leetcode 72

Author: Kondareddy1209

Leetcode/Leetcode_72.py

@@ -0,0 +1,316 @@
+"""
+LeetCode 76. Minimum Window Substring
+=====================================
+
+Problem Statement
+-----------------
+Given two strings `s` and `t`, return the minimum window substring
+of `s` such that every character in `t` (including duplicates)
+is included in the window.
+
+If no such substring exists, return an empty string "".
+
+Examples
+--------
+Input:
+    s = "ADOBECODEBANC"
+    t = "ABC"
+
+Output:
+    "BANC"
+
+Input:
+    s = "a"
+    t = "a"
+
+Output:
+    "a"
+
+Input:
+    s = "a"
+    t = "aa"
+
+Output:
+    ""
+
+Approach
+--------
+Sliding Window + Frequency Counter
+
+1. Store character frequencies of string `t` using Counter.
+2. Expand the right pointer and decrease the required count.
+3. Once all characters are found (count == 0):
+   - Try shrinking the window from the left.
+   - Update the minimum window if a smaller valid window is found.
+4. Continue until the entire string is processed.
+
+Key Observation
+---------------
+Counter values represent how many more occurrences of each character
+are still required.
+
+Positive Value  -> Character still needed.
+Zero Value      -> Exact requirement satisfied.
+Negative Value  -> Extra occurrences present in the current window.
+
+Time Complexity
+---------------
+O(n)
+
+Each character is visited at most twice:
+- Once by the right pointer.
+- Once by the left pointer.
+
+Space Complexity
+----------------
+O(m)
+
+where m = number of unique characters in t.
+
+"""
+
+
+from collections import Counter
+
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+
+        # Edge Case
+        if len(s) < len(t):
+            return ""
+
+        # Frequency map of characters needed
+        n = Counter(t)
+
+        # Left pointer
+        l = 0
+
+        # Total characters still needed
+        count = len(t)
+
+        # Minimum window length
+        ans = float('inf')
+
+        # Starting index of answer
+        start = 0
+
+        # Expand window using right pointer
+        for i in range(len(s)):
+
+            if s[i] in n:
+                n[s[i]] -= 1
+
+                # Required character found
+                if n[s[i]] >= 0:
+                    count -= 1
+
+            # Valid window found
+            while count == 0:
+
+                # Update minimum window
+                if i - l + 1 < ans:
+                    ans = i - l + 1
+                    start = l
+
+                # Remove left character
+                if s[l] in n:
+                    n[s[l]] += 1
+
+                    # Window becomes invalid
+                    if n[s[l]] > 0:
+                        count += 1
+
+                l += 1
+
+        if ans == float('inf'):
+            return ""
+
+        return s[start:start + ans]
+
+
+# ============================================================
+# Dry Run
+# ============================================================
+
+"""
+Input:
+s = "ADOBECODEBANC"
+t = "ABC"
+
+Initial:
+n = {'A':1, 'B':1, 'C':1}
+count = 3
+l = 0
+
+------------------------------------------------------------
+i = 0 -> 'A'
+
+n['A'] = 0
+count = 2
+
+Window = "A"
+
+------------------------------------------------------------
+i = 3 -> 'B'
+
+n['B'] = 0
+count = 1
+
+Window = "ADOB"
+
+------------------------------------------------------------
+i = 5 -> 'C'
+
+n['C'] = 0
+count = 0
+
+Window = "ADOBEC"
+
+Valid Window Found
+
+Length = 6
+ans = 6
+start = 0
+
+Try Shrinking
+
+Remove 'A'
+n['A'] = 1
+
+n['A'] > 0
+count = 1
+
+Stop Shrinking
+
+------------------------------------------------------------
+Continue Expanding
+
+i = 10 -> 'A'
+
+n['A'] = 0
+count = 0
+
+Window Valid Again
+
+Try Shrinking
+
+Remove D
+Remove O
+Remove B
+Remove E
+
+Window = "CODEBA"
+
+Remove C
+
+n['C'] = 1
+count = 1
+
+Stop
+
+------------------------------------------------------------
+i = 12 -> 'C'
+
+n['C'] = 0
+count = 0
+
+Window = "BANC"
+
+Length = 4
+
+ans = 4
+start = 9
+
+Try Shrinking
+
+Remove B
+
+n['B'] = 1
+count = 1
+
+Stop
+
+------------------------------------------------------------
+
+Answer:
+s[9:13]
+
+= "BANC"
+
+Output:
+"BANC"
+"""
+
+# ============================================================
+# Example Usage
+# ============================================================
+
+if __name__ == "__main__":
+    solution = Solution()
+
+    print(solution.minWindow("ADOBECODEBANC", "ABC"))
+    print(solution.minWindow("a", "a"))
+    print(solution.minWindow("a", "aa"))
+
+
+"""
+Interview Explanation
+---------------------
+
+Why do we decrement n[s[i]]?
+
+Because the character has entered the current window.
+
+Example:
+Need:
+A : 1
+
+After finding one A:
+A : 0
+
+Requirement satisfied.
+
+------------------------------------------------
+
+Why check n[s[i]] >= 0 ?
+
+Because only required occurrences should reduce count.
+
+Example:
+
+Need:
+A : 1
+
+Window:
+A A A
+
+Counter values:
+
+0
+-1
+-2
+
+Only the first A contributes toward satisfying t.
+
+------------------------------------------------
+
+Why increment n[s[l]] while shrinking?
+
+Because that character leaves the window.
+
+If its count becomes positive,
+the window no longer contains enough copies of that character.
+
+------------------------------------------------
+
+Why does this work in O(n)?
+
+Each character:
+- enters the window once
+- leaves the window once
+
+Hence total operations are linear.
+
+O(n)
+"""