## Search Insert Position

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binary search
search for a range leetcode
binary search python
insert search
find first and last position of element in sorted array
search in rotated sorted array
insert at position

I'm solving a problem (leetcode 35). My code returns null for test case input: [1,3,5,6], 7. Couldn't find out the bug.

Given a sorted array 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.

You may assume no duplicates in the array.

```Example 1:

Input: [1,3,5,6], 5
Output: 2
Example 2:

Input: [1,3,5,6], 2
Output: 1
Example 3:

Input: [1,3,5,6], 7
Output: 4
Example 4:

Input: [1,3,5,6], 0
Output: 0
```

below my code. i understand this question has many different solutions and my solution is not optimal. but please just help me understand where the bug is instead of giving a completely new solution. thanks!

```class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
try:
return nums.index(target)
except:
for i in range(len(nums)):
print i
if nums[i] - target > 0:
return i
else:
print "hello", len(nums)
return len(nums)
```

Your first part is correct, use `list.index` and catch an exception. However your second part (without prints)

```  for i in range(len(nums)):
if nums[i] - target > 0:
return i  # return if True
else:
return len(nums)  # return if False
```

This means that the first iteration of the `for` loop will always return no matter what.

You need to bring the `else` block out of the for loop; something like this:

```def searchInsert(self, nums, target):
try:
return nums.index(target)
except IndexError:  # best to use explicit except
for index, value in enumerate(nums):  # more pythonic than range(len(nums))
if value > target:
return index
return len(nums)
```

LeetCode – Search Insert Position (Java), Given a sorted array 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. You. LeetCode – Search Insert Position (Java) Given a sorted array 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. You may assume no duplicates in the array. Here are few examples.

Check this out!

```class Solution:
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
nums.append(target)
nums.sort()
for i,j in enumerate(nums):
if target == j:
return i
```

leetcode Search Insert Position · GitHub, leetcode Search Insert Position. SearchInsertPosition.cpp. /*. Given a sorted array and a target value, return the index if the target is found. If not, return the index  I'm solving a problem (leetcode 35). My code returns null for test case input: [1,3,5,6], 7. Couldn't find out the bug. Given a sorted array and a target value, return the index if the target is

Your if else in your for loop is wrong, you should put your else outside of the for loop.

```class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
try:
return nums.index(target)
except:
for i in range(len(nums)):
print i
if nums[i] - target > 0:
return i

print "hello", len(nums)
return len(nums)
```

However, bisect already contains everything you need.

```from bisect import bisect_left
print(bisect_left([1,3,5,6],4))
```

As an exercice, you can try to do this exercise again, without using index, you might want to use enumerate and if else condition.

LeetCode: Search Insert Position » Coder's Cat, If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array. Example 1: Input: [1,3,5,6], 5  C++ Coding Exercise - Search Insert Position We know the array is sorted, so we can reduce the search to O(logN) via binary search. If not found, i>=j then we need to compare target with the nums[i], return i + 1 if the target is bigger (insert after) or i otherwise (insert before).

Search Insert Position, Your first part is correct, use list.index and catch an exception. However your second part (without prints) for i in range(len(nums)): if nums[i]  Flutter Tutorial for Beginners - Build iOS and Android Apps with Google's Flutter & Dart - Duration: 3:22:19. Academind 1,045,809 views

LeetCode Tutorial 35. Search Insert Position, In Java, the most efficient implementation of this will likely be public int searchInsert(int[] numbers, int target) { int index = Arrays. Binary search it is. Thanks to this June challenge, I recall using binary search in Random pick with Weight. I had understood it perfectly, so this one took less than a minute. In my way of writing the binary search, the target is either at the left pointer’s index or the left pointer tells me ‘where the target should have been’.

Search Insert Position, Search Insert Position 60. Question. Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it​  35. Search Insert Position. Java using little modification on binary search to bring the time complexity to O(log N) rahul12 created at: 20 hours ago