

searchRange(A, target, 0, A.length - 1, range);

if (range[0] > range[1]) range[0] = -1;

return range;

}

public void searchRange(int[] A, int target, int left, int right, int[] range) {

if (left > right) return;

int mid = left + (right - left) / 2;

if (A[mid] == target) {

if (mid < range[0]) {

range[0] = mid;

searchRange(A, target, left, mid - 1, range);

}

if (mid > range[1]) {

range[1] = mid;

searchRange(A, target, mid + 1, right, range);

}

} else if (A[mid] < target) {

searchRange(A, target, mid + 1, right, range);

} else {

searchRange(A, target, left, mid - 1, range);

}

}

public int[] searchRange5(int[] nums, int target) {

if (nums == null || nums.length == 0) return new int[]{-1, -1};

return searchRange(nums, target, 0, nums.length-1);

}

public int[] searchRange(int[] nums, int target, int lo, int hi) {

if (lo > hi) return new int[]{-1, -1};

int mid = (lo + hi) / 2;

if (nums[mid] > target) {

return searchRange(nums, target, lo, mid-1);

} else if (nums[mid] < target) {

return searchRange(nums, target, mid+1, hi);

} else {

int[] res = new int[]{mid, mid};

int[] left = searchRange(nums, target, lo, mid-1);

int[] right = searchRange(nums, target, mid+1, hi);

if (left[0] != -1) res[0] = left[0];

if (right[1] != -1) res[1] = right[1];

return res;

}

}

/**

* https://leetcode.com/problems/search-for-a-range/solution/

*/

// returns leftmost (or rightmost) index at which `target` should be

// inserted in sorted array `nums` via binary search.

private int extremeInsertionIndex(int[] nums, int target, boolean left) {

int lo = 0;

int hi = nums.length;

while (lo < hi) {

int mid = (lo+hi)/2;

if (nums[mid] > target || (left && target == nums[mid])) {

hi = mid;

}

else {

lo = mid+1;

}

}

return lo;

}

public int[] searchRange6(int[] nums, int target) {

int[] targetRange = {-1, -1};

int leftIdx = extremeInsertionIndex(nums, target, true);

// assert that `leftIdx` is within the array bounds and that `target`

// is actually in `nums`.

if (leftIdx == nums.length || nums[leftIdx] != target) {

return targetRange;

}

targetRange[0] = leftIdx;

targetRange[1] = extremeInsertionIndex(nums, target, false)-1;

return targetRange;

}

}