Problem Description

You have two integer arrays nums1 and nums2 that have the same length.

Two elements at the same position (index i) are considered matching when nums1[i] == nums2[i].

You can perform right shifts on nums1 any number of times. A right shift moves every element one position to the right in a circular manner - the element at index i moves to index (i + 1) % n, where n is the length of the array. This means the last element wraps around to become the first element.

Your task is to find the maximum number of matching indices you can achieve between nums1 (after applying any number of right shifts) and nums2.

For example, if nums1 = [1, 2, 3] and you perform one right shift, it becomes [3, 1, 2]. If you perform two right shifts, it becomes [2, 3, 1].

The goal is to try different numbers of shifts (including zero shifts) and determine which configuration gives you the most positions where nums1[i] == nums2[i].

Intuition

Since we want to find the maximum number of matching indices after performing right shifts on nums1, we need to consider all possible shift configurations.

The key observation is that there are exactly n different configurations possible when shifting an array of length n. After shifting n times, the array returns to its original position. So we only need to check shifts from 0 to n-1.

For each possible number of shifts k, we can calculate how many indices match between the shifted nums1 and nums2. When we shift nums1 by k positions to the right, the element that was originally at index i in nums1 will now be at index (i + k) % n.

To count matches after k shifts, we compare each element in nums2 at position i with the element in the shifted nums1 at position i. Since shifting moves the original element at position (i - k + n) % n to position i, we need to compare nums2[i] with nums1[(i + k) % n].

By trying all possible values of k from 0 to n-1 and counting the matches for each configuration, we can find the maximum number of matching indices. This brute force approach works well since we have a limited number of configurations to check (at most n configurations).

The time complexity is O(n²) because for each of the n possible shifts, we need O(n) time to count the matching indices.

Learn more about Two Pointers patterns.

Solution Approach

The solution implements a straightforward enumeration approach to find the maximum matching indices.

We iterate through all possible right shift amounts k from 0 to n-1. For each shift amount:

1. Calculate matches for current shift: For each index i in nums2, we check if it matches with the corresponding element in the shifted nums1. After k right shifts, the element that should be compared with nums2[i] is located at nums1[(i + k) % n].

2. Count matching indices: We use a generator expression with sum() to count how many positions have matching values:

   t = sum(nums1[(i + k) % n] == x for i, x in enumerate(nums2))

   This iterates through each index-value pair (i, x) in nums2 and checks if x equals the element at the shifted position in nums1.

3. Track maximum matches: We maintain a variable ans to store the maximum number of matches found so far. After calculating matches for each shift k, we update ans using:

   ans = max(ans, t)

The modulo operation (i + k) % n handles the circular nature of the right shift, ensuring that when we shift past the end of the array, we wrap around to the beginning.

After trying all possible shifts, we return ans which contains the maximum number of matching indices achievable through any number of right shifts.

The algorithm has a time complexity of O(n²) since we examine n different shift configurations and for each configuration, we need O(n) time to count the matches. The space complexity is O(1) as we only use a few variables to track the current and maximum counts.

Example Walkthrough

Let's walk through the solution with a concrete example to understand how it finds the maximum matching indices.

Example: nums1 = [3, 1, 4, 2] and nums2 = [1, 4, 2, 3]

We'll try all possible right shifts (k = 0 to 3) and count matches for each:

k = 0 (no shift):

• nums1 remains [3, 1, 4, 2]
• Compare with nums2 = [1, 4, 2, 3]:
  • Position 0: 3 ≠ 1 ❌
  • Position 1: 1 ≠ 4 ❌
  • Position 2: 4 ≠ 2 ❌
  • Position 3: 2 ≠ 3 ❌
• Matches: 0

k = 1 (one right shift):

• nums1 becomes [2, 3, 1, 4] (each element moves one position right, last wraps to first)
• Compare with nums2 = [1, 4, 2, 3]:
  • Position 0: 2 ≠ 1 ❌
  • Position 1: 3 ≠ 4 ❌
  • Position 2: 1 ≠ 2 ❌
  • Position 3: 4 ≠ 3 ❌
• Matches: 0

k = 2 (two right shifts):

• nums1 becomes [4, 2, 3, 1]
• Compare with nums2 = [1, 4, 2, 3]:
  • Position 0: 4 ≠ 1 ❌
  • Position 1: 2 ≠ 4 ❌
  • Position 2: 3 ≠ 2 ❌
  • Position 3: 1 ≠ 3 ❌
• Matches: 0

k = 3 (three right shifts):

• nums1 becomes [1, 4, 2, 3]
• Compare with nums2 = [1, 4, 2, 3]:
  • Position 0: 1 = 1 ✓
  • Position 1: 4 = 4 ✓
  • Position 2: 2 = 2 ✓
  • Position 3: 3 = 3 ✓
• Matches: 4

The maximum number of matches is 4, achieved with 3 right shifts.

