LeetCode 34. Find First and Last Position of Element in Sorted Array

Description

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 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]

Explanation

Use binary search twice to solve the problem. The first binary search to find the first occurrence of the target. The second binary search to find the second occurrence of the target.

Java Solution

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[2];
        result[0] = -1;
        result[1] = -1;
        
        if (nums == null || nums.length == 0) {
            return result;
        }
        
        // first binary search to find the first occurence of the target
        int start = 0;
        int end = nums.length - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                end = mid;
            } else if (nums[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }            
        }
        
        if (nums[end] == target) {
            result[0] = end;
        }
        if (nums[start] == target) {
            result[0] = start;
        }
        
        // second binary search to find the last occurence of the target
        start = 0;
        end = nums.length - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                start = mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (nums[start] == target) {
            result[1] = start; 
        }
        if (nums[end] == target) {
            result[1] = end;
        }
        
        return result;
    }
}

Python Solution

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        range_start = -1    
        range_end = -1    
        
        if (nums == None or len(nums) == 0):
            return [range_start, range_end]
        
        start = 0
        end = len(nums) - 1
        
        while start + 1 < end:
            mid = start + (end - start) // 2
            if nums[mid] < target:
                start = mid
            elif nums[mid] > target:
                end = mid
            else:
                start = mid
                
        if nums[end] == target:
            range_end = end
        elif nums[start] == target:
            range_end = start
            
        start = 0
        end = len(nums) - 1
        
        while start + 1 < end:
            mid = start + (end - start) // 2
            if nums[mid] < target:
                start = mid
            elif nums[mid] > target:
                end = mid
            else:
                end = mid
                
        if nums[start] == target:
            range_start = start
        elif nums[end] == target:
            range_start = end            
        
        return [range_start, range_end]

• Time complexity: O(log(N)).
• Space complexity: O(1).