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Problem solution · Java

Minimum Time to Make Array Sum At Most x

Minimum Time to Make Array Sum At Most x: a Java solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
35 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

Dynamic programming

For Minimum Time to Make Array Sum At Most x, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 35 lines of Java from the credited upstream file 2809.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 4 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeMinimum Time to Make Array Sum At Most x · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {    final int n = nums1.size();    final int sum1 = nums1.stream().mapToInt(Integer::intValue).sum();    final int sum2 = nums2.stream().mapToInt(Integer::intValue).sum();    // dp[i][j] := the maximum reduced value if we do j operations on the first    // i numbers    int[][] dp = new int[n + 1][n + 1];    List<Pair<Integer, Integer>> sortedNums = new ArrayList<>();     for (int i = 0; i < n; ++i)      sortedNums.add(new Pair<>(nums2.get(i), nums1.get(i)));     sortedNums.sort(Comparator.comparingInt(Pair::getKey));     for (int i = 1; i <= n; ++i) {      final int num2 = sortedNums.get(i - 1).getKey();      final int num1 = sortedNums.get(i - 1).getValue();      for (int j = 1; j <= i; ++j)        dp[i][j] = Math.max(            // the maximum reduced value if we do j ops on the first i - 1 nums            dp[i - 1][j],            // the maximum reduced value if we do j - 1 ops on the first i - 1            // nums + making i-th num of nums1 to 0 at j-th operation            dp[i - 1][j - 1] + num2 * j + num1);    }     for (int op = 0; op <= n; ++op)      if (sum1 + sum2 * op - dp[n][op] <= x)        return op;     return -1;  }}

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