Create 1304-find-n-unique-integers-sum-up-to-zero.java

Author: sanajitjana

Easy/1304-find-n-unique-integers-sum-up-to-zero.java

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+/**
+ * Problem: Generate an array of n unique integers such that their sum is 0.
+ * 
+ * Approach 1:
+ * - Start from 1 and alternate adding -num and +num into the array.
+ * - If n is odd, include 0 in the array.
+ * - This ensures symmetry and guarantees the sum is zero.
+ * 
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ */
+class Solution {
+    public int[] sumZero(int n) {
+        if (n == 1) return new int[n];
+        
+        int[] arr = new int[n];
+        int index = 0;
+        if (n % 2 == 1) {
+            index = 1; // Leave space for 0 if n is odd
+        }
+        
+        boolean flag = true;
+        int num = 1;
+        
+        while (index < n) {
+            if (flag) {
+                arr[index++] = -num;
+                flag = false;
+            } else {
+                arr[index++] = num;
+                flag = true;
+                num++;
+            }
+        }
+        
+        return arr;
+    }
+}
+
+/**
+ * Approach 2 (Optimized and Cleaner):
+ * - Pair numbers symmetrically: -1 & 1, -2 & 2, ..., until n/2 pairs.
+ * - If n is odd, append 0 at the end.
+ * - Cleaner logic without flags or while-loops.
+ * 
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ */
+class Solution {
+    public int[] sumZero(int n) {
+        int[] arr = new int[n];
+        int pairs = n / 2;
+        int index = 0;
+        
+        for (int i = 1; i <= pairs; i++) {
+            arr[index++] = -i;
+            arr[index++] = i;
+        }
+        
+        if (n % 2 == 1) {
+            arr[index] = 0;
+        }
+        
+        return arr;
+    }
+}