Added 8 solutions

General Programming/Basic Programming/Picking Numbers/Solution.java

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+//     Author: Rodney Shaghoulian
+//     Github: github.com/rshaghoulian
+// HackerRank: hackerrank.com/rshaghoulian
+
+import java.util.Scanner;
+
+// Algorithm: 
+//  1 - Notice there is a special restriction on the range of values in the array: values are between 0 and 100.
+//  2 - Since the order of elements in our array does not matter, we can just keep the counts of each number.
+//  3 - Create a new 100-element "counts" array to store the counts of each number.
+//  4 - In our "counts" array, the adjacent pair with the largest sum is our final solution.
+//  5 - This is because the original numbers represented by the 2 adjacent slots in counts[] can be used to create a 
+//      maximal solution
+
+// Time complexity: O(n)
+public class Solution {
+    static final int MAX_VALUE = 100;
+    
+    static int solve(int[] array) {
+        /* Save counts of each number */
+        int[] counts = new int[MAX_VALUE];
+        for (int num : array) {
+            counts[num]++;
+        }
+        
+        /* Find adjacent pair with largest sum */
+        int max = 2;
+        for (int i = 1; i < 100; i++) {
+            int adjacentSum = counts[i] + counts[i-1];
+            max = Math.max(max, adjacentSum);
+        }
+        return max;
+    }
+
+    public static void main(String[] args) {
+        Scanner scan = new Scanner(System.in);
+        int n = scan.nextInt();
+        int[] array = new int[n];
+        for (int i = 0; i < n; i++) {
+            array[i] = scan.nextInt();
+        }
+        scan.close();
+        
+        System.out.println(solve(array));
+    }
+}