{ "data": { "question": { "questionId": "3179", "questionFrontendId": "2920", "boundTopicId": null, "title": "Maximum Points After Collecting Coins From All Nodes", "titleSlug": "maximum-points-after-collecting-coins-from-all-nodes", "content": "
There exists an undirected tree rooted at node 0
with n
nodes labeled from 0
to n - 1
. You are given a 2D integer array edges
of length n - 1
, where edges[i] = [ai, bi]
indicates that there is an edge between nodes ai
and bi
in the tree. You are also given a 0-indexed array coins
of size n
where coins[i]
indicates the number of coins in the vertex i
, and an integer k
.
Starting from the root, you have to collect all the coins such that the coins at a node can only be collected if the coins of its ancestors have been already collected.
\n\nCoins at nodei
can be collected in one of the following ways:
coins[i] - k
points. If coins[i] - k
is negative then you will lose abs(coins[i] - k)
points.floor(coins[i] / 2)
points. If this way is used, then for all the nodej
present in the subtree of nodei
, coins[j]
will get reduced to floor(coins[j] / 2)
.Return the maximum points you can get after collecting the coins from all the tree nodes.
\n\n\n
Example 1:
\n\n\nInput: edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5\nOutput: 11 \nExplanation: \nCollect all the coins from node 0 using the first way. Total points = 10 - 5 = 5.\nCollect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10.\nCollect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11.\nCollect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11.\nIt can be shown that the maximum points we can get after collecting coins from all the nodes is 11. \n\n\n
Example 2:
\n \n\n\nInput: edges = [[0,1],[0,2]], coins = [8,4,4], k = 0\nOutput: 16\nExplanation: \nCoins will be collected from all the nodes using the first way. Therefore, total points = (8 - 0) + (4 - 0) + (4 - 0) = 16.\n\n\n
\n
Constraints:
\n\nn == coins.length
2 <= n <= 105
0 <= coins[i] <= 104
edges.length == n - 1
0 <= edges[i][0], edges[i][1] < n
0 <= k <= 104
dp[x][t]
be the maximum points we can get from the subtree rooted at node x
and the second operation has been used t
times in its ancestors.",
"Note that the value of each node <= 104
, so when t >= 14
dp[x][t]
is always 0
.",
"General equation will be: dp[x][t] = max((coins[x] >> t) - k + sigma(dp[y][t]), (coins[x] >> (t + 1)) + sigma(dp[y][t + 1]))
where nodes denoted by y
in the sigma, are the direct children of node x
."
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