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Solution.java
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Solution.java
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package BINARYSEARCH.firstandlastposition.attempt1;
import java.util.Arrays;
/**
* Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
* <p>
* If target is not found in the array, return [-1, -1].
* <p>
* You must write an algorithm with O(log n) runtime complexity.
* <p>
* <p>
* <p>
* Example 1:
* <p>
* Input: nums = [5,7,7,8,8,10], target = 8
* Output: [3,4]
* Example 2:
* <p>
* Input: nums = [5,7,7,8,8,10], target = 6
* Output: [-1,-1]
* Example 3:
* <p>
* Input: nums = [], target = 0
* Output: [-1,-1]
* <p>
* <p>
* Constraints:
* <p>
* 0 <= nums.length <= 105
* -109 <= nums[i] <= 109
* nums is a non-decreasing array.
* -109 <= target <= 109
*/
class Solution {
/**
* Runtime
* 0 ms
* Beats
* 100%
* Memory
* 45.5 MB
* Beats
* 70.1%
*/
static int[] searchRange(int[] nums, int target) {
int[] ans = {-1, -1};
int leftPos = binSearch(nums, target, true);
if (leftPos == -1) return ans;
int rightPos = binSearch(nums, target, false);
ans = new int[]{leftPos, rightPos};
return ans;
}
static int binSearch(int[] nums, int target, boolean isAsc) {
int start = 0, end = nums.length - 1;
int ans = -1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (target > nums[mid]) start = mid + 1;
else if (target < nums[mid]) end = mid - 1;
else {
// found matching element, now look in left or right half for more occurrences
// save current occurrence index if no more occurrence found
ans = mid;
if (isAsc) {
end = mid - 1;
} else {
start = mid + 1;
}
}
}
return ans;
}
public static void main(String[] args) {
int[] arr = {1, 2, 3, 4, 5, 7, 7, 7, 7, 7, 7, 7, 8, 8, 9, 10};
int[] res = searchRange(arr, 7);
System.out.println(Arrays.toString(res));
}
}