34. Search for a Range

Given a sorted array of integers, 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]

.

Solution 1 Find the leftmost and rightmost bound using binary search

public static int[] searchRange(int[] nums, int target) {
int start = 0;
int end = nums.length - 1;
int range [] = {-1, -1};
while(start < end){
int mid = (start + end) / 2;
if(target > nums[mid]){
start = mid + 1;
}else{
end = mid;
}
}
if(nums[start] != target){
return range;
}else{
range[0] = start;
}
end = nums.length - 1;
while(start < end){
int mid = (start + end) / 2 + 1;
if(nums[mid] > target){
end = mid - 1;
}else{
start = mid;
}
}
range[1] = end;
return range;
}

Solution 2 Recursive

public static int[] searchRange2(int[] nums, int target) {
int [] range = {Integer.MAX_VALUE, -1};
searchRange(nums, target, 0, nums.length - 1, range);
if(range[0] > range[1]){
range[0] = -1;
}
return range;
}

public static void searchRange(int [] nums, int target, int left, int right, int [] range){
if(left > right){
return;
}
int mid = (left + right) / 2;
if(nums[mid] == target){
if(mid < range[0]){
range[0] = mid;
searchRange(nums, target, left, mid - 1, range);
}
if(mid > range[1]){
range[1] = mid;
searchRange(nums, target, mid + 1, right, range);
}
}else if(nums[mid] < target){
searchRange(nums, target, mid + 1, right, range);
}else{
searchRange(nums, target, left, mid - 1, range);
}
}