Create 0976-largest-perimeter-triangle.java

Author: sanajitjana

Easy/0976-largest-perimeter-triangle.java

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+/*
+ * Problem Link:
+ * https://leetcode.com/problems/largest-perimeter-triangle/?envType=daily-question&envId=2025-09-28
+ *
+ * Q: Largest Perimeter Triangle
+ *
+ * You are given an array nums of non-negative integers.
+ * Return the largest perimeter of a triangle with non-zero area, formed from three of these lengths.
+ * If it is impossible to form any triangle of non-zero area, return 0.
+ *
+ * ------------------------------------------------------
+ * Approach 1: Brute Force (O(n^3))
+ * ------------------------------------------------------
+ * Idea:
+ *  - Sort the array.
+ *  - Try every possible triplet (i, j, k).
+ *  - Check the triangle inequality: 
+ *      a + b > c, a + c > b, b + c > a
+ *  - Keep track of the maximum perimeter.
+ *
+ * Dry Run Example:
+ * nums = [2, 1, 2]
+ * After sort = [1, 2, 2]
+ * Pick (1,2,2):
+ *    1 + 2 > 2 (yes)
+ *    1 + 2 > 2 (yes)
+ *    2 + 2 > 1 (yes)
+ * Valid triangle, perimeter = 5
+ * Answer = 5
+ *
+ * Time Complexity: O(n^3)
+ * Space Complexity: O(1)
+ */
+
+import java.util.Arrays;
+
+class SolutionBruteForce {
+    public int largestPerimeter(int[] nums) {
+        Arrays.sort(nums);
+        int max = 0;
+        for (int i = 0; i < nums.length - 2; i++) {
+            int x = nums[i];
+            for (int j = i + 1; j < nums.length - 1; j++) {
+                int y = nums[j];
+                for (int k = j + 1; k < nums.length; k++) {
+                    int z = nums[k];
+                    if (x + y > z && x + z > y && y + z > x) {
+                        max = Math.max(max, x + y + z);
+                    }
+                }
+            }
+        }
+        return max;
+    }
+}
+
+
+/*
+ * ------------------------------------------------------
+ * Approach 2: Optimized Greedy (O(n log n))
+ * ------------------------------------------------------
+ * Idea:
+ *  - Sort the array.
+ *  - The largest perimeter will always come from the three largest numbers
+ *    that can form a valid triangle.
+ *  - Starting from the end of the sorted array, check triplets (i, i-1, i-2).
+ *  - As soon as we find a valid triangle, return its perimeter.
+ *
+ * Why consecutive triplets are enough:
+ *  - After sorting, if nums[i-2] + nums[i-1] > nums[i], then
+ *    (nums[i-2], nums[i-1], nums[i]) is valid and maximizes perimeter.
+ *  - If not, moving left reduces the largest side, so we must keep checking.
+ *
+ * Dry Run Example:
+ * nums = [2, 1, 2]
+ * After sort = [1, 2, 2]
+ * Start from i = 2:
+ *   nums[0] + nums[1] = 1 + 2 = 3 > 2 → valid
+ *   perimeter = 1 + 2 + 2 = 5
+ * Answer = 5
+ *
+ * Time Complexity: O(n log n) (due to sorting)
+ * Space Complexity: O(1)
+ */
+
+class SolutionOptimized {
+    public int largestPerimeter(int[] nums) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        for (int i = n - 1; i >= 2; i--) {
+            if (nums[i - 2] + nums[i - 1] > nums[i]) {
+                return nums[i] + nums[i - 1] + nums[i - 2];
+            }
+        }
+        return 0;
+    }
+}