What is the time complexity of Minimum Operations to Make Array Values Equal to K?

The optimal solution has a time complexity of O(n*m*l).

What data structures are used to solve Minimum Operations to Make Array Values Equal to K?

This problem uses Array, Hash Table. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Operations to Make Array Values Equal to K?

The problem involves iteratively reducing the maximum value in the array towards `k` by selecting valid integers. The key is to efficiently find these valid integers in each iteration. The selection of the valid integer `h` should be done strategically. Choosing the largest valid `h` less than the current maximum value minimizes the number of operations.

Which companies ask LeetCode 3375. Minimum Operations to Make Array Values Equal to K in interviews?

This problem is frequently asked at Lowe's.

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Minimum Operations to Make Array Values Equal to K - LeetCode 3375 Solution

Minimum Operations to Make Array Values Equal to K - Complete Solution Guide

Minimum Operations to Make Array Values Equal to K is LeetCode problem 3375, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array nums and an integer k . An integer h is called valid if all values in the array that are strictly greater than h are identical . For example, if nums = [10, 8, 10, 8] , a valid integer is h = 9 because all nums[i] > 9 are equal to 10, but 5 is not a valid integer. You are allowed to perform the following operation on nums : Select an integer h that is valid for the current values in nums . For each index i where nums[i] > h , set nums[i] to h . Return the minimum n

Detailed Explanation

The problem asks you to find the minimum number of operations needed to make all elements in an integer array equal to a target value `k`. In each operation, you can select a *valid* integer `h`. A valid `h` is an integer such that all elements in the array strictly greater than `h` are identical. After selecting `h`, all elements greater than `h` are set to `h`. The goal is to reach a state where all elements are equal to `k`. If it's impossible to reach this state, return -1.

Solution Approach

The provided solutions use an iterative approach. They repeatedly identify the largest valid integer `h` and update the array elements greater than `h` to `h`. This process continues until either all elements are equal to `k` or no valid `h` can be found, in which case -1 is returned. The solutions differ slightly in how they find the largest valid `h` and handle edge cases.

Step-by-Step Algorithm

Check if `k` is present in the array or if any elements are larger than `k`. If neither is true, return -1 (as it's impossible).
Iterate until all elements are equal to `k`:
Find the largest valid integer `h` (an integer such that all elements > `h` are identical).
If no valid `h` is found, return -1.
Update all elements greater than `h` to `h`.
Increment the operation counter.

Key Insights

The problem involves iteratively reducing the maximum value in the array towards `k` by selecting valid integers. The key is to efficiently find these valid integers in each iteration.
The selection of the valid integer `h` should be done strategically. Choosing the largest valid `h` less than the current maximum value minimizes the number of operations.
Edge cases need to be handled carefully: situations where `k` is not present in the array, or when no valid `h` can be found at a given step. These scenarios indicate the impossibility of reaching the target state.

Complexity Analysis

Time Complexity: O(n*m*l)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Lowe's.

Minimum Operations to Make Array Values Equal to K | LeetCode 3375

Easy

Array Hash Table

Time: O(n*m*l)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minOperations(int[] nums, int k) {
        int operations = 0;
        while (true) {
            boolean allEqual = true;
            for (int num : nums) {
                if (num != k) {
                    allEqual = false;
                    break;
                }
            }
            if (allEqual) {
                return operations;
            }

            int maxVal = -1;
            for (int num : nums) {
                maxVal = Math.max(maxVal, num);
            }

            if (k > maxVal) {
                return -1;
            }
            
            int validH = -1;
            for(int h = maxVal - 1; h >= k; h--){
                boolean isValid = true;
                int greaterVal = -1;
                boolean foundGreater = false;
                for(int num : nums){
                    if(num > h){
                        if(!foundGreater){
                            greaterVal = num;
                            foundGreater = true;
                        } else if(num != greaterVal){
                            isValid = false;
                            break;
                        }
                    }
                }
                if(isValid){
                    validH = h;
                    break;
                }
            }

            if (validH == -1) {
                if(k == maxVal){
                    return -1;
                }
                
                
                int count = 0;
                for(int i = 0; i < nums.length; i++){
                    if(nums[i] > k){
                        nums[i] = k;
                        count++;
                    }
                }
                 if (count > 0) {
                    return operations + count;
                 }else{
                    return -1;
                 }
            } else {
                for (int i = 0; i < nums.length; i++) {
                    if (nums[i] > validH) {
                        nums[i] = validH;
                    }
                }
                operations++;
            }
        }
    }
}

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Problem Info

Description

Problem 3375: Minimum Operations to Make Array Values Equal to K - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Lowe's

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3375

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Check if `k` is present in the array or if any elements are larger than `k`. If neither is true, return -1 (as it's impossible).
2
Iterate until all elements are equal to `k`:
3
Find the largest valid integer `h` (an integer such that all elements > `h` are identical).
4
If no valid `h` is found, return -1.
5
Update all elements greater than `h` to `h`.
6
Increment the operation counter.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m*l)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates until the array is all k. Each iteration involves finding a valid `h` (which is O(n*m) in the worst case because it involves checking for each possible h up to the maximum value. Then it updates the array (O(n)).The Java and C solutions have a slightly better time complexity because the loop to find a valid h starts at the maximum value and goes downwards, terminating early once a valid h is found. The provided python solution has redundant while loops which do not improve the outcome and only add complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Greedy Approach with Sorted Array

Sort the array. Iterate from the largest element down to `k`. For each element greater than `k`, check if all larger elements are equal. If so, update them to the current value. This approach is more efficient but requires sorting.

Time: O(n log n) for sorting + O(n) for iteration, dominated by sorting. Overall O(n log n).Space: O(1) or O(n) depending on the sorting algorithm used.

Trade-offs: This approach is more efficient than the provided solution but requires extra space for some sorting algorithms. It's preferable if the array size is relatively large.

✨

Key Insights

1The problem involves iteratively reducing the maximum value in the array towards `k` by selecting valid integers. The key is to efficiently find these valid integers in each iteration.
2The selection of the valid integer `h` should be done strategically. Choosing the largest valid `h` less than the current maximum value minimizes the number of operations.
3Edge cases need to be handled carefully: situations where `k` is not present in the array, or when no valid `h` can be found at a given step. These scenarios indicate the impossibility of reaching the target state.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Case 1: `k` is not in `nums` and there are numbers greater than `k`. In this case, it's impossible to reach the target state, and -1 should be returned. All solutions correctly handle this.
Case 2: No valid `h` exists at a certain iteration. This signifies the impossibility of reaching the target state and requires returning -1. The solutions handle this by checking if the set of valid `h` is empty.

⚠️

Common Mistakes to Avoid

✗Incorrectly identifying valid integers `h`. Failing to handle cases where no valid `h` exists. Many submissions might incorrectly loop through all possible values from 1 to 100 when they should be checking only numbers between k and the current maximum value.

✗Inefficiently searching for valid `h`. Not optimizing the search for the largest valid `h` leading to unnecessary iterations. An optimized approach would be to start searching downwards from the maximum element.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Data normalization or cleaning: Imagine scenarios where you need to bring data points closer to a target value while maintaining consistency within groups of similar values. This is very common in database operations.

🏦Resource allocation: In resource management, you might need to adjust resource levels to a target value, respecting certain constraints or group characteristics, similar to this problem's rules.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Operations to Make Array Values Equal to K"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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