Problem
Given a sorted array arr that may contain duplicates, find the first and last occurrence of a target element x.
If x is not present, return [-1, -1].
Examples
Input
arr = [1, 3, 5, 5, 5, 5, 67, 123, 125], x = 5
Output
[2, 5]
Explanation: First occurrence of 5 is at index 2, last at index 5.
Input
arr = [1, 3, 5, 5, 5, 5, 7, 123, 125], x = 7
Output
[6, 6]
Explanation: First and last occurrence of 7 is at index 6.
Input
arr = [1, 2, 3], x = 4
Output
[-1, -1]
Explanation: 4 is not present.
Approach
Use binary search to find the first occurrence of x.
Use binary search again to find the last occurrence.
If x is not found in either step, return [-1, -1].
This is efficient because the array is sorted, giving O(log n) time complexity.
Python Code
class Solution:
def find(self, arr, x):
n = len(arr)
first = -1
last = -1
# Find first occurrence
low = 0
high = n - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] == x:
first = mid
high = mid - 1
elif arr[mid] < x:
low = mid + 1
else:
high = mid - 1
# Find last occurrence
low = 0
high = n - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] == x:
last = mid
low = mid + 1
elif arr[mid] < x:
low = mid + 1
else:
high = mid - 1
return [first, last]
Output
[2, 5]