# AUG!!! 0ms, beat 70%, O(log n), JAVA BS, basic and prunning.

• ``````public class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = {-1, -1};
if(nums == null || nums.length < 1)
return result;

return searchRangeHelper(nums,0, nums.length - 1, target, true, true);
}
public int[] searchRangeHelper(int[] nums, int start, int end, int target, boolean searchLeft, boolean searchRight)
{
int[] result = {-1, -1};
int left = start;
int right = end;
int mid = 0;
// Step 1, we need to find at least one value in range [left, right] equals to target
while(left <= right)
{
mid = (left + right) / 2;
if(nums[mid] == target)
{
// Step2, if we find that number and searchLeft is true, we need to keep search
//range [left, mid-1] to find if there exists a even lefter bond.
//meanwhile set searchRight to false because we do not need to find a right bond
// in the next search
if(mid > left && searchLeft)
{
int[] leftRes = searchRangeHelper(nums,start, mid - 1, target, true, false);
result[0] = leftRes[0] == -1? mid:leftRes[0];
}
else
result[0] = mid;
//Step3, just like step 2, we need to find the biggest right bond.
if(mid < right && searchRight)
{
int[] rightRes = searchRangeHelper(nums,mid + 1, end, target, false, true);
result[1] = rightRes[1] == -1? mid:rightRes[1];
}
else
result[1] = mid;

return result;
}
else
{
if(nums[mid] < target)
left = mid + 1;
else
right = mid - 1;
}
}
return result;
}
}
``````

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