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Problem solution · Java

Difference Between Maximum and Minimum Price Sum

Difference Between Maximum and Minimum Price Sum: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
63 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Difference Between Maximum and Minimum Price Sum, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 63 lines of Java from the credited upstream file 2538.java.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDifference Between Maximum and Minimum Price Sum · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long maxOutput(int n, int[][] edges, int[] price) {    List<Integer>[] tree = new List[n];    // maxSums[i] := the maximum the sum of path rooted at i    int[] maxSums = new int[n];     for (int i = 0; i < n; ++i)      tree[i] = new ArrayList<>();     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      tree[u].add(v);      tree[v].add(u);    }     // Precalculate `maxSums`.    maxSum(tree, 0, /*prev=*/-1, maxSums, price);    reroot(tree, 0, /*prev=*/-1, /*parentSum=*/0, maxSums, price);    return (long) ans;  }   private int ans = 0;   private int maxSum(List<Integer>[] tree, int u, int prev, int[] maxSums, int[] price) {    int maxChildSum = 0;    for (final int v : tree[u])      if (v != prev)        maxChildSum = Math.max(maxChildSum, maxSum(tree, v, u, maxSums, price));    return maxSums[u] = price[u] + maxChildSum;  }   private void reroot(List<Integer>[] tree, int u, int prev, int parentSum, int[] maxSums,                      int[] price) {    // Get top two sums and top one node index.    int maxSubtreeSum1 = 0;    int maxSubtreeSum2 = 0;    int maxNode = -1;    for (final int v : tree[u]) {      if (v == prev)        continue;      if (maxSums[v] > maxSubtreeSum1) {        maxSubtreeSum2 = maxSubtreeSum1;        maxSubtreeSum1 = maxSums[v];        maxNode = v;      } else if (maxSums[v] > maxSubtreeSum2) {        maxSubtreeSum2 = maxSums[v];      }    }     if (tree[u].size() == 1)      ans = Math.max(ans, Math.max(parentSum, maxSubtreeSum1));     for (final int v : tree[u]) {      if (v == prev)        continue;      final int nextParentSum = (v == maxNode ? price[u] + Math.max(parentSum, maxSubtreeSum2)                                              : price[u] + Math.max(parentSum, maxSubtreeSum1));      reroot(tree, v, u, nextParentSum, maxSums, price);    }  }}

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