Is Maximum Number of Matching Indices After Right Shifts easy or hard?

Maximum Number of Matching Indices After Right Shifts is generally classified as a Medium difficulty problem. The brute-force idea is straightforward, but recognizing that matching pairs imply a unique rotation shift is the key insight that leads to the optimal linear-time solution.

Maximum Number of Matching Indices After Right Shifts Python/Java solution

Typical implementations iterate through indices and compute the shift needed to align matching elements. A frequency map tracks how often each shift occurs. The maximum frequency represents the best rotation. The same logic works in Python, Java, C++, Go, and TypeScript with O(n) average time complexity.

What is the best approach for Maximum Number of Matching Indices After Right Shifts?

The most efficient approach counts how many element pairs align for each possible rotation shift. For every pair where nums1[i] equals nums2[j], compute the shift k = (j - i + n) % n and increment its frequency. The shift with the highest count gives the maximum matching indices. This method runs in O(n) average time with O(n) extra space.

Is Maximum Number of Matching Indices After Right Shifts asked at Google/Amazon/Meta?

Rotation alignment and cyclic array matching problems appear frequently in interviews at large tech companies such as Google, Amazon, and Meta. Variants often test understanding of array rotations, modular arithmetic, and counting strategies to reduce brute-force simulations.

What data structure is used in Maximum Number of Matching Indices After Right Shifts?

The optimized solution uses arrays or hash maps to count how many element pairs correspond to each rotation shift. The problem itself revolves around array traversal and modular index calculations, sometimes combined with frequency counting.

What is the time complexity of Maximum Number of Matching Indices After Right Shifts?

The brute-force simulation checks all n rotations and compares n elements per rotation, resulting in O(n^2) time. An optimized shift-counting approach reduces the complexity to O(n) on average by computing the rotation that aligns matching values and counting frequencies.

Description Hints 1 Solution Video 10:59

3400. Maximum Number of Matching Indices After Right Shifts

Q: How to solve Maximum Number of Matching Indices After Right Shifts in O(n)?

Iterate through elements of nums1 and nums2. When nums1[i] equals nums2[j], compute the shift required to align them using k = (j - i + n) % n. Track how many pairs produce the same shift using a hash map or frequency array. The shift with the highest frequency yields the maximum number of matching indices.

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You are given two integer arrays, nums1 and nums2, of the same length.

An index i is considered matching if nums1[i] == nums2[i].

Return the maximum number of matching indices after performing any number of right shifts on nums1.

A right shift is defined as shifting the element at index i to index (i + 1) % n, for all indices.

Example 1:

Input: nums1 = [3,1,2,3,1,2], nums2 = [1,2,3,1,2,3]

Output: 6

Explanation:

If we right shift nums1 2 times, it becomes [1, 2, 3, 1, 2, 3]. Every index matches, so the output is 6.

Example 2:

Input: nums1 = [1,4,2,5,3,1], nums2 = [2,3,1,2,4,6]

Output: 3

Explanation:

If we right shift nums1 3 times, it becomes [5, 3, 1, 1, 4, 2]. Indices 1, 2, and 4 match, so the output is 3.

Constraints:

nums1.length == nums2.length
1 <= nums1.length, nums2.length <= 3000
1 <= nums1[i], nums2[i] <= 10⁹

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Hint 1

Simulate the right shifts and check number of equal elements.

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3400. Maximum Number of Matching Indices After Right Shifts

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