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| 1 | +/* |
| 2 | +122. Best Time to Buy and Sell Stock II |
| 3 | +
|
| 4 | +You are given an integer array prices where prices[i] is the price of a given stock on the ith day. |
| 5 | +On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day. |
| 6 | +Find and return the maximum profit you can achieve. |
| 7 | +
|
| 8 | +Example 1: |
| 9 | +Input: prices = [7,1,5,3,6,4] |
| 10 | +Output: 7 |
| 11 | +Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. |
| 12 | +Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. |
| 13 | +Total profit is 4 + 3 = 7. |
| 14 | +
|
| 15 | +Example 2: |
| 16 | +Input: prices = [1,2,3,4,5] |
| 17 | +Output: 4 |
| 18 | +Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. |
| 19 | +Total profit is 4. |
| 20 | +
|
| 21 | +Example 3: |
| 22 | +Input: prices = [7,6,4,3,1] |
| 23 | +Output: 0 |
| 24 | +Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0. |
| 25 | +
|
| 26 | +Constraints: |
| 27 | +1 <= prices.length <= 3 * 104 |
| 28 | +0 <= prices[i] <= 104 |
| 29 | +*/ |
| 30 | +class Solution { |
| 31 | + static int memo[][] =new int[30001][2]; |
| 32 | + public int maxProfit(int[] prices) { |
| 33 | + //Recursion::) |
| 34 | + //return solveRec(prices,0,1); |
| 35 | + |
| 36 | + // Arrays.stream(memo).forEach(a->Arrays.fill(a,-1)); |
| 37 | + // return solveMemo(prices,0,1); |
| 38 | + |
| 39 | + return solveTab(prices); |
| 40 | + |
| 41 | + } |
| 42 | + public static int solveRec(int[] prices ,int indx,int buy){ |
| 43 | + //Base Case :: |
| 44 | + if(indx==prices.length) return 0; |
| 45 | + int profit=0; |
| 46 | + if(buy==1){ |
| 47 | + profit = -prices[indx]+solveRec(prices,indx+1,0); |
| 48 | + }else{ |
| 49 | + profit =prices[indx]+solveRec(prices,indx+1,1); |
| 50 | + } |
| 51 | + int skip =solveRec(prices,indx+1,buy); |
| 52 | + return Math.max(profit,skip); |
| 53 | + } |
| 54 | + public static int solveMemo(int[] prices,int indx,int buy){ |
| 55 | + //Base Case :: |
| 56 | + if(indx==prices.length) return 0; |
| 57 | + if(memo[indx][buy]!=-1){ |
| 58 | + return memo[indx][buy]; |
| 59 | + } |
| 60 | + int profit=0; |
| 61 | + if(buy==1){ |
| 62 | + profit = -prices[indx]+solveMemo(prices,indx+1,0); |
| 63 | + }else{ |
| 64 | + profit =prices[indx]+solveMemo(prices,indx+1,1); |
| 65 | + } |
| 66 | + int skip =solveMemo(prices,indx+1,buy); |
| 67 | + return memo[indx][buy]=Math.max(profit,skip); |
| 68 | + } |
| 69 | + public static int solveTab(int[] prices){ |
| 70 | + int n=prices.length; |
| 71 | + int tab[][] =new int[n+1][2]; |
| 72 | + Arrays.stream(tab).forEach(a->Arrays.fill(a,-1)); |
| 73 | + tab[n][0]=0; |
| 74 | + tab[n][1]=0; |
| 75 | + |
| 76 | + int profit=0; |
| 77 | + for(int i=n-1;i>=0;i--){ |
| 78 | + for(int j=0;j<=1;j++){ |
| 79 | + if(j==1){ |
| 80 | + profit = Math.max(prices[i]+tab[i+1][0],tab[i+1][1]); |
| 81 | + } |
| 82 | + if(j==0){ |
| 83 | + profit = Math.max(-prices[i]+tab[i+1][1],tab[i+1][0]); |
| 84 | + } |
| 85 | + tab[i][j]=profit; |
| 86 | + } |
| 87 | + } |
| 88 | + return tab[0][0]; |
| 89 | + } |
| 90 | +} |
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