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content/tracks/algorithms-101/leetcode/easy/205/index.en.md

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+---
+title: 205. Isomorphic Strings
+seoTitle: LeetCode 205. Isomorphic Strings | Python solution and explanation
+description: 205. Isomorphic Strings
+toc: true
+tags: []
+categories: [Algorithms, LeetCode]
+date: 2024-04-02
+lastMod: 2024-01-01
+featuredImage: https://picsum.photos/700/155?grayscale
+weight: 205
+---
+
+[LeetCode problem 205](https://leetcode.com/problems/isomorphic-strings/)
+
+Use two dictionaries to track the mappings from `s` to `t` and from `t` to `s`. This helps in ensuring no two characters map to the same character.
+
+- **Key Idea**: To verify if two strings are isomorphic, we need to ensure that each character in s maps to a unique character in t, and vice versa. This bi-directional mapping is crucial to maintain the isomorphism property.
+- **Data Structures**: Two hash maps (or dictionaries in Python) are ideal for maintaining these mappings efficiently.
+
+## Approach
+
+1. Initial Checks: If the lengths of `s` and `t` are different, they cannot be isomorphic. Return false immediately.
+2. Create Two Mappings: Initialize two dictionaries. One for mapping characters from `s` to `t` (`s_to_t`) and another from `t` to `s` (`t_to_s`).
+3. Iterate Over Characters:
+
+- For each character pair (`s_char`, `t_char`) in `s` and `t`, check:
+  - If `s_char` maps to a different `t_char` in `s_to_t`, or `t_char` maps to a different `s_char` in `t_to_s`, return false.
+  - Update `s_to_t[s_char] = t_char` and `t_to_s[t_char] = s_char`.
+
+1. Return True: If the loop completes without returning false, then `s` and `t` are isomorphic.
+
+```python
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        s_to_t, t_to_s = {}, {}
+
+        for s_char, t_char in zip(s, t):
+            if (s_char in s_to_t and s_to_t[s_char] != t_char) or \
+            (t_char in t_to_s and t_to_s[t_char] != s_char):
+                return False
+            s_to_t[s_char] = t_char
+            t_to_s[t_char] = s_char
+
+        return True
+```

content/tracks/algorithms-101/leetcode/easy/205/index.ru.md

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+---
+title: 205. Isomorphic Strings
+seoTitle: LeetCode 205. Isomorphic Strings | Python solution and explanation
+description: 205. Isomorphic Strings
+toc: true
+tags: []
+categories: [Algorithms, LeetCode]
+date: 2024-04-02
+lastMod: 2024-01-01
+featuredImage: https://picsum.photos/700/155?grayscale
+weight: 205
+---
+
+[LeetCode problem 205](https://leetcode.com/problems/isomorphic-strings/)
+
+Use two dictionaries to track the mappings from `s` to `t` and from `t` to `s`. This helps in ensuring no two characters map to the same character.
+
+- **Key Idea**: To verify if two strings are isomorphic, we need to ensure that each character in s maps to a unique character in t, and vice versa. This bi-directional mapping is crucial to maintain the isomorphism property.
+- **Data Structures**: Two hash maps (or dictionaries in Python) are ideal for maintaining these mappings efficiently.
+
+## Approach
+
+1. Initial Checks: If the lengths of `s` and `t` are different, they cannot be isomorphic. Return false immediately.
+2. Create Two Mappings: Initialize two dictionaries. One for mapping characters from `s` to `t` (`s_to_t`) and another from `t` to `s` (`t_to_s`).
+3. Iterate Over Characters:
+
+- For each character pair (`s_char`, `t_char`) in `s` and `t`, check:
+  - If `s_char` maps to a different `t_char` in `s_to_t`, or `t_char` maps to a different `s_char` in `t_to_s`, return false.
+  - Update `s_to_t[s_char] = t_char` and `t_to_s[t_char] = s_char`.
+
+1. Return True: If the loop completes without returning false, then `s` and `t` are isomorphic.
+
+```python
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        s_to_t, t_to_s = {}, {}
+
+        for s_char, t_char in zip(s, t):
+            if (s_char in s_to_t and s_to_t[s_char] != t_char) or \
+            (t_char in t_to_s and t_to_s[t_char] != s_char):
+                return False
+            s_to_t[s_char] = t_char
+            t_to_s[t_char] = s_char
+
+        return True
+```

content/tracks/algorithms-101/leetcode/easy/58/index.en.md

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+---
+title: 58. Length of Last Word
+seoTitle: LeetCode 58. Length of Last Word | Python solution and explanation
+description: 58. Length of Last Word
+toc: true
+tags: []
+categories: [Algorithms, LeetCode]
+date: 2024-01-01
+lastMod: 2024-01-01
+featuredImage: https://picsum.photos/700/155?grayscale
+weight: 58
+---
+
+[LeetCode problem 58](https://leetcode.com/problems/length-of-last-word/)
+
+```python
+class Solution:
+    def lengthOfLastWord(self, s: str) -> int:
+        i = len(s) - 1
+        while i >= 0 and s[i] == ' ':
+            i -= 1
+        j = i
+        while j >= 0 and s[j] != ' ':
+            j -= 1
+        return i - j
+```
+
+```python
+class Solution:
+    def lengthOfLastWord(self, s: str) -> int:
+        ar = s.split()
+        return len(ar[-1])
+```

content/tracks/algorithms-101/leetcode/easy/58/index.ru.md

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+---
+title: 58. Length of Last Word
+seoTitle: LeetCode 58. Length of Last Word | Python solution and explanation
+description: 58. Length of Last Word
+toc: true
+tags: []
+categories: [Algorithms, LeetCode]
+date: 2024-01-01
+lastMod: 2024-01-01
+featuredImage: https://picsum.photos/700/155?grayscale
+weight: 58
+---
+
+[LeetCode problem 58](https://leetcode.com/problems/length-of-last-word/)
+
+```python
+class Solution:
+    def lengthOfLastWord(self, s: str) -> int:
+        i = len(s) - 1
+        while i >= 0 and s[i] == ' ':
+            i -= 1
+        j = i
+        while j >= 0 and s[j] != ' ':
+            j -= 1
+        return i - j
+```
+
+```python
+class Solution:
+    def lengthOfLastWord(self, s: str) -> int:
+        ar = s.split()
+        return len(ar[-1])
+```