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1537. Get the Maximum Score 👍

  • Time: $O(|\texttt{nums1}| + |\texttt{nums2}|)$
  • Space: $O(1)$
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class Solution {
 public:
  int maxSum(vector<int>& nums1, vector<int>& nums2) {
    constexpr int kMod = 1'000'000'007;
    // Keep the running the sum of `nums1` and `nums2` before the next
    // rendezvous. Since `nums1` and `nums2` are increasing, move forward on the
    // smaller one to ensure we don't miss any rendezvous. When meet rendezvous,
    // choose the better path.
    long ans = 0;
    // sum(nums1) in (the prevoious rendezvous, the next rendezvous)
    long sum1 = 0;
    // sum(nums2) in (the prevoious rendezvous, the next rendezvous)
    long sum2 = 0;
    int i = 0;  // nums1's index
    int j = 0;  // nums2's index

    while (i < nums1.size() && j < nums2.size())
      if (nums1[i] < nums2[j]) {
        sum1 += nums1[i++];
      } else if (nums1[i] > nums2[j]) {
        sum2 += nums2[j++];
      } else {  // An rendezvous happens.
        ans += max(sum1, sum2) + nums1[i];
        sum1 = 0;
        sum2 = 0;
        ++i;
        ++j;
      }

    while (i < nums1.size())
      sum1 += nums1[i++];

    while (j < nums2.size())
      sum2 += nums2[j++];

    return (ans + max(sum1, sum2)) % kMod;
  }
};

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class Solution {
  public int maxSum(int[] nums1, int[] nums2) {
    final int kMod = 1_000_000_007;
    // Keep the running the sum of `nums1` and `nums2` before the next rendezvous.
    // Since `nums1` and `nums2` are increasing, move forward on the smaller one
    // to ensure we don't miss any rendezvous. When meet rendezvous, choose the
    // better path.
    long ans = 0;
    // sum(nums1) in (the prevoious rendezvous, the next rendezvous)
    long sum1 = 0;
    // sum(nums2) in (the prevoious rendezvous, the next rendezvous)
    int i = 0; // nums1's index
    int j = 0; // nums2's index

    while (i < nums1.length && j < nums2.length)
      if (nums1[i] < nums2[j]) {
        sum1 += nums1[i++];
      } else if (nums1[i] > nums2[j]) {
        sum2 += nums2[j++];
      } else { // An rendezvous happens.
        ans += Math.max(sum1, sum2) + nums1[i];
        sum1 = 0;
        sum2 = 0;
        ++i;
        ++j;
      }

    while (i < nums1.length)
      sum1 += nums1[i++];

    while (j < nums2.length)
      sum2 += nums2[j++];

    return (int) ((ans + Math.max(sum1, sum2)) % kMod);
  }
}

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class Solution:
  def maxSum(self, nums1: list[int], nums2: list[int]) -> int:
    # Keep the running the sum of `nums1` and `nums2` before the next rendezvous.
    # Since `nums1` and `nums2` are increasing, move forward on the smaller one
    # to ensure we don't miss any rendezvous. When meet rendezvous, choose the
    # better path.
    ans = 0
    sum1 = 0  # sum(nums1) in (the prevoious rendezvous, the next rendezvous)
    sum2 = 0  # sum(nums2) in (the prevoious rendezvous, the next rendezvous)
    i = 0  # nums1's index
    j = 0  # nums2's index

    while i < len(nums1) and j < len(nums2):
      if nums1[i] < nums2[j]:
        sum1 += nums1[i]
        i += 1
      elif nums1[i] > nums2[j]:
        sum2 += nums2[j]
        j += 1
      else:  # An rendezvous happens.
        ans += max(sum1, sum2) + nums1[i]
        sum1 = 0
        sum2 = 0
        i += 1
        j += 1

    while i < len(nums1):
      sum1 += nums1[i]
      i += 1

    while j < len(nums2):
      sum2 += nums2[j]
      j += 1

    return (ans + max(sum1, sum2)) % (10**9 + 7)