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| 1 | +// https://leetcode.com/problems/house-robber-ii/ |
| 2 | + |
| 3 | +class Solution { |
| 4 | + |
| 5 | + public int robbing(int[] houses, int startIdx, int endIdx) { |
| 6 | + |
| 7 | + // OBJECTIVE: Iterate array linearly with 2 different iterators. Return the highest profitable iterator |
| 8 | + |
| 9 | + // Add money from 1st house to 3rd house |
| 10 | + houses[2] += houses[startIdx]; |
| 11 | + |
| 12 | + // Iterate houses |
| 13 | + for (int i=3; i<endIdx; i++) { |
| 14 | + houses[i] += Math.max(houses[i - 2], houses[i - 3]); |
| 15 | + } |
| 16 | + |
| 17 | + // Return the most profitable robbing spree |
| 18 | + return Math.max(houses[endIdx - 1], houses[endIdx - 2]); |
| 19 | + } |
| 20 | + |
| 21 | + public void sliceArray(int[] source, int[] dest, int startIdx, int endIdx) { |
| 22 | + |
| 23 | + // OBJECTIVE: Add elements from start to end to dest array |
| 24 | + |
| 25 | + for (int i=startIdx; i<endIdx; i++) { |
| 26 | + dest[i] = source[i]; |
| 27 | + } |
| 28 | + } |
| 29 | + |
| 30 | + public int rob(int[] nums) { |
| 31 | + |
| 32 | + // Get length of house |
| 33 | + int n = nums.length; |
| 34 | + |
| 35 | + // If there are less than 4 houses, return the house with the most money |
| 36 | + // NOTE: Can't rob adjacent houses! |
| 37 | + if (n < 4) { |
| 38 | + |
| 39 | + int maximumElem = 0; |
| 40 | + for (int i=0; i<n; i++) { |
| 41 | + maximumElem = Math.max(maximumElem, nums[i]); |
| 42 | + } |
| 43 | + |
| 44 | + return maximumElem; |
| 45 | + } |
| 46 | + |
| 47 | + // If there are exactly 4 houses, return the pair with the most money |
| 48 | + else if (n == 4) { |
| 49 | + return Math.max(nums[0] + nums[2], nums[1] + nums[3]); |
| 50 | + } |
| 51 | + |
| 52 | + // Create 2 new arrays |
| 53 | + int[] sliceA = new int[n]; |
| 54 | + int[] sliceB = new int[n]; |
| 55 | + |
| 56 | + // Slice nums[] and add elements to sliceA and sliceB |
| 57 | + sliceArray(nums, sliceA, 0, n - 1); |
| 58 | + sliceArray(nums, sliceB, 1, n); |
| 59 | + |
| 60 | + // If there are more than 4 houses, return the most profitable robbing spree |
| 61 | + return Math.max(robbing(sliceA, 0, sliceA.length), robbing(sliceB, 0, sliceB.length)); |
| 62 | + } |
| 63 | +} |
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