Added 10 solutions

General Programming/Basic Programming/Electronics Shop/Solution.java

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+//     Author: Rodney Shaghoulian
+//     Github: github.com/rshaghoulian
+// HackerRank: hackerrank.com/rshaghoulian
+
+import java.util.Scanner;
+import java.util.Arrays;
+
+// Algorithm: 
+//     1. Sort both arrays.
+//     2. Linearly walk through both arrays (at the same time).
+//        If the combined price is unaffordable, find a cheaper harddrive (j--).
+//        If the combined price is too low, update "max", and find a more expensive keyboard (i++)
+
+// Time complexity: O(n log n + m log m)
+public class Solution {
+    static int getMoneySpent(int[] keyboards, int[] drives, int s) {
+        Arrays.sort(keyboards); // O(n log n) time complexity
+        Arrays.sort(drives);    // O(m log m) time complexity
+        int max = -1;
+        int i = 0;
+        int j = drives.length - 1;
+        while (i < keyboards.length && j >= 0) { // O(n + m) time complexity
+            int cost = keyboards[i] + drives[j];
+            if (cost > s) {
+                j--; // look for a cheaper hard drive
+            } else {
+                if (cost > max) {
+                    max = cost;
+                }
+                i++; // look for a more expensive keyboard
+            }
+        }
+        return max;
+    }
+
+    public static void main(String[] args) {
+        /* Save input */
+        Scanner scan = new Scanner(System.in);
+        int s = scan.nextInt();
+        int n = scan.nextInt();
+        int m = scan.nextInt();
+        int[] keyboards = new int[n];
+        for (int i = 0; i < n; i++) {
+            keyboards[i] = scan.nextInt();
+        }
+        int[] drives = new int[m];
+        for (int i = 0; i < m; i++) {
+            drives[i] = scan.nextInt();
+        }
+        scan.close();
+        
+        /* Calculate result */
+        int moneySpent = getMoneySpent(keyboards, drives, s);
+        System.out.println(moneySpent);
+    }
+}