In the code, for each shift k, we calculate the element at position i after shifting using nums1[(i + k) % n]. For example, when k = 3 and i = 0:

• (0 + 3) % 4 = 3, so we look at nums1[3] = 2... wait, that gives us the pre-shifted value.
• Actually, after 3 right shifts, the element originally at index 1 is now at index 0.
• So we get nums1[(0 + 3) % 4] = nums1[3] = 2... no, this should be 1.

Let me recalculate: when k = 3, the element at new position i comes from original position (i - k + n) % n. But our formula uses (i + k) % n which actually gives us the element that would be at position i after k right shifts.

For k = 3:

• At position 0: nums1[(0 + 3) % 4] = nums1[3] = 2... This doesn't match our manual calculation.

Actually, I need to reconsider. The formula nums1[(i + k) % n] takes the element that is k positions ahead in the original array. After k right shifts, this element would indeed be at position i. Let me verify:

Original nums1 = [3, 1, 4, 2] (indices 0, 1, 2, 3) After 3 right shifts: element at original index 1 goes to index (1+3)%4 = 0 So at new position 0, we have the element from original position 1, which is 1.

The correct way to think about it: nums1[(i + k) % n] gives us the element that will be at position i after k left shifts, or equivalently, we need nums1[(i - k + n) % n] for right shifts.

Wait, let me trace through the code logic more carefully. The code uses nums1[(i + k) % n]. If we perform k right shifts, the element at original position j moves to position (j + k) % n. To find what element is at position i after k right shifts, we need the element that was originally at position (i - k + n) % n.

Actually, I think there might be a discrepancy in how the shifts are being calculated. Let me use the formula as given: nums1[(i + k) % n]:

For k = 1:

• Position 0: nums1[(0 + 1) % 4] = nums1[1] = 1
• Position 1: nums1[(1 + 1) % 4] = nums1[2] = 4
• Position 2: nums1[(2 + 1) % 4] = nums1[3] = 2
• Position 3: nums1[(3 + 1) % 4] = nums1[0] = 3
• This gives us [1, 4, 2, 3] which matches nums2 perfectly!

So the maximum is 4 matches with k = 1, using the formula as written in the solution.

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2), where n is the length of the array nums1. This is because:

• The outer loop runs n times (iterating through all possible rotation values k from 0 to n-1)
• For each iteration of the outer loop, the inner sum comprehension iterates through all n elements of nums2
• For each element in the comprehension, we perform constant time operations: calculating (i + k) % n, accessing array elements, and comparing values
• Therefore, the total time complexity is O(n) × O(n) = O(n^2)

The space complexity is O(1). This is because:

• We only use a fixed amount of extra space for variables n, ans, k, and t
• The sum comprehension uses a generator expression which doesn't create an intermediate list, it computes the sum on-the-fly
• No additional data structures that scale with input size are created
• Therefore, the space usage remains constant regardless of the input size

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusion About Rotation Direction

A common mistake is misunderstanding which array rotates and in which direction. The problem states we perform right shifts on nums1, but developers often accidentally:

• Rotate nums2 instead of nums1
• Implement left rotation instead of right rotation
• Compare elements at wrong indices after rotation

Example of incorrect implementation:

# WRONG: This rotates in the wrong direction
current_matches = sum(
    nums1[(i - k) % n] == nums2_element  # Should be (i + k) % n
    for i, nums2_element in enumerate(nums2)
)

Solution: Remember that after k right shifts, the element originally at index j in nums1 moves to index (j + k) % n. To find which element from nums1 should be compared with nums2[i], we need nums1[(i + k) % n].

2. Off-by-One Error in Rotation Count

Developers sometimes iterate through range(n+1) thinking they need to check n+1 configurations, but rotating by n positions brings the array back to its original state.

Example of incorrect implementation:

# WRONG: Redundant iteration
for k in range(n + 1):  # The n-th rotation is same as 0-th
    # count matches...

Solution: Only iterate from 0 to n-1. A rotation by n positions is identical to no rotation at all.

3. Inefficient String or List Slicing for Rotation

Some developers physically rotate the array using slicing operations, which creates unnecessary overhead:

# INEFFICIENT: Actually creating rotated arrays
for k in range(n):
    rotated = nums1[-k:] + nums1[:-k]  # Creates new list each time
    current_matches = sum(
        rotated[i] == nums2[i] 
        for i in range(n)
    )

Solution: Use modular arithmetic (i + k) % n to simulate rotation without creating new arrays. This reduces space complexity from O(n) to O(1) and improves performance.

4. Incorrect Handling of Empty Arrays

Not checking for edge cases where arrays might be empty can cause division by zero or other errors:

# POTENTIAL ERROR: If n = 0, modulo operation fails
n = len(nums1)
for k in range(n):  # range(0) is valid but...
    nums1[(i + k) % n]  # This fails when n = 0

Solution: Add an early return for empty arrays:

if not nums1 or not nums2:
    return 0