Problem Description

You have a circular array nums where the last element connects back to the first element. Given this array and a list of queries, for each query index queries[i], you need to find the minimum distance from position queries[i] to any other position j in the circular array where nums[j] has the same value as nums[queries[i]].

The distance in a circular array can be calculated in two directions:

• Moving forward (clockwise): from position i to position j
• Moving backward (counterclockwise): from position i to position j

The minimum of these two distances is the answer for that query.

Key points:

• The array is circular, meaning after the last element, you wrap around to the first element
• You're looking for other positions with the same value, not including the query position itself
• If no other position contains the same value as nums[queries[i]], return -1 for that query
• Return an array answer where answer[i] is the result for queries[i]

Example understanding: If nums = [1, 2, 3, 2] and queries = [1]:

• At index 1, we have value 2
• Another 2 exists at index 3
• The forward distance is 2 (from index 1 to 3)
• The backward distance (going circular) is 2 (from index 1 to 3 via 0, 4, 3)
• The minimum distance is 2

Intuition

The key insight is that in a circular array, finding the minimum distance to another identical element requires checking both directions - forward and backward. However, dealing with circular wraparound can be tricky.

To handle the circular nature elegantly, we can double the array. If our original array is [1, 2, 3, 2], we create [1, 2, 3, 2, 1, 2, 3, 2]. This transformation allows us to treat the circular distance problem as a linear one - any circular path in the original array becomes a straight path in the doubled array.

For each position in this doubled array, we need to find the closest identical element. We can approach this from two directions:

1. Looking left: For each position, find the closest identical element that appeared before it
2. Looking right: For each position, find the closest identical element that will appear after it

We use hash tables to track:

• left: stores the most recent position where each value appeared as we scan from left to right
• right: stores the nearest future position where each value will appear as we scan from right to left

For each position i in the doubled array:

• When scanning left to right, if we've seen nums[i] before at position left[nums[i]], the distance is i - left[nums[i]]
• When scanning right to left, if we'll see nums[i] again at position right[nums[i]], the distance is right[nums[i]] - i
• We take the minimum of these two distances

Finally, since we doubled the array, each original index i now appears at both position i and position i + n. We take the minimum distance calculated for both positions. If this minimum distance is greater than or equal to n (the original array length), it means the nearest identical element is actually the same element itself (after going full circle), so we return -1.

Learn more about Binary Search patterns.

Solution Approach

The implementation follows the intuition of doubling the array and tracking positions with hash tables:

Step 1: Initialize the data structures

• Create a distance array d of size m = 2n (where n is the original array length), initialized with m (a large value)
• Create two hash tables: left and right to track element positions

Step 2: Left-to-right scan to find previous occurrences

for i in range(m):
    x = nums[i % n]  # Get element from circular array
    if x in left:
        d[i] = min(d[i], i - left[x])  # Update minimum distance
    left[x] = i  # Update last seen position

• For each position i in the doubled array, we check if the current element x has appeared before
• If yes, calculate the distance i - left[x] and update d[i] if this distance is smaller
• Update the hash table to record this position as the new "last seen" position for element x

Step 3: Right-to-left scan to find next occurrences

for i in range(m - 1, -1, -1):
    x = nums[i % n]
    if x in right:
        d[i] = min(d[i], right[x] - i)  # Update minimum distance
    right[x] = i  # Update next occurrence position

• Similar to the left scan, but moving from right to left
• For each position, check if the element will appear again to the right
• Calculate distance right[x] - i and keep the minimum

Step 4: Combine results for original positions

for i in range(n):
    d[i] = min(d[i], d[i + n])

• Since each original index appears twice in the doubled array (at positions i and i + n), we take the minimum of both
• This ensures we capture the true minimum distance considering the circular nature

Step 5: Process queries and return results

return [-1 if d[i] >= n else d[i] for i in queries]

• For each query index i, check the minimum distance stored in d[i]
• If the distance is greater than or equal to n, it means no other identical element exists (the closest one is itself after a full circle), so return -1
• Otherwise, return the actual minimum distance

The time complexity is O(n + q) where n is the array length and q is the number of queries. The space complexity is O(n) for the hash tables and distance array.

