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3540. Minimum Time to Visit All Houses

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  long long minTotalTime(vector<int>& forward, vector<int>& backward,
                         vector<int>& queries) {
    const int n = forward.size();
    const long sum = accumulate(backward.begin(), backward.end(), 0L);
    long ans = 0;
    int pos = 0;
    vector<long> prefixF(n + 1);
    vector<long> prefixB(n + 1);

    for (int i = 0; i < n; ++i) {
      prefixF[i + 1] = prefixF[i] + forward[i];
      prefixB[i] = (i == 0 ? 0 : prefixB[i - 1]) + backward[i];
    }

    for (const int q : queries) {
      const long r = (q < pos ? prefixF[n] : 0) + prefixF[q] - prefixF[pos];
      const long l = (q > pos ? sum : 0) + prefixB[pos] - prefixB[q];
      ans += min(l, r);
      pos = q;
    }

    return ans;
  }
};

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class Solution {
  public long minTotalTime(int[] forward, int[] backward, int[] queries) {
    final int n = forward.length;
    final long sum = Arrays.stream(backward).asLongStream().sum();
    long ans = 0;
    int pos = 0;
    long[] prefixF = new long[n + 1];
    long[] prefixB = new long[n + 1];

    for (int i = 0; i < n; i++) {
      prefixF[i + 1] = prefixF[i] + forward[i];
      prefixB[i] = (i == 0 ? 0 : prefixB[i - 1]) + backward[i];
    }

    for (final int q : queries) {
      final long r = (q < pos ? prefixF[n] : 0) + prefixF[q] - prefixF[pos];
      final long l = (q > pos ? sum : 0) + prefixB[pos] - prefixB[q];
      ans += Math.min(l, r);
      pos = q;
    }

    return ans;
  }
}

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class Solution:
  def minTotalTime(
      self,
      forward: list[int],
      backward: list[int],
      queries: list[int]
  ) -> int:
    summ = sum(backward)
    ans = 0
    pos = 0
    prefixF = [0] + list(itertools.accumulate(forward))
    prefixB = list(itertools.accumulate(backward)) + [0]

    for q in queries:
      r = int(q < pos) * prefixF[-1] + prefixF[q] - prefixF[pos]
      l = int(q > pos) * summ + prefixB[pos] - prefixB[q]
      ans += min(l, r)
      pos = q

    return ans