|
| 1 | +// Time O(nlogn) and space O(n) using max heap |
| 2 | +classSolution { |
| 3 | +funconstrainedSubsetSum(nums:IntArray,k:Int):Int { |
| 4 | +var res= nums[0] |
| 5 | +val maxHeap=PriorityQueue<IntArray>() { a, b-> b[0]- a[0] } |
| 6 | + maxHeap.add(intArrayOf(nums[0],0)) |
| 7 | + |
| 8 | +for (iin1 until nums.size) { |
| 9 | +while (i- maxHeap.peek()[1]> k) |
| 10 | + maxHeap.poll() |
| 11 | + |
| 12 | +var curMax= maxOf(nums[i], nums[i]+ maxHeap.peek()[0]) |
| 13 | + res= maxOf(res, curMax) |
| 14 | + maxHeap.add(intArrayOf(curMax, i)) |
| 15 | + } |
| 16 | + |
| 17 | +return res |
| 18 | + } |
| 19 | +} |
| 20 | + |
| 21 | +// Time O(n) and space O(n) using monotonic array/list |
| 22 | +classSolution { |
| 23 | +funconstrainedSubsetSum(nums:IntArray,k:Int):Int { |
| 24 | +var win=LinkedList<Int>() |
| 25 | + |
| 26 | +var res=Integer.MIN_VALUE |
| 27 | +for (iin0 until nums.size) { |
| 28 | + nums[i]+= (win.peekFirst()?:0) |
| 29 | + res= maxOf(res, nums[i]) |
| 30 | + |
| 31 | +while (win.isNotEmpty()&& win.peekLast()< nums[i]) |
| 32 | + win.removeLast() |
| 33 | +if (nums[i]>0) |
| 34 | + win.addLast(nums[i]) |
| 35 | +if (i>= k&& win.isNotEmpty()&& win.peekFirst()== nums[i- k]) |
| 36 | + win.removeFirst() |
| 37 | + } |
| 38 | + |
| 39 | +return res |
| 40 | + } |
| 41 | +} |