The challenge

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

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

Follow up: Could you write an algorithm with O(log n) runtime complexity?

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

Constraints:

  • 0 <= nums.length <= 10<sup>5</sup>
  • -10<sup>9</sup>&nbsp;<= nums[i]&nbsp;<= 10<sup>9</sup>
  • nums is a non-decreasing array.
  • -10<sup>9</sup>&nbsp;<= target&nbsp;<= 10<sup>9</sup>

The solution in Java code

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class Solution {
    public int[] searchRange(int[] nums, int target) {
        int first = -1, last = -1;
        
        for (int i=0; i<nums.length; i++) {
            if (nums[i]==target) first = i;
            continue;
        }
        
        for (int i=nums.length-1; i>=0; i--) {
            if (nums[i]==target) last = i;
            continue;
        }
        
        return new int[] {last, first};
    }
}