What is the time complexity of Minimum Total Cost to Make Arrays Unequal?

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

What data structures are used to solve Minimum Total Cost to Make Arrays Unequal?

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

What are the key insights for solving Minimum Total Cost to Make Arrays Unequal?

Insight 1: First determine if it is impossible to make nums1 and nums2 unequal, returning -1. A necessary condition is that for no value `v` does its count in both arrays exceed the array size `n`. This can be checked using frequency counts. Insight 2: Calculate the initial cost by summing the indices where `nums1[i] == nums2[i]`. These indices are the ones that require swaps.

Which companies ask LeetCode 2499. Minimum Total Cost to Make Arrays Unequal in interviews?

This problem is frequently asked at Flipkart, razorpay.

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Minimum Total Cost to Make Arrays Unequal - LeetCode 2499 Solution

Q: How to solve Minimum Total Cost to Make Arrays Unequal?

The solution first identifies indices where `nums1[i] == nums2[i]`, as these require swaps. It then calculates the cost of swapping all such indices assuming they can be made unequal. If there is a value which appears too frequently among these indices ('major' value), additional swaps from indices where `nums1[i] != nums2[i]` are required to ensure the condition `nums1[i] != nums2[i]` is satisfied for all `i`. If at any point, the number of occurences of a single value in both `nums1` and `nums2` combined exceeds `n`, it's impossible to achieve the desired result, and -1 is returned.

Minimum Total Cost to Make Arrays Unequal - Complete Solution Guide

Minimum Total Cost to Make Arrays Unequal is LeetCode problem 2499, a Hard 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 two 0-indexed integer arrays nums1 and nums2 , of equal length n . In one operation, you can swap the values of any two indices of nums1 . The cost of this operation is the sum of the indices. Find the minimum total cost of performing the given operation any number of times such that nums1[i] != nums2[i] for all 0 <= i <= n - 1 after performing all the operations. Return the minimum total cost such that nums1 and nums2 satisfy the above condition . In case it is not possible, retur

Detailed Explanation

The problem asks us to find the minimum total cost to make two integer arrays, `nums1` and `nums2`, unequal at every index. An array is considered unequal if `nums1[i] != nums2[i]` for all `i`. We can swap values at any two indices in `nums1`. The cost of a swap is the sum of the indices involved. If it's impossible to make the arrays unequal, return -1. The input arrays are of equal length `n`, with elements ranging from 1 to `n`. The goal is to minimize the sum of indices involved in the swaps.

Solution Approach

The solution first identifies indices where `nums1[i] == nums2[i]`, as these require swaps. It then calculates the cost of swapping all such indices assuming they can be made unequal. If there is a value which appears too frequently among these indices ('major' value), additional swaps from indices where `nums1[i] != nums2[i]` are required to ensure the condition `nums1[i] != nums2[i]` is satisfied for all `i`. If at any point, the number of occurences of a single value in both `nums1` and `nums2` combined exceeds `n`, it's impossible to achieve the desired result, and -1 is returned.

Step-by-Step Algorithm

Step 1: Initialize variables. Store the indices where `nums1[i] == nums2[i]` in a list `swaps_needed_indices`. Also, create a dictionary `same_val_counts` to count the occurrences of each value at these indices.
Step 2: Check if it's impossible to make the arrays unequal. Iterate through all values in `nums1` and `nums2`. If the combined count of any value exceeds `n`, return -1.
Step 3: Calculate the initial cost. Sum all indices in the `swaps_needed_indices` list.
Step 4: Find the major value. Determine the value that appears most frequently in `same_val_counts` and its frequency (`max_freq`). This is the value that's causing the most conflicts.
Step 5: Check if additional swaps are needed. If `2 * max_freq <= num_swaps_needed`, no additional swaps are needed, and the initial cost is the minimum cost.
Step 6: Calculate the number of additional swaps required. Calculate `must_swap_out = 2 * max_freq - num_swaps_needed`. This represents the number of indices containing the major value that need to be swapped out from locations where they're already equal.
Step 7: Perform additional swaps. Iterate through the array and find indices `i` that are not already in `swaps_needed_indices` and for which `nums1[i] != major_val && nums2[i] != major_val`. Add these indices to the cost until `must_swap_out` becomes 0.
Step 8: Return the total cost.

