Create 0830-positions-of-large-groups.java

Author: sanajitjana

Easy/0830-positions-of-large-groups.java

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+/**
+ * Title: Positions of Large Groups
+ * LeetCode Link: https://leetcode.com/problems/positions-of-large-groups/
+ *
+ * Question Info:
+ * In a string s of lowercase letters, a group is a consecutive run of the same character.
+ * A large group is defined as a group that has length >= 3.
+ * Return the intervals [start, end] of every large group sorted in increasing order of start index.
+ *
+ * Example:
+ * Input: s = "abbxxxxzzy"
+ * Output: [[3,6]]
+ *
+ * Approach:
+ * - Use two pointers to track the start of a group and the current index.
+ * - Iterate through the string from i = 1 to n.
+ * - If we reach the end OR the current char differs from the start char:
+ *     - Check if the group length >= 3.
+ *     - If yes, record the interval [start, i-1].
+ *     - Move start = i to begin a new group.
+ * - Return the list of all intervals.
+ *
+ * Dry Run (s = "abbxxxxzzy"):
+ * i=1: s[1]='b', s[0]='a' => different, group size=1 (<3), skip, start=1
+ * i=2: s[2]='b', s[1]='b' => same, continue
+ * i=3: s[3]='x', s[1]='b' => different, group size=2 (<3), skip, start=3
+ * i=4,5,6: still 'x', continue
+ * i=7: s[7]='z', s[3]='x' => different, group size=4 (>=3), record [3,6], start=7
+ * i=8: s[8]='z', s[7]='z' => same
+ * i=9: s[9]='y', s[7]='z' => different, group size=2 (<3), skip, start=9
+ * i=10: reached end, group size=1 (<3), skip.
+ * Result: [[3,6]]
+ *
+ * Time Complexity: O(n), where n = length of string (single pass).
+ * Space Complexity: O(1), ignoring output list.
+ */
+
+class Solution {
+    public List<List<Integer>> largeGroupPositions(String s) {
+        List<List<Integer>> ans = new ArrayList<>();
+        int n = s.length();
+        int start = 0;
+
+        for (int i = 1; i <= n; i++) {
+            if (i == n || s.charAt(i) != s.charAt(start)) {
+                if (i - start >= 3) {
+                    ans.add(Arrays.asList(start, i - 1));
+                }
+                start = i;
+            }
+        }
+
+        return ans;
+    }
+}