feat: add Partition Labels problem

- Greedy algorithm implementation
- Multiple approaches: Two Pointers, Sliding Window, While loop
- O(n) time, O(1) space complexity
- Comprehensive explanations and edge cases

Author: TechnoBlogger14o3

neetcode-150/greedy/partition-labels/explanation.md

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+# Partition Labels
+
+## Problem Statement
+
+You are given a string `s`. We want to partition the string into as many parts as possible so that each letter appears in at most one part.
+
+Note that the partition is done so that after concatenating all the parts in order, the resultant string should be `s`.
+
+Return a list of integers representing the size of these parts.
+
+## Examples
+
+**Example 1:**
+```
+Input: s = "ababcbacadefegdehijhklij"
+Output: [9,7,8]
+Explanation:
+The partition is "ababcbaca", "defegde", "hijhklij".
+This is a partition so that each letter appears in at most one part.
+A partition like "ababcbacadefegde", "hijhklij" is incorrect, because the letters "a" and "b" appear in both parts.
+```
+
+## Approach
+
+### Method 1: Greedy Algorithm (Recommended)
+1. Find the last occurrence of each character
+2. Use greedy approach to extend partition as far as possible
+3. When we reach the end of current partition, add its length
+4. Most efficient approach
+
+**Time Complexity:** O(n) - Two passes
+**Space Complexity:** O(1) - At most 26 characters
+
+### Method 2: Two Pointers
+1. Use two pointers to track partition boundaries
+2. Extend partition until all characters are included
+3. Less efficient than greedy approach
+
+**Time Complexity:** O(n) - Two passes
+**Space Complexity:** O(1) - At most 26 characters
+
+## Algorithm
+
+```
+1. Find last occurrence of each character
+2. Initialize start = 0, end = 0
+3. For i from 0 to n-1:
+   a. end = max(end, lastOccurrence[s[i]])
+   b. If i == end:
+      c. Add partition length to result
+      d. start = i + 1
+4. Return result
+```
+
+## Key Insights
+
+- **Greedy Choice**: Extend partition as far as possible
+- **Local Optimum**: Include all occurrences of current characters
+- **Global Optimum**: Minimize number of partitions
+- **Space Optimization**: Use only necessary space
+
+## Alternative Approaches
+
+1. **Two Pointers**: Use two pointers for boundaries
+2. **Sliding Window**: Use sliding window technique
+3. **Brute Force**: Try all possible partitions
+
+## Edge Cases
+
+- Empty string: Return empty list
+- Single character: Return [1]
+- All same characters: Return [n]
+- No repeated characters: Return [1, 1, ..., 1]
+
+## Applications
+
+- Greedy algorithms
+- String partitioning
+- Algorithm design patterns
+- Interview preparation
+- System design
+
+## Optimization Opportunities
+
+- **Greedy Algorithm**: Most efficient approach
+- **Space Optimization**: O(1) space complexity
+- **Two Passes**: O(n) time complexity
+- **No Extra Space**: Use only necessary space

neetcode-150/greedy/partition-labels/solution.java

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+/**
+ * Time Complexity: O(n) - Two passes
+ * Space Complexity: O(1) - At most 26 characters
+ */
+class Solution {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        // Find last occurrence of each character
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int start = 0;
+        int end = 0;
+        
+        for (int i = 0; i < s.length(); i++) {
+            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}
+
+// Alternative approach using two pointers
+class SolutionTwoPointers {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int left = 0;
+        int right = 0;
+        
+        for (int i = 0; i < s.length(); i++) {
+            right = Math.max(right, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == right) {
+                result.add(right - left + 1);
+                left = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}
+
+// Alternative approach using sliding window
+class SolutionSlidingWindow {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int windowStart = 0;
+        int windowEnd = 0;
+        
+        for (int i = 0; i < s.length(); i++) {
+            windowEnd = Math.max(windowEnd, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == windowEnd) {
+                result.add(windowEnd - windowStart + 1);
+                windowStart = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}
+
+// Alternative approach using iterative
+class SolutionIterative {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int start = 0;
+        int end = 0;
+        
+        for (int i = 0; i < s.length(); i++) {
+            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}
+
+// Alternative approach using while loop
+class SolutionWhileLoop {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int start = 0;
+        int end = 0;
+        int i = 0;
+        
+        while (i < s.length()) {
+            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+            
+            i++;
+        }
+        
+        return result;
+    }
+}
+
+// Alternative approach using enhanced for loop
+class SolutionEnhancedForLoop {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int start = 0;
+        int end = 0;
+        
+        for (int i = 0; i < s.length(); i++) {
+            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
+            
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}
+
+// More concise version
+class SolutionConcise {
+    public List<Integer> partitionLabels(String s) {
+        List<Integer> result = new ArrayList<>();
+        int[] lastOccurrence = new int[26];
+        
+        for (int i = 0; i < s.length(); i++) {
+            lastOccurrence[s.charAt(i) - 'a'] = i;
+        }
+        
+        int start = 0, end = 0;
+        for (int i = 0; i < s.length(); i++) {
+            end = Math.max(end, lastOccurrence[s.charAt(i) - 'a']);
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+        }
+        
+        return result;
+    }
+}