Example Walkthrough

Let's work through a concrete example with nums = [2, 1, 3, 1, 2] and queries = [0, 3].

Step 1: Double the array

• Original: [2, 1, 3, 1, 2] (indices 0-4)
• Doubled: [2, 1, 3, 1, 2, 2, 1, 3, 1, 2] (indices 0-9)
• Initialize distance array d = [10, 10, 10, 10, 10, 10, 10, 10, 10, 10]

Step 2: Left-to-right scan

• i=0: x=2, left={}, no previous 2, d[0]=10, left={2:0}
• i=1: x=1, left={2:0}, no previous 1, d[1]=10, left={2:0, 1:1}
• i=2: x=3, left={2:0, 1:1}, no previous 3, d[2]=10, left={2:0, 1:1, 3:2}
• i=3: x=1, left={2:0, 1:1, 3:2}, previous 1 at index 1, d[3]=min(10, 3-1)=2, left={2:0, 1:3, 3:2}
• i=4: x=2, left={2:0, 1:3, 3:2}, previous 2 at index 0, d[4]=min(10, 4-0)=4, left={2:4, 1:3, 3:2}
• i=5: x=2, left={2:4, 1:3, 3:2}, previous 2 at index 4, d[5]=min(10, 5-4)=1, left={2:5, 1:3, 3:2}
• i=6: x=1, left={2:5, 1:3, 3:2}, previous 1 at index 3, d[6]=min(10, 6-3)=3, left={2:5, 1:6, 3:2}
• i=7: x=3, left={2:5, 1:6, 3:2}, previous 3 at index 2, d[7]=min(10, 7-2)=5, left={2:5, 1:6, 3:7}
• i=8: x=1, left={2:5, 1:6, 3:7}, previous 1 at index 6, d[8]=min(10, 8-6)=2, left={2:5, 1:8, 3:7}
• i=9: x=2, left={2:5, 1:8, 3:7}, previous 2 at index 5, d[9]=min(10, 9-5)=4, left={2:9, 1:8, 3:7}

After left scan: d = [10, 10, 10, 2, 4, 1, 3, 5, 2, 4]

Step 3: Right-to-left scan

• i=9: x=2, right={}, no next 2, d[9]=4, right={2:9}
• i=8: x=1, right={2:9}, no next 1, d[8]=2, right={2:9, 1:8}
• i=7: x=3, right={2:9, 1:8}, no next 3, d[7]=5, right={2:9, 1:8, 3:7}
• i=6: x=1, right={2:9, 1:8, 3:7}, next 1 at index 8, d[6]=min(3, 8-6)=2, right={2:9, 1:6, 3:7}
• i=5: x=2, right={2:9, 1:6, 3:7}, next 2 at index 9, d[5]=min(1, 9-5)=1, right={2:5, 1:6, 3:7}
• i=4: x=2, right={2:5, 1:6, 3:7}, next 2 at index 5, d[4]=min(4, 5-4)=1, right={2:4, 1:6, 3:7}
• i=3: x=1, right={2:4, 1:6, 3:7}, next 1 at index 6, d[3]=min(2, 6-3)=2, right={2:4, 1:3, 3:7}
• i=2: x=3, right={2:4, 1:3, 3:7}, next 3 at index 7, d[2]=min(10, 7-2)=5, right={2:4, 1:3, 3:2}
• i=1: x=1, right={2:4, 1:3, 3:2}, next 1 at index 3, d[1]=min(10, 3-1)=2, right={2:4, 1:1, 3:2}
• i=0: x=2, right={2:4, 1:1, 3:2}, next 2 at index 4, d[0]=min(10, 4-0)=4, right={2:0, 1:1, 3:2}

After right scan: d = [4, 2, 5, 2, 1, 1, 2, 5, 2, 4]