Key Insights

Insight 1: First determine if it is impossible to make nums1 and nums2 unequal, returning -1. A necessary condition is that for no value `v` does its count in both arrays exceed the array size `n`. This can be checked using frequency counts.
Insight 2: Calculate the initial cost by summing the indices where `nums1[i] == nums2[i]`. These indices are the ones that require swaps.
Insight 3: If there exists a 'major' value, which appears more than half the times needed to be swapped, then additional swaps will need to be performed. Efficiently search for indices where swapping will increase the total cost the least.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Greedy, Counting.

Companies

Asked at: Flipkart, razorpay.

Minimum Total Cost to Make Arrays Unequal | LeetCode 2499

Hard

Array Hash Table Greedy Counting

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long minimumTotalCost(int[] nums1, int[] nums2) {
        int n = nums1.length;
        
        List<Integer> swaps_needed_indices = new ArrayList<>();
        Map<Integer, Integer> same_val_counts = new HashMap<>();
        
        Map<Integer, Integer> c1 = new HashMap<>();
        Map<Integer, Integer> c2 = new HashMap<>();
        
        for (int i = 0; i < n; i++) {
            c1.put(nums1[i], c1.getOrDefault(nums1[i], 0) + 1);
            c2.put(nums2[i], c2.getOrDefault(nums2[i], 0) + 1);
            if (nums1[i] == nums2[i]) {
                swaps_needed_indices.add(i);
                same_val_counts.put(nums1[i], same_val_counts.getOrDefault(nums1[i], 0) + 1);
            }
        }

        Set<Integer> allVals = new HashSet<>();
        allVals.addAll(c1.keySet());
        allVals.addAll(c2.keySet());

        for (int val : allVals) {
            int count1 = c1.getOrDefault(val, 0);
            int count2 = c2.getOrDefault(val, 0);
            if (count1 + count2 > n) {
                return -1;
            }
        }

        int num_swaps_needed = swaps_needed_indices.size();
        if (num_swaps_needed == 0) {
            return 0;
        }

        int max_freq = 0;
        int major_val = -1;
        if (!same_val_counts.isEmpty()) {
            Map.Entry<Integer, Integer> maxEntry = null;
            for (Map.Entry<Integer, Integer> entry : same_val_counts.entrySet()) {
                if (maxEntry == null || entry.getValue() > maxEntry.getValue()) {
                    maxEntry = entry;
                }
            }
            major_val = maxEntry.getKey();
            max_freq = maxEntry.getValue();
        }
        
        long cost = 0;
        for (int index : swaps_needed_indices) {
            cost += index;
        }
        
        if (2 * (long)max_freq <= num_swaps_needed) {
            return cost;
        }

        int must_swap_out = 2 * max_freq - num_swaps_needed;
        
        Set<Integer> swaps_needed_set = new HashSet<>(swaps_needed_indices);
        
        for (int i = 0; i < n; i++) {
            if (must_swap_out == 0) {
                break;
            }
            if (!swaps_needed_set.contains(i)) {
                if (nums1[i] != major_val && nums2[i] != major_val) {
                    cost += i;
                    must_swap_out -= 1;
                }
            }
        }
        
        return cost;
    }
}

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Problem Info

Description

Problem 2499: Minimum Total Cost to Make Arrays Unequal - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Counting

