update sliding window pattern

Author: javydevx

src/data/patterns.json

@@ -16,12 +16,15 @@
     },
     {
       "title": "Sliding Window",
-      "description": "Two major forms of Sliding Window exist: (1) Fixed-size window (size = k), used for sums/averages and constant-length subarrays; and (2) Dynamic (variable-size) window, used when the window expands and shrinks based on a condition (e.g., character frequency, distinct characters, constraints).",
+      "description": "Sliding Window is a technique to efficiently handle contiguous subarrays or substrings. Two major forms exist: (1) Fixed-size window (size = k), used for sums/averages and constant-length subarrays; (2) Dynamic (variable-size) window, used when the window expands and shrinks based on a condition (e.g., character frequency, distinct characters, constraints). For binary arrays or situations where only one counter is needed, an optimized dynamic window without a hashmap can be used.",
       "templates": {
-        "python": "# ---------------------------\n# Version 1: Fixed-Size Window (size = k)\n# ---------------------------\nleft = 0\nwindow_sum = 0\nbest = 0\n\nfor right in range(len(arr)):\n    window_sum += arr[right]\n\n    # When window reaches size k\n    if right - left + 1 == k:\n        best = max(best, window_sum)\n        window_sum -= arr[left]\n        left += 1\n\n\n# ---------------------------\n# Version 2: Dynamic Window (variable size)\n# ---------------------------\nleft = 0\nmax_len = 0\ncount = {}\n\nfor right in range(len(s)):\n    char = s[right]\n    count[char] = count.get(char, 0) + 1\n\n    # Shrink window while it violates a condition\n    while window_invalid:\n        count[s[left]] -= 1\n        if count[s[left]] == 0:\n            del count[s[left]]\n        left += 1\n\n    max_len = max(max_len, right - left + 1)",
-        "javascript": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nlet left = 0;\nlet windowSum = 0;\nlet best = 0;\n\nfor (let right = 0; right < arr.length; right++) {\n    windowSum += arr[right];\n\n    if (right - left + 1 === k) {\n        best = Math.max(best, windowSum);\n        windowSum -= arr[left];\n        left++;\n    }\n}\n\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size)\n// ---------------------------\nlet left2 = 0;\nlet maxLen = 0;\nlet count = {};\n\nfor (let right = 0; right < s.length; right++) {\n    const char = s[right];\n    count[char] = (count[char] || 0) + 1;\n\n    while (windowInvalid) {\n        count[s[left2]]--;\n        if (count[s[left2]] === 0) delete count[s[left2]];\n        left2++;\n    }\n\n    maxLen = Math.max(maxLen, right - left2 + 1);\n}",
-        "java": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nint left = 0;\nint windowSum = 0;\nint best = 0;\n\nfor (int right = 0; right < arr.length; right++) {\n    windowSum += arr[right];\n\n    if (right - left + 1 == k) {\n        best = Math.max(best, windowSum);\n        windowSum -= arr[left];\n        left++;\n    }\n}\n\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size)\n// ---------------------------\nint left2 = 0;\nint maxLen = 0;\nMap<Character, Integer> count = new HashMap<>();\n\nfor (int right = 0; right < s.length(); right++) {\n    char c = s.charAt(right);\n    count.put(c, count.getOrDefault(c, 0) + 1);\n\n    while (windowInvalid) {\n        char lc = s.charAt(left2);\n        count.put(lc, count.get(lc) - 1);\n        if (count.get(lc) == 0) count.remove(lc);\n        left2++;\n    }\n\n    maxLen = Math.max(maxLen, right - left2 + 1);\n}",
-        "go": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nleft := 0\nwindowSum := 0\nbest := 0\n\nfor right := 0; right < len(arr); right++ {\n    windowSum += arr[right]\n\n    if right-left+1 == k {\n        if windowSum > best {\n            best = windowSum\n        }\n        windowSum -= arr[left]\n        left++\n    }\n}\n\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size)\n// ---------------------------\nleft2 := 0\nmaxLen := 0\ncount := make(map[rune]int)\n\nfor right, char := range s {\n    count[char]++\n\n    for windowInvalid {\n        leftChar := rune(s[left2])\n        count[leftChar]--\n        if count[leftChar] == 0 {\n            delete(count, leftChar)\n        }\n        left2++\n    }\n\n    if right-left2+1 > maxLen {\n        maxLen = right-left2+1\n    }\n}"
+        "python": "# ---------------------------\n# Version 1: Fixed-Size Window (size = k)\n# ---------------------------\nleft = 0\nwindow_sum = 0\nbest = 0\n\nfor right in range(len(arr)):\n    window_sum += arr[right]\n\n    # When window reaches size k\n    if right - left + 1 == k:\n        best = max(best, window_sum)\n        window_sum -= arr[left]\n        left += 1\n\n\n# ---------------------------\n# Version 2: Dynamic Window (variable size, with hashmap)\n# ---------------------------\nleft = 0\nmax_len = 0\ncount = {}\n\nfor right in range(len(s)):\n    char = s[right]\n    count[char] = count.get(char, 0) + 1\n\n    # Shrink window while it violates a condition\n    while window_invalid:\n        count[s[left]] -= 1\n        if count[s[left]] == 0:\n            del count[s[left]]\n        left += 1\n\n    max_len = max(max_len, right - left + 1)\n\n\n# ---------------------------\n# Version 3: Dynamic Window (optimized counters, no hashmap)\n# ---------------------------\nleft = 0\nmax_len = 0\ncounter = 0  # counts \"bad\" elements in the window (e.g., zeros)\n\nfor right in range(len(nums)):\n    if nums[right] == bad_element:  # e.g., 0 in binary arrays\n        counter += 1\n\n    # Shrink window if it violates the allowed limit\n    while counter > limit:\n        if nums[left] == bad_element:\n            counter -= 1\n        left += 1\n\n    max_len = max(max_len, right - left + 1)  # adjust -1 if problem requires deletion"
