Minimum Seconds to Equalize a Circular Array - Problem
Minimum Seconds to Equalize a Circular Array
Imagine you have a circular array where elements can "spread" their values to adjacent positions simultaneously. At each second, every element in the array can choose to keep its current value or adopt the value from either of its neighbors (left or right, wrapping around circularly).
🎯 Your Goal: Find the minimum number of seconds needed for all elements to become equal.
The Process:
• At each second, all elements update simultaneously
• Each element can choose: its current value, left neighbor's value, or right neighbor's value
• The array is
• Values "spread" optimally to minimize equalization time
Example:
Imagine you have a circular array where elements can "spread" their values to adjacent positions simultaneously. At each second, every element in the array can choose to keep its current value or adopt the value from either of its neighbors (left or right, wrapping around circularly).
🎯 Your Goal: Find the minimum number of seconds needed for all elements to become equal.
The Process:
• At each second, all elements update simultaneously
• Each element can choose: its current value, left neighbor's value, or right neighbor's value
• The array is
circular - position 0's left neighbor is position n-1• Values "spread" optimally to minimize equalization time
Example:
[1,2,1,2] → After 1 second, all elements can become 1 or 2 → Answer: 1 Input & Output
example_1.py — Basic Case
$
Input:
nums = [1,2,1,2]
›
Output:
1
💡 Note:
Value 1 is at positions [0,2] and value 2 is at positions [1,3]. Both have maximum gaps of 2, so spreading takes 2/2 = 1 second.
example_2.py — All Same Elements
$
Input:
nums = [2,2,2,2]
›
Output:
0
💡 Note:
All elements are already equal, so no time is needed.
example_3.py — Single Element Spread
$
Input:
nums = [1,2,2,2,1]
›
Output:
2
💡 Note:
Value 1 at positions [0,4] has gap 1, while value 2 at [1,2,3] has maximum gap 3. Min(1/2, 3/2) = min(0, 1) but since we need ceiling, it's min(1, 2) = 1. Actually, value 1 positions [0,4] have circular gap 1, time=1/2=0.5→1. Value 2 positions [1,2,3] have max gap 3 (from 3 to 1), time=3/2=1.5→2. Answer is min(1,2)=1. Wait, let me recalculate: gap from pos 3 to pos 1 going around is (1+5-3)=3, so max gap for value 2 is 3, time needed is 1 (since it can spread from both 2's). For value 1, gap is 1 (4 to 0 circularly), time needed is 2 (since 1's are far apart). So answer is min(1,2)=1.
Constraints
- 1 ≤ nums.length ≤ 105
- 1 ≤ nums[i] ≤ 109
- Array elements can be any positive integers
- Circular array: position 0 is adjacent to position n-1
Visualization
Tap to expand
Understanding the Visualization
1
Identify Sources
Each unique value has specific source positions in the circular array
2
Calculate Gaps
Find the largest distance between consecutive same values (considering wrap-around)
3
Determine Spread Time
Each gap can be filled from both ends, so time = maxGap / 2
4
Find Optimal
Choose the value with minimum spreading time
Key Takeaway
🎯 Key Insight: The optimal strategy recognizes that values spread bidirectionally in circular arrays, making the time proportional to the largest gap divided by 2.
💡
Explanation
AI Ready
💡 Suggestion
Tab
to accept
Esc
to dismiss
// Output will appear here after running code