Step 4: Combine for original positions

• d[0] = min(d[0], d[5]) = min(4, 1) = 1
• d[1] = min(d[1], d[6]) = min(2, 2) = 2
• d[2] = min(d[2], d[7]) = min(5, 5) = 5
• d[3] = min(d[3], d[8]) = min(2, 2) = 2
• d[4] = min(d[4], d[9]) = min(1, 4) = 1

Final d for original indices: [1, 2, 5, 2, 1]

Step 5: Process queries

• Query 0: d[0] = 1, which is < 5, so answer is 1 (distance from index 0 to index 4)
• Query 3: d[3] = 2, which is < 5, so answer is 2 (distance from index 3 to index 1)

Result: [1, 2]

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums.

The algorithm performs the following operations:

• First loop: iterates through 2n elements (from 0 to m-1 where m = 2n), performing constant time operations for each element - O(n)
• Second loop: iterates through 2n elements in reverse (from m-1 to 0), performing constant time operations for each element - O(n)
• Third loop: iterates through n elements to update d[i] with the minimum of d[i] and d[i+n] - O(n)
• Final list comprehension: iterates through len(queries) elements, which in the worst case could be O(n), performing constant time operations for each query

Total time complexity: O(n) + O(n) + O(n) + O(n) = O(n)

The space complexity is O(n), where n is the length of the array nums.

The algorithm uses:

• Array d of size 2n - O(n)
• Dictionary left which stores at most n unique elements from nums - O(n)
• Dictionary right which stores at most n unique elements from nums - O(n)
• Output list of size equal to the number of queries, which could be at most O(n)

Total space complexity: O(n) + O(n) + O(n) + O(n) = O(n)

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

Common Pitfalls

1. Confusing Array Doubling with Simple Concatenation

A common mistake is thinking that doubling the array means simply concatenating it once (nums + nums). However, the algorithm requires treating indices beyond n with modulo operations (i % n) to properly simulate the circular behavior.

Incorrect approach:

# Wrong: This creates an actual doubled array in memory
doubled_array = nums + nums
for i in range(len(doubled_array)):
    current_value = doubled_array[i]  # Direct access without modulo

Correct approach:

# Right: Use modulo to simulate doubling without extra space
for i in range(2 * n):
    current_value = nums[i % n]  # Modulo ensures circular access

2. Incorrect Distance Calculation in Circular Array

Many developers forget that in a circular array, the minimum distance between two positions isn't always the direct difference. You must consider both clockwise and counterclockwise distances.

Pitfall example:

# Wrong: Only considering forward distance
distance = (j - i) % n  # This misses the backward circular distance

Solution: The algorithm handles this by scanning in both directions (left-to-right and right-to-left) and taking the minimum of both distances.

3. Not Handling the "No Other Occurrence" Case Properly

A subtle bug occurs when checking if another occurrence exists. Simply checking if min_distances[i] == doubled_length might miss edge cases.

Pitfall:

# Potentially problematic
if min_distances[i] == doubled_length:
    return -1

Better approach:

# More robust check
if min_distances[i] >= n:  # Any distance >= n means no valid other occurrence
    return -1

4. Forgetting to Merge Results from Both Halves

The most critical step that's often overlooked is merging the minimum distances from positions i and i + n:

Missing step (common mistake):

# Wrong: Directly using distances without merging
return [-1 if min_distances[i] >= n else min_distances[i] for i in queries]

Correct implementation:

# Essential merging step
for i in range(n):
    min_distances[i] = min(min_distances[i], min_distances[i + n])

# Then process queries
return [-1 if min_distances[i] >= n else min_distances[i] for i in queries]

5. Hash Table Key Overwriting Issues

When updating the hash tables (last_position_left and next_position_right), developers might incorrectly handle multiple occurrences of the same value.

Potential issue: Not realizing that the hash table should always store the most recent (for left scan) or nearest upcoming (for right scan) position, which naturally happens with direct assignment but could be misunderstood if trying to store all positions.

Key insight: The algorithm correctly overwrites previous positions because we only need the closest occurrence in each direction, not all occurrences.