Minimum Operations to Collect Elements solution and explanation
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The problem
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 <= 501 <= nums[i] <= nums.length1 <= k <= nums.length- The input is generated such that you can collect elements
1, 2, ..., k.
Approach overview
Problem Overview: You are given an integer array nums and an integer k. Starting from the end of the array, you remove elements one by one. The goal is to collect every value from 1 to k at least once. The task is to return the minimum number of operations required to achieve this.
Approach 1: Reverse Collection (O(n) time, O(k) space)
The most direct strategy is to scan the array from right to left while collecting values between 1 and k. Maintain a set (or hash structure) that stores which required elements have been seen so far. Each step represents one operation since you are effectively removing an element from the end. When you encounter a value <= k, add it to the set. As soon as the set size becomes k, you have collected all required elements and can return the number of processed elements.
This approach works because removing elements from the end is equivalent to iterating the array in reverse. Hash lookups and insertions are constant time, so the overall complexity remains linear. The method relies on basic array traversal combined with fast membership checks using a hash table.
Approach 2: Decremental Loop Through the Array (O(n) time, O(k) space)
Another implementation tracks how many required elements are still missing instead of storing them all explicitly. Start with a counter remaining = k and a boolean array (or bitmask) to mark which values from 1..k have been collected. Traverse the array from the end and check each value. If the number is within the required range and hasn't been seen before, mark it and decrement remaining.
Once remaining reaches zero, you have collected all elements and can return the number of operations performed. This version replaces a dynamic set with a fixed-size tracking structure, which can also be implemented using bit manipulation for very compact state tracking.
The logic remains the same as the previous approach: iterate backward, track required values, and stop once every number from 1 to k appears. Both solutions run in linear time and scan the array only once.
Recommended for interviews: The reverse traversal with a hash set is the clearest explanation and easiest to implement under time pressure. It demonstrates understanding of array traversal and constant‑time lookups. The decremental tracking version is slightly more optimized and shows stronger control over state management, but interviewers usually accept either once the reverse‑scan insight is explained.
Approach 1: Reverse Collection
This approach processes the array from the end to the start and uses a set to track the unique elements collected. We iterate from the last element backward, updating the set, until we've collected all elements from 1 through k.
def min_operations(nums, k):
collected = set()
ops = 0
for num in reversed(nums):
ops += 1
if num <= k:
collected.add(num)
if len(collected) == k:
return ops
return ops
# Example usage
nums = [3, 1, 5, 4, 2]
k = 2
print(min_operations(nums, k)) # Output: 4import java.util.HashSet;
import java.util.Set;
public class Solution {
public static int minOperations(int[] nums, int k) {
Set<Integer> collected = new HashSet<>();
int ops = 0;
for (int i = nums.length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
collected.add(nums[i]);
if (collected.size() == k) {
return ops;
}
}
}
return ops;
}
public static void main(String[] args) {
int[] nums = {3, 1, 5, 4, 2};
int k = 2;
System.out.println(minOperations(nums, k)); // Output: 4
}
}#include <vector>
#include <unordered_set>
#include <iostream>
int minOperations(std::vector<int>& nums, int k) {
std::unordered_set<int> collected;
int n = nums.size();
for (int i = n - 1, ops = 0; i >= 0; --i, ++ops) {
if (nums[i] <= k) {
collected.insert(nums[i]);
if (collected.size() == static_cast<size_t>(k)) {
return ops + 1;
}
}
}
return n;
}
int main() {
std::vector<int> nums = {3, 1, 5, 4, 2};
int k = 2;
std::cout << minOperations(nums, k) << std::endl;
return 0;
}function minOperations(nums, k) {
const collected = new Set();
let ops = 0;
for (let i = nums.length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
collected.add(nums[i]);
if (collected.size === k) {
return ops;
}
}
}
return ops;
}
const nums = [3, 1, 5, 4, 2];
const k = 2;
console.log(minOperations(nums, k)); // Output: 4#include <stdio.h>
#include <stdbool.h>
#include <string.h>
// Function to find minimum operations
int minOperations(int* nums, int numsSize, int k) {
bool collected[k + 1];
memset(collected, false, sizeof(collected));
int collectCount = 0;
int ops = 0;
for (int i = numsSize - 1; i >= 0; i--) {
int num = nums[i];
if (num <= k && !collected[num]) {
collected[num] = true;
collectCount++;
if (collectCount == k) {
return ops + 1;
}
}
ops++;
}
return ops;
}
int main() {
int nums[] = {3, 1, 5, 4, 2};
int k = 2;
printf("%d\n", minOperations(nums, sizeof(nums) / sizeof(nums[0]), k));
return 0;
}using System;
using System.Collections.Generic;
class Solution {
public int MinOperations(int[] nums, int k) {
HashSet<int> collected = new HashSet<int>();
int ops = 0;
for (int i = nums.Length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
collected.Add(nums[i]);
if (collected.Count == k) {
return ops;
}
}
}
return ops;
}
static void Main(string[] args) {
int[] nums = { 3, 1, 5, 4, 2 };
int k = 2;
Console.WriteLine((new Solution()).MinOperations(nums, k)); // Output: 4
}
}The Python solution uses a set to track collected elements. It iterates over the array in reverse and stops when all elements from 1 to k are collected, returning the total operations.
Time Complexity: O(n), where n is the length of nums.
