Leetcode #2499: Minimum Total Cost to Make Arrays Unequal
In this guide, we solve Leetcode #2499 Minimum Total Cost to Make Arrays Unequal in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Array, Hash Table, Counting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums1 = [1,2,3,4,5], nums2 = [1,2,3,4,5]
Output: 10
Explanation:
One of the ways we can perform the operations is:
- Swap values at indices 0 and 3, incurring cost = 0 + 3 = 3. Now, nums1 = [4,2,3,1,5]
- Swap values at indices 1 and 2, incurring cost = 1 + 2 = 3. Now, nums1 = [4,3,2,1,5].
- Swap values at indices 0 and 4, incurring cost = 0 + 4 = 4. Now, nums1 =[5,3,2,1,4].
We can see that for each index i, nums1[i] != nums2[i]. The cost required here is 10.
Note that there are other ways to swap values, but it can be proven that it is not possible to obtain a cost less than 10.
Python Solution
class Solution:
def minimumTotalCost(self, nums1: List[int], nums2: List[int]) -> int:
ans = same = 0
cnt = Counter()
for i, (a, b) in enumerate(zip(nums1, nums2)):
if a == b:
same += 1
ans += i
cnt[a] += 1
m = lead = 0
for k, v in cnt.items():
if v * 2 > same:
m = v * 2 - same
lead = k
break
for i, (a, b) in enumerate(zip(nums1, nums2)):
if m and a != b and a != lead and b != lead:
ans += i
m -= 1
return -1 if m else ans
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.