#179

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Largest Number

The problem asks to arrange integers to form the largest possible number, focusing on greedy algorithms and string handling.

array string greedy sorting

Problem Statement

You are given a list of non-negative integers nums. Your task is to arrange them such that they form the largest possible number and return it as a string.

Since the result may be very large, the return value should be a string rather than an integer. Pay attention to the order of digits when forming the number.

Examples

Example 1

Input: nums = [10,2]

Output: "210"

Example details omitted.

Example 2

Input: nums = [3,30,34,5,9]

Output: "9534330"

Example details omitted.

Constraints

1 <= nums.length <= 100
0 <= nums[i] <= 109

Solution Approach

Greedy Approach with Sorting

The primary approach is to use a greedy strategy where numbers are compared to determine their order based on string concatenation. By sorting the numbers using a custom comparator, we can ensure that each pair of numbers forms the largest concatenated result.

Custom Sorting Comparator

The key to the solution is designing a custom comparator for sorting. For any two numbers, the comparator will decide their order by comparing the concatenated results of both possible orders. This ensures that larger numbers always appear in front.

Edge Case Handling

After sorting the numbers, we must handle edge cases like leading zeros. If the first number is zero, we return '0' since the result will be zero. This check prevents invalid outputs like '000'.

Complexity Analysis

Metric	Value
Time	O(n \log n)
Space	O(n)

The time complexity of the solution is O(n log n), where n is the number of elements in the input list, due to the sorting step. The space complexity is O(n) as we need space for the sorted list and some additional space for string manipulations.

What Interviewers Usually Probe

Can the candidate efficiently explain why sorting is key in forming the largest number?
Does the candidate address edge cases such as leading zeros or large number sizes?
How well does the candidate handle performance considerations when discussing sorting and comparisons?

Common Pitfalls or Variants

Common pitfalls

Overlooking the need for a custom sorting function for concatenation comparisons.
Failing to handle edge cases, especially when all elements are zeros or very large numbers.
Returning an integer instead of a string when the output can be very large.

Follow-up variants

Use a different sorting method such as a priority queue or bucket sort to optimize performance.
Consider a variation where you need to handle both positive and negative numbers in the input.
Solve the problem while restricting to a certain time complexity or memory limit.

FAQ

How does greedy strategy help solve the Largest Number problem?

Greedy strategy helps by ensuring that numbers are ordered in a way that maximizes the concatenated result. The custom sorting based on concatenated strings guarantees the largest number formation.

What is the time complexity of the solution?

The time complexity of the solution is O(n log n) due to the sorting step where n is the number of elements in the list.

How do we handle cases where all numbers are zero?

If all numbers are zero, the result should be '0' to avoid an invalid output like '000'. This is checked after sorting.

Can we use other sorting algorithms for this problem?

While other sorting algorithms could be used, the key is using a custom comparator to ensure correct order based on concatenation. A heap or bucket sort could be used in certain cases.

What happens if we return an integer instead of a string?

Returning an integer instead of a string will cause issues with large numbers that exceed integer limits. The problem explicitly requires the result to be a string.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def largestNumber(self, nums: List[int]) -> str:
        nums = [str(v) for v in nums]
        nums.sort(key=cmp_to_key(lambda a, b: 1 if a + b < b + a else -1))
        return "0" if nums[0] == "0" else "".join(nums)

Continue Practicing

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