Companies

Flipkartrazorpay

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2499

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

8 steps

1
Step 1: Initialize variables. Store the indices where `nums1[i] == nums2[i]` in a list `swaps_needed_indices`. Also, create a dictionary `same_val_counts` to count the occurrences of each value at these indices.
2
Step 2: Check if it's impossible to make the arrays unequal. Iterate through all values in `nums1` and `nums2`. If the combined count of any value exceeds `n`, return -1.
3
Step 3: Calculate the initial cost. Sum all indices in the `swaps_needed_indices` list.
4
Step 4: Find the major value. Determine the value that appears most frequently in `same_val_counts` and its frequency (`max_freq`). This is the value that's causing the most conflicts.
5
Step 5: Check if additional swaps are needed. If `2 * max_freq <= num_swaps_needed`, no additional swaps are needed, and the initial cost is the minimum cost.
6
Step 6: Calculate the number of additional swaps required. Calculate `must_swap_out = 2 * max_freq - num_swaps_needed`. This represents the number of indices containing the major value that need to be swapped out from locations where they're already equal.
7
Step 7: Perform additional swaps. Iterate through the array and find indices `i` that are not already in `swaps_needed_indices` and for which `nums1[i] != major_val && nums2[i] != major_val`. Add these indices to the cost until `must_swap_out` becomes 0.
8
Step 8: Return the total cost.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are the iterations through the input arrays, which are O(n). The hash map operations (insert, get, etc.) take O(1) average time. The space is determined by hash maps which can potentially store O(n) distinct elements from the input arrays.

📖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 Priority Queue

Instead of first calculating the total cost and then adjusting for the 'major' value, we could use a priority queue to greedily select the swaps with the lowest cost while avoiding creating more instances of the 'major' value being equal at the same index. Requires more complex data structure handling.

Time: O(n log n) in the worst case if priority queue operations take log n time.Space: O(n) to store the indices and the priority queue.

Trade-offs: This approach might be more intuitive for some, but the priority queue operations increase the time complexity. The main solution is generally preferred due to its simpler implementation and linear time complexity.

✨

Key Insights

1Insight 1: First determine if it is impossible to make nums1 and nums2 unequal, returning -1. A necessary condition is that for no value `v` does its count in both arrays exceed the array size `n`. This can be checked using frequency counts.
2Insight 2: Calculate the initial cost by summing the indices where `nums1[i] == nums2[i]`. These indices are the ones that require swaps.
3Insight 3: If there exists a 'major' value, which appears more than half the times needed to be swapped, then additional swaps will need to be performed. Efficiently search for indices where swapping will increase the total cost the least.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `nums1` and `nums2` are already unequal at all indices. The algorithm correctly returns 0 since no swaps are needed.
Edge case 2: It's impossible to make the arrays unequal due to a value appearing too frequently. The algorithm correctly returns -1.
Edge case 3: When num_swap_needed is 0. The algorithm will return 0 without processing other steps.
Edge case 4: A situation where no major value is present( `2 * max_freq <= num_swaps_needed`). The algorithm correctly determines that no additional swaps are needed, and the initial cost is the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if it's impossible to make the arrays unequal. Failing to account for this edge case will lead to incorrect results.

✗Common mistake 2: Not handling the 'major' value correctly. The most frequent conflicting element significantly affects the final cost and requires careful consideration.

✗Common mistake 3: Integer overflow. When the indices are large, the sum of the indices can exceed the capacity of a 32-bit integer. Using a 64-bit integer (long long in C++, long in Java, int in Python) is necessary.

✗Common mistake 4: Incorrectly calculating `must_swap_out`. Ensure that `must_swap_out = 2 * max_freq - num_swaps_needed` is calculated correctly to avoid over or under correcting the initial solution.

✅

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?

💼Real-world scenario 1: Resource allocation. Suppose we have two lists representing resource assignments, and we need to re-allocate resources to minimize conflicts. This problem models the scenario where the cost of re-allocation is based on the resource index.

🏦Real-world scenario 2: Scheduling. In a scheduling problem, we might need to adjust the schedule to avoid conflicts. The 'cost' of rescheduling an event can be based on its original position in the schedule, and we want to minimize the total disruption.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

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: November 7, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Total Cost to Make Arrays Unequal"?

For this hard-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.

🤖 Need help with this problem?