LeetCode #2808 — MEDIUM

Minimum Seconds to Equalize a Circular Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed array nums containing n integers.

At each second, you perform the following operation on the array:

For every index i in the range [0, n - 1], replace nums[i] with either nums[i], nums[(i - 1 + n) % n], or nums[(i + 1) % n].

Note that all the elements get replaced simultaneously.

Return the minimum number of seconds needed to make all elements in the array nums equal.

Example 1:

Input: nums = [1,2,1,2]
Output: 1
Explanation: We can equalize the array in 1 second in the following way:
- At 1^st second, replace values at each index with [nums[3],nums[1],nums[3],nums[3]]. After replacement, nums = [2,2,2,2].
It can be proven that 1 second is the minimum amount of seconds needed for equalizing the array.

Example 2:

Input: nums = [2,1,3,3,2]
Output: 2
Explanation: We can equalize the array in 2 seconds in the following way:
- At 1^st second, replace values at each index with [nums[0],nums[2],nums[2],nums[2],nums[3]]. After replacement, nums = [2,3,3,3,3].
- At 2^nd second, replace values at each index with [nums[1],nums[1],nums[2],nums[3],nums[4]]. After replacement, nums = [3,3,3,3,3].
It can be proven that 2 seconds is the minimum amount of seconds needed for equalizing the array.

Example 3:

Input: nums = [5,5,5,5]
Output: 0
Explanation: We don't need to perform any operations as all elements in the initial array are the same.

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums containing n integers. At each second, you perform the following operation on the array: For every index i in the range [0, n - 1], replace nums[i] with either nums[i], nums[(i - 1 + n) % n], or nums[(i + 1) % n]. Note that all the elements get replaced simultaneously. Return the minimum number of seconds needed to make all elements in the array nums equal.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,2,1,2]

Example 2

[2,1,3,3,2]

Example 3

[5,5,5,5]

Step 02

Core Insight

What unlocks the optimal approach

For every possible x - the final value of the array, calculate the number of seconds needed to make all elements equal to x.
Notice that if you take two consecutive occurrences (i, j) of x, then the number of operations to make segment [i + 1, j - 1] equal to x is floor((j - i) / 2)

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2808: Minimum Seconds to Equalize a Circular Array
class Solution {
    public int minimumSeconds(List<Integer> nums) {
        Map<Integer, List<Integer>> d = new HashMap<>();
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            d.computeIfAbsent(nums.get(i), k -> new ArrayList<>()).add(i);
        }
        int ans = 1 << 30;
        for (List<Integer> idx : d.values()) {
            int m = idx.size();
            int t = idx.get(0) + n - idx.get(m - 1);
            for (int i = 1; i < m; ++i) {
                t = Math.max(t, idx.get(i) - idx.get(i - 1));
            }
            ans = Math.min(ans, t / 2);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2808: Minimum Seconds to Equalize a Circular Array
func minimumSeconds(nums []int) int {
	d := map[int][]int{}
	for i, x := range nums {
		d[x] = append(d[x], i)
	}
	ans := 1 << 30
	n := len(nums)
	for _, idx := range d {
		m := len(idx)
		t := idx[0] + n - idx[m-1]
		for i := 1; i < m; i++ {
			t = max(t, idx[i]-idx[i-1])
		}
		ans = min(ans, t/2)
	}
	return ans
}

# Accepted solution for LeetCode #2808: Minimum Seconds to Equalize a Circular Array
class Solution:
    def minimumSeconds(self, nums: List[int]) -> int:
        d = defaultdict(list)
        for i, x in enumerate(nums):
            d[x].append(i)
        ans = inf
        n = len(nums)
        for idx in d.values():
            t = idx[0] + n - idx[-1]
            for i, j in pairwise(idx):
                t = max(t, j - i)
            ans = min(ans, t // 2)
        return ans

// Accepted solution for LeetCode #2808: Minimum Seconds to Equalize a Circular Array
/**
 * [2808] Minimum Seconds to Equalize a Circular Array
 */
pub struct Solution {}

// submission codes start here

use std::collections::HashMap;

impl Solution {
    pub fn minimum_seconds(nums: Vec<i32>) -> i32 {
        let mut map = HashMap::new();

        for (index, num) in nums.iter().enumerate() {
            let entry = map.entry(*num).or_insert(Vec::new());
            entry.push(index);
        }

        let mut result = i32::MAX;

        for (_, pos) in map {
            let mut max_distance = pos[0] + nums.len() - pos.last().unwrap();

            for i in 1..pos.len() {
                max_distance = max_distance.max(pos[i] - pos[i - 1]);
            }

            result = result.min(max_distance as i32 / 2);
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2808() {
        assert_eq!(Solution::minimum_seconds(vec![1, 2, 1, 2]), 1);
        assert_eq!(Solution::minimum_seconds(vec![2, 1, 3, 3, 2]), 2);
    }
}

// Accepted solution for LeetCode #2808: Minimum Seconds to Equalize a Circular Array
function minimumSeconds(nums: number[]): number {
    const d: Map<number, number[]> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        if (!d.has(nums[i])) {
            d.set(nums[i], []);
        }
        d.get(nums[i])!.push(i);
    }
    let ans = 1 << 30;
    for (const [_, idx] of d) {
        const m = idx.length;
        let t = idx[0] + n - idx[m - 1];
        for (let i = 1; i < m; ++i) {
            t = Math.max(t, idx[i] - idx[i - 1]);
        }
        ans = Math.min(ans, t >> 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.