# Easy understanding Java solution using Binary Search.

• ``````public class SearchForRange {
public int[] searchRange(int[] nums, int target) {
int[] result = new int[]{-1, -1};
int hi = nums.length - 1;
int lo = 0;
while (lo <= hi) {
int mid = (hi + lo) / 2;
if (nums[mid] > target) {
hi = mid - 1;
} else if (nums[mid] < target) {
lo = mid + 1;
} else {
//move forward to find the first
int temp = mid;

for (; temp >= 0 && nums[temp] == target; temp--) {
result[0] = temp;
}
//move backward to find the last
temp = mid;
for (; temp < nums.length && nums[temp] == target; temp++) {
result[1] = temp;
}

if (result[1] == -1) {
result[1] = result[0];
}
return result;
}
}
return result;
}

public static void main(String[] args) {
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{5, 7, 7, 8, 8, 8, 10}, 8)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{2, 2}, 3)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{1}, 1)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{1, 2}, 2)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{3, 3}, 3)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{1, 2, 3}, 3)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{1, 1, 2}, 1)));
System.out.println(Arrays.toString(new SearchForRange().searchRange(new int[]{2, 2}, 2)));
}
``````

}

• In the worse case, this is O(n) solution. For example, the input could be [3,3,3,3,3,3] and target is 3. Actually, you could also apply binary search in your second "else" block.

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