updated readme

2017-04-30 18:30:26 -07:00
parent 9b6ef0f13b
commit 7145449310
57 changed files with 287 additions and 268 deletions
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+//     Author: Rodney Shaghoulian
+//     Github: github.com/rshaghoulian
+// HackerRank: hackerrank.com/rshaghoulian
+
+/*
+
+ Time Complexity: O(n + k)
+Space Complexity: O(k)
+
+We keep track of the mod values in buckets for each number we read. We can do this 
+by creating k buckets where bucket "i" counts each number "n" where n % k = i.
+
+Notice that each bucket has a corresponding "complement" bucket. That is, if we take 
+the mod value of one bucket and add it to the mod value of another bucket, we get "k".
+
+Bucket   Complement Bucket
+------   -----------------
+   0            0
+   1           k-1
+   2           k-2
+   3           k-3
+        ...
+  k-3           3
+  k-2           2
+  k-1           1
+
+As we come across each number, we find its corresponding complement bucket. Each number in 
+this complement bucket can be paired with our original number to create a sum divisible by *k*
+
+*/
+
+import java.util.Scanner;
+
+public class Solution {
+    public static void main(String [] args) {
+        Scanner scan = new Scanner(System.in);
+        int n = scan.nextInt();
+        int k = scan.nextInt();
+
+        int [] bucket = new int[k];
+        int count = 0;
+        for (int i = 0; i < n; i++) {
+            int modValue = scan.nextInt() % k;
+            count += bucket[(k - modValue) % k]; // adds # of elements in complement bucket
+            bucket[modValue]++;                  // saves in bucket
+        }
+       
+        scan.close();
+        System.out.println(count);
+    }
+}