From 5f017537f49d469d1b728d4323a8752cde932ab7 Mon Sep 17 00:00:00 2001
From: Jumanazar Saidov <jsr1611@outlook.com>
Date: Thu, 17 Aug 2023 16:50:31 +0500
Subject: [PATCH] added java solution that has O(1) time and space complexity

---
 solutions/015.md | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)

diff --git a/solutions/015.md b/solutions/015.md
index 2b2c9e6..a2655f8 100644
--- a/solutions/015.md
+++ b/solutions/015.md
@@ -65,6 +65,32 @@ function sumOfMultiples(n: number): number {
 }
 ```
 
+
+#### Java (with O(1) complexity solution)
+```java
+class Day008 {
+    /* 
+        Calculate sum of multiples using arithmetic progression technique
+     */
+    public int sumOfMultiples(int n) {
+        return calcArithmetic(3, n) + calcArithmetic(5, n) + calcArithmetic(7, n) + calcArithmetic(3 * 5 * 7, n)
+                - calcArithmetic(3 * 5, n) - calcArithmetic(3 * 7, n) - calcArithmetic(7 * 5, n);
+    }
+    
+    /*
+        Return arithmetic of numbers starting from a1 with a1 difference up to num using
+        the formula: s = n/2 * (a1 + an) where
+        - a1: first term
+        - an: last term
+        - n: number of terms
+     */
+    public int calcArithmetic(int a1, int num) {
+        int n = num / a1;
+        int an = n * a1;
+        return n * (a1 + an) / 2;
+    }
+}
+```
 The given solution solves the problem by performing the following steps:
 1. Initialize a variable `total` to 0 to keep track of the sum.
 2. Iterate over the range from 1 to `n+1`.