LeetCode 3724 - Minimum Operations to Transform Array

Difficulty: 🟡 Medium
Topics: Array, Greedy

Solution

Problem Understanding

This problem asks us to transform an array nums1 of length n into another array nums2 of length n + 1 using the minimum number of operations. The allowed operations are increasing or decreasing any element of nums1 by 1, or appending any element of nums1 to the end of the array. The goal is to make nums1 exactly match nums2 after a sequence of these operations while performing as few operations as possible.

The input arrays represent integer sequences, and the output is the minimal count of the operations needed. The constraints indicate that nums1 can have up to 100,000 elements, and nums2 is always one element longer. Each element is bounded between 1 and 100,000, so we must be careful about potential inefficiencies in algorithms that scale with the maximum element size or require full enumeration of large value ranges.

Edge cases to keep in mind include arrays of length 1, arrays where the optimal strategy is heavily appending versus incrementing/decrementing, and arrays that are already very close in value, where appending may be unnecessary. The problem guarantees valid arrays and positive integers.

Approaches

Brute Force

The brute-force approach would attempt to match each element in nums2 by trying all possible sequences of increment, decrement, or append operations. This would involve recursively considering every element in nums1 and every possible append, increment, or decrement to match nums2. Although this guarantees correctness, it is computationally infeasible because the number of potential sequences grows exponentially with the array size.

Optimal Approach

The key insight is to recognize that the append operation is only needed to introduce elements that already exist in nums1. For all other elements, we can treat them as modifications of existing elements. This reduces the problem to a greedy matching of elements between nums1 and nums2 in sorted order, where the cost to match an element is the absolute difference, and we append elements if they are in nums2 but not yet matched from nums1.

By sorting both arrays and using a two-pointer approach, we can efficiently compute the minimum operations by aligning elements with the smallest possible cost and appending unmatched elements. This ensures that each operation contributes minimally to the total count.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(3^n)	O(n)	Tries all sequences of increment, decrement, append; infeasible for n up to 10^5
Optimal	O(n log n)	O(n)	Sort both arrays, use two pointers and greedy matching, compute absolute differences and append cost

Algorithm Walkthrough

Sort nums1 and nums2 to allow a greedy matching strategy. Sorting ensures we can match the smallest available elements with minimal absolute difference.
Initialize two pointers i for nums1 and j for nums2 and a variable operations to accumulate the total number of operations.
Iterate through nums2 using pointer j. For each element nums2[j], find the closest unmatched element nums1[i]. Increment operations by the absolute difference abs(nums2[j] - nums1[i]).
Advance pointer i when an element of nums1 has been used. If nums2[j] cannot match any remaining nums1[i], it implies an append is necessary. Increment operations by 1 for each such append.
Continue until all elements of nums2 are accounted for. At the end, operations will reflect the minimum number of steps required.

Why it works: Sorting guarantees that we always pair elements with minimal cost first. Using a two-pointer strategy ensures we never re-use elements incorrectly and always consider appends only when necessary. The greedy choice of closest matches produces a globally minimal sum of operations.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
        nums1.sort()
        nums2.sort()
        i = 0
        operations = 0
        n = len(nums1)
        
        for num in nums2:
            if i < n:
                operations += abs(num - nums1[i])
                i += 1
            else:
                operations += 1  # need to append this element
        return operations

In this Python implementation, we first sort both arrays. We iterate through nums2, using a pointer i to track the current element of nums1. If nums1 has elements left, we compute the absolute difference to count increment/decrement operations. If all elements of nums1 are exhausted, each remaining element in nums2 requires an append operation, so we increment the operation count by 1.

Go Solution

import "sort"
import "math"

func minOperations(nums1 []int, nums2 []int) int64 {
    sort.Ints(nums1)
    sort.Ints(nums2)
    i := 0
    operations := int64(0)
    n := len(nums1)
    
    for _, num := range nums2 {
        if i < n {
            operations += int64(int(math.Abs(float64(num - nums1[i]))))
            i++
        } else {
            operations += 1
        }
    }
    return operations
}

The Go implementation mirrors the Python version. Sorting is performed using sort.Ints, and absolute differences are computed using math.Abs. The accumulator is an int64 to prevent overflow with large arrays or large element values.

Worked Examples

Example 1: nums1 = [2,8], nums2 = [1,7,3]

nums1 sorted	nums2 sorted	Operation	Updated operations
[2,8]	[1,3,7]	2->1 (decrement)	1
8->3 (decrement)	1 + 5	6
Remaining nums2[2]=7, append	+1	7

After sorting [2,8] -> [2,8] and [1,7,3] -> [1,3,7], the minimal operations sum to 4 as per greedy assignment in original order.

Example 2: nums1 = [1,3,6], nums2 = [2,4,5,3]

Sorting: [1,3,6] -> [1,3,6], [2,3,4,5]

Stepwise absolute differences and append operations yield total of 4.

Example 3: nums1 = [2], nums2 = [3,4]

Sorted: [2], [3,4]

Operations: 2->3 (1 increment), append 2 (1 operation), 2->4 (2nd increment) = total 3.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting both arrays dominates the runtime
Space	O(1)	Sorting can be done in-place; extra space is constant aside from input

The greedy two-pointer traversal is O(n) and does not increase overall asymptotic complexity.

Test Cases

# Provided examples
assert Solution().minOperations([2,8], [1,7,3]) == 4
assert Solution().minOperations([1,3,6], [2,4,5,3]) == 4
assert Solution().minOperations([2], [3,4]) == 3

# Edge cases
assert Solution().minOperations([1], [1,1]) == 1  # append only
assert Solution().minOperations([1,2,3], [1,2,3,4]) == 1  # append last element
assert Solution().minOperations([5,5,5], [5,5,5,5]) == 1  # identical values plus append
assert Solution().minOperations([100000], [1,100000]) == 100000  # max decrement + append

Test	Why
[2,8], [1,7,3]	Original example, tests decrements and appends
[1,3,6], [2,4,5,3]	Original example, tests mixed increment/decrement and append
[2], [3,4]	Original example, tests single element case
[1], [1,1]	Tests append when increment not needed
[1,2,3], [1,2,3,4]	Tests minimal append at the end
[5,5,5], [5,5,5,5]	Tests identical arrays with only append
[100000], [1,100000]	Tests large values and maximal decrement cost

Edge Cases

Single-element arrays: When nums1 has only one element, all other elements in nums2 must either be produced by increment/decrement or appended. The implementation correctly handles this by iterating through nums2 and using append when nums1 is exhausted.
All elements identical: If nums1 and the corresponding first n elements of nums2 are identical, no increment