Find First and Last Position of Element in Sorted Array

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Question

Given an array of integers nums 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].

Example

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

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

Solution

Use binary search approach. Finding the first one and should check the left first and then right pointer. For the second one, we should find the first number that is greater than our target and then check the right first.

Code

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// time:O(logn)
// space:O(1)
public int[] searchRange(int[] nums, int target) {
        if (nums == null || nums.length == 0) return new int[]{-1,-1};
        int first = findFirst(nums, target);
        int second = findSecond(nums, target);
        if (first != -1 && second != -1) return new int[]{first, second};
        else return new int[]{-1,-1};
    }
    
    public int findFirst(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] >= target) right = mid;
            else left = mid;
        }
        if (nums[left] == target) return left;
        if (nums[right] == target) return right;
        return -1;
    }
    
    public int findSecond(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target) right = mid;
            else left = mid;
        }
        if (nums[right] == target) return right;
        if (nums[left] == target) return left;
        return -1;
    }