Python - Algorithm #11

Python Algorithm

LeetCode

2903. Find Indices With Index and Value Difference I

• You are given a 0-indexed integer array [nums] having length n, an integer indexDifference, and an integer valueDifference.
• Your task is to find two indices i and j, both in the range [0, n - 1], that satisfy the following conditions:
  • abs(i - j) ≥ indexDifference, and abs(nums[i] - nums[j]) ≥ valueDifference
• Return an integer array answer, where answer = [i, j] if there are two such indices, and answer = [-1, -1] otherwise. If there are multiple choices for the two indices, return any of them.
• Note:
  • i and j may be equal.
• Constraints:
  • 1 ≤ n == nums.length ≤ 100
  • 0 ≤ nums[i] ≤ 50
  • 0 ≤ indexDifference ≤ 100
  • 0 ≤ valueDifference ≤ 50

class Solution:
    def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:
        # 2중 for문을 이용해서 풀어볼 것
        for i in range(len(nums)):
            for j in range(len(nums)):
                # abs(i - j) >= indexDifference
                # abs(nums[i] - nums[j]) >= valueDifference
                # 두 가지 조건을 만족시키면 answer에 i, j append, break
                if abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference:
                    return [i, j]
        return [-1, -1]

class Solution:
    def findIndices(self, nums: List[int], indexDifference: int, valueDifference: int) -> List[int]:
        for i in range(len(nums)):
            for j in range(len(nums)):
                if abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference:
                    return [i, j]
        return [-1, -1]

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2000. Reverse Prefix of Word

• Given a 0-indexed string [word] and a character ch, reverse the segment of [word] that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.
• For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".
• Return the resulting string.
• Constraints:
  
  • 1 ≤ word.length ≤ 250
  • word consists of lowercase English letters.
  • ch is a lowercase English letter.

class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        loc = word.find(ch)
        return word[:loc+1][::-1] + word[loc+1:]

class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        n = word.find(ch)
        if n != -1:
            return word[n::-1] + word[n+1:]
        else:
            return word

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https://leetcode.com/problems/find-indices-with-index-and-value-difference-i/

https://leetcode.com/problems/reverse-prefix-of-word/