Space Complexity: O(k), needed for the set collecting elements up to k.
Approach 2: Decremental Loop Through the Array
Instead of starting at the end and using a set, consider decrementing k directly and track how many of these elements ('1' to current 'k') have been collected as you iterate from the end.
def min_operations(nums, k):
ops = 0
count = k
for num in reversed(nums):
ops += 1
if num <= k:
count -= 1
if count == 0:
return ops
return ops
# Example usage
nums = [3, 1, 5, 4, 2]
k = 2
print(min_operations(nums, k)) # Output: 4public class Solution2 {
public static int minOperations(int[] nums, int k) {
int ops = 0;
for (int i = nums.length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
k--;
if (k == 0) {
return ops;
}
}
}
return ops;
}
public static void main(String[] args) {
int[] nums = {3, 1, 5, 4, 2};
int k = 2;
System.out.println(minOperations(nums, k)); // Output: 4
}
}#include <vector>
#include <iostream>
int minOperations(std::vector<int>& nums, int k) {
int ops = 0;
int count = k;
for (int i = nums.size() - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
count--;
if (count == 0) {
return ops;
}
}
}
return ops;
}
int main() {
std::vector<int> nums = {3, 1, 5, 4, 2};
int k = 2;
std::cout << minOperations(nums, k) << std::endl;
return 0;
}function minOperations(nums, k) {
let ops = 0;
for (let i = nums.length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
k--;
if (k === 0) {
return ops;
}
}
}
return ops;
}
const nums = [3, 1, 5, 4, 2];
const k = 2;
console.log(minOperations(nums, k)); // Output: 4#include <stdio.h>
// Function to find minimum operations
int minOperations(int* nums, int numsSize, int k) {
int ops = 0;
for (int i = numsSize - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
k--;
if (k == 0) {
return ops;
}
}
}
return ops;
}
int main() {
int nums[] = {3, 1, 5, 4, 2};
int k = 2;
printf("%d\n", minOperations(nums, sizeof(nums) / sizeof(nums[0]), k));
return 0;
}using System;
class Solution2 {
public int MinOperations(int[] nums, int k) {
int ops = 0;
for (int i = nums.Length - 1; i >= 0; i--) {
ops++;
if (nums[i] <= k) {
k--;
if (k == 0) {
return ops;
}
}
}
return ops;
}
static void Main(string[] args) {
int[] nums = { 3, 1, 5, 4, 2 };
int k = 2;
Console.WriteLine((new Solution2()).MinOperations(nums, k)); // Output: 4
}
}The Python version iterates backwards, decrementing a count of k as needed. It finishes when all numbers up to k are collected.
Time Complexity: O(n), traversing the nums array.
Space Complexity: O(1), no additional space is utilized.
Approach 3: Traverse in Reverse Order
We can traverse the array in reverse order. For each element encountered during the traversal that is less than or equal to k and has not been added to the set yet, we add it to the set until the set contains elements from 1 to k.
The time complexity is O(n), where n is the length of the array nums. The space complexity is O(k).
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;
}
}
}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
}
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Reverse Collection | Time Complexity: O(n), where n is the length of nums. |
| Decremental Loop Through the Array | Time Complexity: O(n), traversing the nums array. |
| Traverse in Reverse Order | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Reverse Collection with Hash Set | O(n) | O(k) | General case; easiest to implement and explain in interviews |
| Decremental Loop with Boolean/Bit Tracking | O(n) | O(k) | When you want tighter control over memory or prefer array/bit tracking instead of a hash set |
Frequently asked questions
Is Minimum Operations to Collect Elements easy or hard?
Minimum Operations to Collect Elements is classified as an Easy problem. The key idea is recognizing that removing elements from the end is equivalent to reverse iteration and tracking required values with a simple data structure.
Minimum Operations to Collect Elements Python/Java solution
In Python or Java, iterate the array from the last index to the first while maintaining a set or boolean array for values 1..k. Increment an operation counter at each step and add valid elements to the tracking structure. When all k numbers are collected, return the counter. The complexity remains O(n) time and O(k) space.
How to solve Minimum Operations to Collect Elements in O(n)?
Iterate through the array in reverse and keep a set or boolean array of collected numbers between 1 and k. Every step counts as one operation. When a new required value appears, record it and decrease the remaining count. Once all k numbers are found, return the number of processed elements.
What is the best approach for Minimum Operations to Collect Elements?
The optimal approach scans the array from right to left while tracking values from 1 to k that have already been collected. A hash set or boolean array stores which numbers have appeared. Once all k values are seen, the current step count is the answer. This solution runs in O(n) time and O(k) space.
Is Minimum Operations to Collect Elements asked at Google/Amazon/Meta?
Problems involving reverse traversal, hash sets, and element tracking frequently appear in interviews at companies like Amazon, Google, and Meta. While this exact problem may vary, the underlying pattern of scanning from the end and tracking required values is a common interview technique.
What data structure is used in Minimum Operations to Collect Elements?
The typical implementation uses a hash set to store which numbers from 1 to k have been collected so far. Some solutions replace the set with a boolean array or a bitmask for faster constant-time checks and slightly lower overhead.
What is the time complexity of Minimum Operations to Collect Elements?
The optimal solution runs in O(n) time because the array is traversed once from the end. Each element is processed with constant-time checks using a hash set or boolean array. The auxiliary space complexity is O(k) for tracking collected values from 1 to k.