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34-find-first-last-position-of-element-in-sorted-array.java
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80 lines (61 loc) · 2.03 KB
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/*
34. Find First and Last Position of Element in Sorted Array
Medium - 42.9%
Given an array of integers nums sorted in non-decreasing order, find the starting and ending
position of a given target value.
If target is not found in the array, return [-1, -1].
You must 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]
*/
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = {-1, -1};
if (nums.length == 0) return result;
// Find first position
int first = findFirst(nums, target);
if (first == -1) return result;
// Find last position
int last = findLast(nums, target);
return new int[]{first, last};
}
private int findFirst(int[] nums, int target) {
int left = 0, right = nums.length - 1;
int first = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
first = mid;
right = mid - 1; // Continue searching left for first occurrence
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return first;
}
private int findLast(int[] nums, int target) {
int left = 0, right = nums.length - 1;
int last = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
last = mid;
left = mid + 1; // Continue searching right for last occurrence
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return last;
}
}