Create 2348-number-of-zero-filled-subarrays.java

Author: sanajitjana

Medium/2348-number-of-zero-filled-subarrays.java

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+/**
+ * 2348. Number of Zero-Filled Subarrays
+ * Difficulty: Medium
+ * URL: https://leetcode.com/problems/number-of-zero-filled-subarrays/description/
+ *
+ * ---------------------
+ * Approach 1: Brute Force
+ *
+ * - Generate all possible subarrays using two loops.
+ * - For each subarray, check if all elements are zero.
+ * - If yes, increase the count.
+ *
+ * - In the given code, this is done by converting the subarray into a string
+ *   and checking if all characters are '0'.
+ *
+ * Time Complexity: O(n^3) → (O(n^2) subarrays × O(n) checking each)
+ * Space Complexity: O(n) due to string building.
+ *
+ * Not practical for large inputs, but helps in understanding.
+ *
+ */
+class Solution {
+    public long zeroFilledSubarray(int[] nums) {
+        int count = 0;
+        for (int i = 0; i < nums.length; i++) {
+            String str = "";
+            for (int j = i; j < nums.length; j++) {
+                str += nums[j] + "";
+                boolean flag = true;
+                for (int k = 0; k < str.length(); k++) {
+                    if (str.charAt(k) != '0') {
+                        flag = false;
+                        break;
+                    }
+                }
+                if (flag) {
+                    count++;
+                }
+            }
+        }
+        return count;
+    }
+}
+
+
+/**
+ * ---------------------
+ * Approach 2: Optimized (Mathematical Counting)
+ *
+ * - Instead of generating all subarrays, notice:
+ *   For a streak of `k` consecutive zeros, number of zero-subarrays is:
+ *        k * (k + 1) / 2
+ *   (Formula of sum of first n natural numbers).
+ *
+ * Example: nums = [0,0,1,0]
+ * - Streak1 = "00" → 2*(2+1)/2 = 3 subarrays → [0], [0], [0,0]
+ * - Streak2 = "0"  → 1*(1+1)/2 = 1 subarray → [0]
+ * Total = 4
+ *
+ * Algorithm:
+ * 1. Traverse array and keep track of consecutive zeros.
+ * 2. When a non-zero is encountered, add contribution of that streak.
+ * 3. After loop ends, add last streak if any.
+ *
+ * Time Complexity: O(n)
+ * Space Complexity: O(1)
+ *
+ */
+class Solution {
+    public long zeroFilledSubarray(int[] nums) {
+        long count = 0;
+        long streak = 0;
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] == 0) {
+                streak++;
+            } else {
+                count += (streak * (streak + 1)) / 2;
+                streak = 0;
+            }
+        }
+        // Handle streak at the end
+        if (streak > 0) {
+            count += (streak * (streak + 1)) / 2;
+        }
+        return count;
+    }
+}