LeetCode #2735 — MEDIUM

Collecting Chocolates

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

The Problem

Problem Statement

You are given a 0-indexed integer array nums of size n representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index i is nums[i]. Each chocolate is of a different type, and initially, the chocolate at the index i is of i^th type.

In one operation, you can do the following with an incurred cost of x:

Simultaneously change the chocolate of i^th type to ((i + 1) mod n)^th type for all chocolates.

Return the minimum cost to collect chocolates of all types, given that you can perform as many operations as you would like.

Example 1:

Input: nums = [20,1,15], x = 5
Output: 13
Explanation: Initially, the chocolate types are [0,1,2]. We will buy the 1^st type of chocolate at a cost of 1.
Now, we will perform the operation at a cost of 5, and the types of chocolates will become [1,2,0]. We will buy the 2^ndtype of chocolate at a cost of 1.
Now, we will again perform the operation at a cost of 5, and the chocolate types will become [2,0,1]. We will buy the 0^thtype of chocolate at a cost of 1. 
Thus, the total cost will become (1 + 5 + 1 + 5 + 1) = 13. We can prove that this is optimal.

Example 2:

Input: nums = [1,2,3], x = 4
Output: 6
Explanation: We will collect all three types of chocolates at their own price without performing any operations. Therefore, the total cost is 1 + 2 + 3 = 6.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁹
1 <= x <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums of size n representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index i is nums[i]. Each chocolate is of a different type, and initially, the chocolate at the index i is of ith type. In one operation, you can do the following with an incurred cost of x: Simultaneously change the chocolate of ith type to ((i + 1) mod n)th type for all chocolates. Return the minimum cost to collect chocolates of all types, given that you can perform as many operations as you would like.

Baseline thinking

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

Pattern signal: Array

Example 1

[20,1,15]
5

Example 2

[1,2,3]
4

Step 02

Core Insight

What unlocks the optimal approach

How many maximum rotations will be needed?
The array will be rotated for a max of N times, so try all possibilities as N = 1000.

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 #2735: Collecting Chocolates
class Solution {
    public long minCost(int[] nums, int x) {
        int n = nums.length;
        int[][] f = new int[n][n];
        for (int i = 0; i < n; ++i) {
            f[i][0] = nums[i];
            for (int j = 1; j < n; ++j) {
                f[i][j] = Math.min(f[i][j - 1], nums[(i - j + n) % n]);
            }
        }
        long ans = 1L << 60;
        for (int j = 0; j < n; ++j) {
            long cost = 1L * x * j;
            for (int i = 0; i < n; ++i) {
                cost += f[i][j];
            }
            ans = Math.min(ans, cost);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2735: Collecting Chocolates
func minCost(nums []int, x int) int64 {
	n := len(nums)
	f := make([][]int, n)
	for i, v := range nums {
		f[i] = make([]int, n)
		f[i][0] = v
		for j := 1; j < n; j++ {
			f[i][j] = min(f[i][j-1], nums[(i-j+n)%n])
		}
	}
	ans := 1 << 60
	for j := 0; j < n; j++ {
		cost := x * j
		for i := 0; i < n; i++ {
			cost += f[i][j]
		}
		ans = min(ans, cost)
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2735: Collecting Chocolates
class Solution:
    def minCost(self, nums: List[int], x: int) -> int:
        n = len(nums)
        f = [[0] * n for _ in range(n)]
        for i, v in enumerate(nums):
            f[i][0] = v
            for j in range(1, n):
                f[i][j] = min(f[i][j - 1], nums[(i - j) % n])
        return min(sum(f[i][j] for i in range(n)) + x * j for j in range(n))

// Accepted solution for LeetCode #2735: Collecting Chocolates
impl Solution {
    pub fn min_cost(nums: Vec<i32>, x: i32) -> i64 {
        let n = nums.len();
        let mut f = vec![vec![0; n]; n];
        for i in 0..n {
            f[i][0] = nums[i];
            for j in 1..n {
                f[i][j] = f[i][j - 1].min(nums[(i - j + n) % n]);
            }
        }
        let mut ans = i64::MAX;
        for j in 0..n {
            let mut cost = (x as i64) * (j as i64);
            for i in 0..n {
                cost += f[i][j] as i64;
            }
            ans = ans.min(cost);
        }
        ans
    }
}

// Accepted solution for LeetCode #2735: Collecting Chocolates
function minCost(nums: number[], x: number): number {
    const n = nums.length;
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = 0; i < n; ++i) {
        f[i][0] = nums[i];
        for (let j = 1; j < n; ++j) {
            f[i][j] = Math.min(f[i][j - 1], nums[(i - j + n) % n]);
        }
    }
    let ans = Infinity;
    for (let j = 0; j < n; ++j) {
        let cost = x * j;
        for (let i = 0; i < n; ++i) {
            cost += f[i][j];
        }
        ans = Math.min(ans, cost);
    }
    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^2)

Space

O(n^2)

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.