+        ,
+        "javascript": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nlet left = 0;\nlet windowSum = 0;\nlet best = 0;\n\nfor (let right = 0; right < arr.length; right++) {\n    windowSum += arr[right];\n\n    if (right - left + 1 === k) {\n        best = Math.max(best, windowSum);\n        windowSum -= arr[left];\n        left++;\n    }\n}\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size, with hashmap)\n// ---------------------------\nlet left2 = 0;\nlet maxLen = 0;\nlet count = {};\n\nfor (let right = 0; right < s.length; right++) {\n    const char = s[right];\n    count[char] = (count[char] || 0) + 1;\n\n    while (windowInvalid) {\n        count[s[left2]]--;\n        if (count[s[left2]] === 0) delete count[s[left2]];\n        left2++;\n    }\n\n    maxLen = Math.max(maxLen, right - left2 + 1);\n}\n\n// ---------------------------\n// Version 3: Dynamic Window (optimized counters, no hashmap)\n// ---------------------------\nlet left3 = 0;\nlet maxLen2 = 0;\nlet counter = 0; // counts bad elements\n\nfor (let right = 0; right < nums.length; right++) {\n    if (nums[right] === badElement) counter++;\n\n    while (counter > limit) {\n        if (nums[left3] === badElement) counter--;\n        left3++;\n    }\n\n    maxLen2 = Math.max(maxLen2, right - left3 + 1);\n}"
+        ,
+        "java": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nint left = 0;\nint windowSum = 0;\nint best = 0;\n\nfor (int right = 0; right < arr.length; right++) {\n    windowSum += arr[right];\n\n    if (right - left + 1 == k) {\n        best = Math.max(best, windowSum);\n        windowSum -= arr[left];\n        left++;\n    }\n}\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size, with hashmap)\n// ---------------------------\nint left2 = 0;\nint maxLen = 0;\nMap<Character, Integer> count = new HashMap<>();\n\nfor (int right = 0; right < s.length(); right++) {\n    char c = s.charAt(right);\n    count.put(c, count.getOrDefault(c, 0) + 1);\n\n    while (windowInvalid) {\n        char lc = s.charAt(left2);\n        count.put(lc, count.get(lc) - 1);\n        if (count.get(lc) == 0) count.remove(lc);\n        left2++;\n    }\n\n    maxLen = Math.max(maxLen, right - left2 + 1);\n}\n\n// ---------------------------\n// Version 3: Dynamic Window (optimized counters, no hashmap)\n// ---------------------------\nint left3 = 0;\nint maxLen2 = 0;\nint counter = 0;\n\nfor (int right = 0; right < nums.length; right++) {\n    if (nums[right] == badElement) counter++;\n\n    while (counter > limit) {\n        if (nums[left3] == badElement) counter--;\n        left3++;\n    }\n\n    maxLen2 = Math.max(maxLen2, right - left3 + 1);\n}"
+        ,
+        "go": "// ---------------------------\n// Version 1: Fixed-Size Window (size = k)\n// ---------------------------\nleft := 0\nwindowSum := 0\nbest := 0\n\nfor right := 0; right < len(arr); right++ {\n    windowSum += arr[right]\n\n    if right-left+1 == k {\n        if windowSum > best {\n            best = windowSum\n        }\n        windowSum -= arr[left]\n        left++\n    }\n}\n\n// ---------------------------\n// Version 2: Dynamic Window (variable size, with hashmap)\n// ---------------------------\nleft2 := 0\nmaxLen := 0\ncount := make(map[rune]int)\n\nfor right, char := range s {\n    count[char]++\n\n    for windowInvalid {\n        leftChar := rune(s[left2])\n        count[leftChar]--\n        if count[leftChar] == 0 {\n            delete(count, leftChar)\n        }\n        left2++\n    }\n\n    if right-left2+1 > maxLen {\n        maxLen = right-left2+1\n    }\n}\n\n// ---------------------------\n// Version 3: Dynamic Window (optimized counters, no hashmap)\n// ---------------------------\nleft3 := 0\nmaxLen2 := 0\ncounter := 0\n\nfor right := 0; right < len(nums); right++ {\n    if nums[right] == badElement {\n        counter++\n    }\n\n    for counter > limit {\n        if nums[left3] == badElement {\n            counter--\n        }\n        left3++\n    }\n\n    if right-left3+1 > maxLen2 {\n        maxLen2 = right-left3+1\n    }\n}"
       },
       "problems": [
         "Maximum Sum Subarray of Size K",
@@ -30,7 +33,9 @@
         "Longest Substring with K Distinct Characters",
         "Minimum Window Substring",
         "Permutation in String",
-        "Smallest Subarray With Sum ≥ Target"
+        "Smallest Subarray With Sum ≥ Target",
+        "Longest Subarray of 1's After Deleting One Element",
+        "Max Consecutive Ones III"
       ]
     },
     {
@@ -85,7 +90,6 @@
       },
       "problems": ["Subarray Sum Equals K", "Range Sum Query", "Continuous Subarray Sum"]
     },
-
     {
       "title": "Backtracking",
       "description": "Used for generating combinations/permutations.",