Added best time stock 4

Author: FreeTymeKiyan

Hard/BestTimeStock4.java

@@ -0,0 +1,90 @@
+import java.util.*;
+
+/**
+ * Say you have an array for which the ith element is the price of a given 
+ * stock on day i.
+ * 
+ * Design an algorithm to find the maximum profit. You may complete at most k 
+ * transactions.
+ * 
+ * Note:
+ * You may not engage in multiple transactions at the same time (ie, you must
+ * sell the stock before you buy again).
+ * 
+ * Tags: DP
+ */
+class BestTimeStock4 {
+    
+    public static void main(String[] args) {
+        // 2, [3,2,6,5,0,3]
+        // 2, [3,3,5,0,0,3,1,4]
+        BestTimeStock4 b = new BestTimeStock4();
+        int[] A = new int[]{3,3,5,0,0,3,1,4};
+        int[] B = new int[]{3,2,6,5,0,3};
+        System.out.println(b.maxProfitOpt(2, A));
+        System.out.println(b.maxProfit(2, A));
+        System.out.println(b.maxProfitOpt(2, B));
+        System.out.println(b.maxProfit(2, B));
+    }
+    
+    /**
+     * DP, bottom-up, O(kn) Time, O(n) Space
+     * If k >= n/2, we can have transactions any time, O(n).
+     * dp[k][i+1] represents the max profit of using [0, i] and k transactions
+     * It can be dp[k-1][i+1](add 1 more transaction changes nothing)
+     * It can be dp[k][i](prices[i] changes nothing)
+     * It can be prices[i] + max(dp[k-1][j] - prices[j]), 0 <= j < i
+     * means prices[i] will change the max profit, find the biggest from k-1 
+     * transacions and add prices[i]
+     * dp[k][i+1] = max(dp[k-1][i+1], dp[k][i], prices[i] + max(dp[k-1][j] - 
+     * prices[j])), (0 <= j < i)
+     */
+    public int maxProfitOpt(int k, int[] prices) {
+        if (prices == null || prices.length < 2 || k == 0) return 0;
+        int n = prices.length;
+        int res = 0;
+        if (k >= n / 2) { // as many transactions as possible
+            for (int i = 1; i < n; i++) {
+                if (prices[i] > prices[i - 1]) res += prices[i] - prices[i - 1];
+            }
+            return res;
+        }
+        int[] cur = new int[n+1];
+        for (int i = 1; i <= k; i++) {
+            int curMax = Integer.MIN_VALUE;
+            for (int j = 0; j < n; j++) {
+                int temp = cur[j+1];
+                cur[j+1] = Math.max(Math.max(cur[j+1], cur[j]), prices[j] + curMax);
+                System.out.print(curMax + "|");
+                curMax = Math.max(curMax, temp - prices[j]);
+                System.out.print(curMax + "\n");
+            }
+            System.out.println(Arrays.toString(cur));
+        }
+        return cur[n];
+    }
+    
+    /**
+     * DP, bottom-up, O(kn) Time, O(kn) Space
+     */
+    public int maxProfit(int k, int[] prices) {
+        if (prices == null || prices.length < 2 || k == 0) return 0;
+        int n = prices.length;
+        int res = 0;
+        if (k >= n / 2) {
+            for (int i = 1; i < n; i++) {
+                if (prices[i] > prices[i - 1]) res += prices[i] - prices[i - 1];
+            }
+            return res;
+        }
+        int[][] dp = new int[k+1][n+1];
+        for (int i = 1; i <= k; i++) {
+            int curMax = Integer.MIN_VALUE;
+            for (int j = 0; j < n; j++) {
+                dp[i][j+1] = Math.max(Math.max(dp[i-1][j+1], dp[i][j]), prices[j] + curMax);
+                curMax = Math.max(curMax, dp[i-1][j] - prices[j]);
+            }
+        }
+        return dp[k][n];
+    }
+}