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3400. Maximum Number of Matching Indices After Right Shifts 👍

  • Time: $O(n^2)$
  • Space: $O(1)$
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class Solution {
 public:
  int maximumMatchingIndices(vector<int>& nums1, vector<int>& nums2) {
    const int n = nums1.size();
    int ans = 0;

    for (int shift = 0; shift < n; ++shift) {
      int matches = 0;
      for (int i = 0; i < n; ++i)
        if (nums1[(i + shift) % n] == nums2[i])
          ++matches;
      ans = max(ans, matches);
    }

    return ans;
  }
};

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class Solution {
  public int maximumMatchingIndices(int[] nums1, int[] nums2) {
    final int n = nums1.length;
    int ans = 0;

    for (int shift = 0; shift < n; ++shift) {
      int matches = 0;
      for (int i = 0; i < n; ++i)
        if (nums1[(i + shift) % n] == nums2[i])
          ++matches;
      ans = Math.max(ans, matches);
    }

    return ans;
  }
}

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class Solution:
  def maximumMatchingIndices(self, nums1: list[int], nums2: list[int]) -> int:
    n = len(nums1)
    ans = 0

    for shift in range(n):
      matches = 0
      for i, num2 in enumerate(nums2):
        if nums1[(i + shift) % n] == num2:
          matches += 1
      ans = max(ans, matches)

    return ans