Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
Example 1:
Input: nums = [1,3,5,6], target = 5 Output: 2
Example 2:
Input: nums = [1,3,5,6], target = 2 Output: 1
Example 3:
Input: nums = [1,3,5,6], target = 7 Output: 4
Example 4:
Input: nums = [1,3,5,6], target = 0 Output: 0
Example 5:
Input: nums = [1], target = 0 Output: 0
Constraints:
1 <= nums.length <= 104-104 <= nums[i] <= 104numscontains distinct values sorted in ascending order.-104 <= target <= 104
Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
/**
* Time Complexity: O(log N) : N is length of input array .
* Space Complexity: O(C) : Constant
* Insert position is nothing but the 'start' index in the iterative search.
* ]0[[[ > start will be left of 0 or right of 0 in case of single element.
*/
class Solution {
public int searchInsert(int[] a, int target) {
int start = 0;
int end = a.length -1 ;
int mid = 0;
while(start <= end) {
System.out.println(start + ":" + end + ":" + mid);
mid = start + (end - start) / 2;
if(target == a[mid]) {
return mid;
}
if(target < a[mid]) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return start;
}
}
