LeetCode #2869 — EASY

Minimum Operations to Collect Elements

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given an array nums of positive integers and an integer k.

In one operation, you can remove the last element of the array and add it to your collection.

Return the minimum number of operations needed to collect elements 1, 2, ..., k.

Example 1:

Input: nums = [3,1,5,4,2], k = 2
Output: 4
Explanation: After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4.

Example 2:

Input: nums = [3,1,5,4,2], k = 5
Output: 5
Explanation: After 5 operations, we collect elements 2, 4, 5, 1, and 3, in this order. Our collection contains elements 1 through 5. Hence, the answer is 5.

Example 3:

Input: nums = [3,2,5,3,1], k = 3
Output: 4
Explanation: After 4 operations, we collect elements 1, 3, 5, and 2, in this order. Our collection contains elements 1 through 3. Hence, the answer is 4.

Constraints:

1 <= nums.length <= 50
1 <= nums[i] <= nums.length
1 <= k <= nums.length
The input is generated such that you can collect elements 1, 2, ..., k.

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of positive integers and an integer k. In one operation, you can remove the last element of the array and add it to your collection. Return the minimum number of operations needed to collect elements 1, 2, ..., k.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

[3,1,5,4,2]
2

Example 2

[3,1,5,4,2]
5

Example 3

[3,2,5,3,1]
3

Core Insight

What unlocks the optimal approach

Use an occurrence array.
Iterate over the elements in reverse order.
If the current element <code>nums[i]</code> is not marked in the occurrence array and <code>nums[i] <= k</code>, mark <code>nums[i]</code>.
Keep track of how many integers you have marked.
Return the current index as soon as the number of marked integers becomes equal to <code>k</code>.

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 #2869: Minimum Operations to Collect Elements
class Solution {
    public int minOperations(List<Integer> nums, int k) {
        boolean[] isAdded = new boolean[k];
        int n = nums.size();
        int count = 0;
        for (int i = n - 1;; i--) {
            if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
                continue;
            }
            isAdded[nums.get(i) - 1] = true;
            count++;
            if (count == k) {
                return n - i;
            }
        }
    }
}

// Accepted solution for LeetCode #2869: Minimum Operations to Collect Elements
func minOperations(nums []int, k int) int {
	isAdded := make([]bool, k)
	count := 0
	n := len(nums)
	for i := n - 1; ; i-- {
		if nums[i] > k || isAdded[nums[i]-1] {
			continue
		}
		isAdded[nums[i]-1] = true
		count++
		if count == k {
			return n - i
		}
	}
}

# Accepted solution for LeetCode #2869: Minimum Operations to Collect Elements
class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        is_added = [False] * k
        count = 0
        n = len(nums)
        for i in range(n - 1, -1, -1):
            if nums[i] > k or is_added[nums[i] - 1]:
                continue
            is_added[nums[i] - 1] = True
            count += 1
            if count == k:
                return n - i

// Accepted solution for LeetCode #2869: Minimum Operations to Collect Elements
fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
    let mut nums = nums;
    let mask = (1u64 << k as u64) - 1;

    let mut ret = 0;
    let mut bits = 0u64;
    while let Some(n) = nums.pop() {
        ret += 1;
        bits |= 1u64 << (n as u64 - 1);
        if (bits & mask) == mask {
            break;
        }
    }

    ret
}

fn main() {
    let nums = vec![3, 1, 5, 4, 2];
    let ret = min_operations(nums, 2);
    println!("ret={ret}");
}

#[test]
fn test_min_operations() {
    {
        let nums = vec![3, 1, 5, 4, 2];
        let ret = min_operations(nums, 2);
        assert_eq!(ret, 4);
    }
    {
        let nums = vec![3, 1, 5, 4, 2];
        let ret = min_operations(nums, 5);
        assert_eq!(ret, 5);
    }
    {
        let nums = vec![3, 2, 5, 3, 1];
        let ret = min_operations(nums, 3);
        assert_eq!(ret, 4);
    }
    {
        let nums = vec![3,28,33,26,34,20,27,5,21,23,4,21,37,35,32,15,14,1,7,2,9,6,38,17,30,18,16,13,24,29,12,14,8,36,11,31,25,22,10,19];
        let ret = min_operations(nums, 38);
        assert_eq!(ret, 40);
    }
}

// Accepted solution for LeetCode #2869: Minimum Operations to Collect Elements
function minOperations(nums: number[], k: number): number {
    const n = nums.length;
    const isAdded = Array(k).fill(false);
    let count = 0;
    for (let i = n - 1; ; --i) {
        if (nums[i] > k || isAdded[nums[i] - 1]) {
            continue;
        }
        isAdded[nums[i] - 1] = true;
        ++count;
        if (count === k) {
            return n - i;
        }
    }
}

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(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.