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| 1 | +class Solution { |
| 2 | + public int maxProfit(int[] prices) { |
| 3 | + |
| 4 | + int n = prices.length; |
| 5 | + |
| 6 | + if(n==0) return 0; |
| 7 | + |
| 8 | + int[] left = new int[n]; |
| 9 | + int[] right = new int[n]; |
| 10 | + |
| 11 | + //Bidirectional DP Logic |
| 12 | + //Left to right traversal |
| 13 | + int min = prices[0]; |
| 14 | + for(int i = 1; i < n; i++) { |
| 15 | + //1. Update Buy price |
| 16 | + if(prices[i] < min) min=prices[i]; |
| 17 | + //2. Calculate current profit |
| 18 | + int profit = prices[i]-min; |
| 19 | + //3. Fill left array (based on condition) |
| 20 | + left[i] = Math.max(left[i-1], profit); |
| 21 | + } |
| 22 | + |
| 23 | + |
| 24 | + //Right to left traversal |
| 25 | + int max = prices[n-1]; |
| 26 | + for(int i = n-2; i>=0; i--) { |
| 27 | + //1. Update Sell price |
| 28 | + if(prices[i] > max) max=prices[i]; |
| 29 | + //2. Calculate current profit |
| 30 | + int profit = max-prices[i]; |
| 31 | + //3. Update right array |
| 32 | + right[i] = Math.max(right[i+1], profit); |
| 33 | + } |
| 34 | + |
| 35 | + int maxProfit = 0; |
| 36 | + for(int i = 0; i<n; i++) { |
| 37 | + maxProfit = Math.max(maxProfit, left[i]+right[i]); |
| 38 | + } |
| 39 | + |
| 40 | + return maxProfit; |
| 41 | + } |
| 42 | +} |
| 43 | + |
| 44 | + |
| 45 | +class Solution { |
| 46 | + public int maxProfit(int[] prices) { |
| 47 | + |
| 48 | + // c1, c2, p1, p2 |
| 49 | + int c1 = Integer.MAX_VALUE; |
| 50 | + int c2 = Integer.MAX_VALUE; |
| 51 | + int p1 = 0; |
| 52 | + int p2 = 0; |
| 53 | + |
| 54 | + //Iterate over prices |
| 55 | + for(int price : prices) { |
| 56 | + c1 = Math.min(c1, price); |
| 57 | + p1 = Math.max(p1, price-c1); |
| 58 | + c2 = Math.min(c2, price-p1); |
| 59 | + p2 = Math.max(p2, price-c2); |
| 60 | + } |
| 61 | + |
| 62 | + return p2; |
| 63 | + } |
| 64 | +} |
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