Find & Index about List & String

String

Find

The find method checks if an element exists in the string and returns the lowest index of the substring if found. If not found, it returns -1.

text = "Hello, world!"
result = text.find("world")
print(result)  # Output: 7

Index

The index method retrieves the index of an element in the string. If the element is not found, it raises an error. Therefore, it is recommended to use find or in beforehand.

text = "Hello, world!"
result = text.index("world")
print(result)  # Output: 7

# Uncommenting the following line will raise a ValueError
# result = text.index("Python")  # Raises ValueError

List

Find

In a list, you cannot use find to check if an element exists. You can use in to perform the check.

list1 = ['a', 'b', 'test']

if 'test' in list1:
    print(True)  # Output: True

Index

The index method returns the index of an element in the list, but if the element is not present, it raises an error.

list1 = ['a', 'b', 'test']

result = list1.index('test')
print(result)  # Output: 2

# Uncommenting the following line will raise a ValueError
# result = list1.index('Python')  # Raises ValueError

Example

Leetcode 219. Contains Duplicate II

Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.

Example 1:
Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:
Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:
Input: nums = [1,2,3,1,2,3], k = 2
Output: false

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        temp_site = {}
        for site in range(len(nums)):
            if nums[site] in temp_site:
                if abs(site - temp_site[nums[site]]) <= k:
                    return True
                else:
                    temp_site[nums[site]] = site
            else:
                temp_site[nums[site]] = site
        return False