Create 1995-count-special-quadruplets.java

Author: sanajitjana

Easy/1995-count-special-quadruplets.java

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+/**
+ * LeetCode Problem: 1995. Count Special Quadruplets
+ *
+ * Difficulty: Easy
+ *
+ * Problem Statement:
+ * You are given a 0-indexed integer array nums. Return the number of distinct quadruplets (a, b, c, d) such that:
+ *
+ *   - nums[a] + nums[b] + nums[c] == nums[d], and
+ *   - a < b < c < d
+ *
+ * Example 1:
+ * Input: nums = [1,2,3,6]
+ * Output: 1
+ * Explanation: The only quadruplet is (0,1,2,3) because 1 + 2 + 3 == 6.
+ *
+ * Example 2:
+ * Input: nums = [3,3,6,4,5]
+ * Output: 0
+ * Explanation: No such quadruplet exists.
+ *
+ * Example 3:
+ * Input: nums = [1,1,1,3,5]
+ * Output: 4
+ * Explanation: The quadruplets are (0,1,2,3), (0,1,2,4), (0,2,1,3), (0,2,1,4).
+ *
+ * Constraints:
+ *  - 4 <= nums.length <= 50
+ *  - 1 <= nums[i] <= 100
+ *
+ * Approach:
+ *  We need to count all ordered quadruplets (a, b, c, d) where:
+ *      nums[a] + nums[b] + nums[c] == nums[d] and a < b < c < d
+ *
+ *  Brute force solution:
+ *    - Iterate through all possible quadruplets using 4 nested loops.
+ *    - For each (i, j, k, l), compute sum = nums[i] + nums[j] + nums[k].
+ *    - If sum == nums[l], increment the count.
+ *
+ *  Optimization:
+ *    - Since nums[i] <= 100, if sum > 100 we can break early (small pruning).
+ *    - This helps avoid unnecessary checks.
+ *
+ *  Complexity:
+ *    - Time Complexity: O(n^4) in the worst case (since we check all quadruplets).
+ *      With n <= 50, this is acceptable (50^4 = 6.25 million iterations).
+ *    - Space Complexity: O(1) since we only use a counter.
+ */
+
+class Solution {
+    public int countQuadruplets(int[] nums) {
+        int count = 0;
+        for (int i = 0; i < nums.length; i++) {
+            for (int j = i + 1; j < nums.length; j++) {
+                for (int k = j + 1; k < nums.length; k++) {
+                    int sum = nums[i] + nums[j] + nums[k];
+                    // small pruning: since nums[d] ≤ 100, skip if sum > 100
+                    if (sum > 100) continue; 
+                    for (int l = k + 1; l < nums.length; l++) {
+                        if (sum == nums[l]) {
+                            count++;
+                        }
+                    }
+                }
+            }
+        }
+        return count;
+    }
+}