Walmart 面试经验分享：解决算法挑战与编程技巧 | 面试代面, OA代做, 面试辅导

Walmart 面试经验分享：解决算法挑战与编程技巧 | 面试代面, OA代做, 面试辅导

在这次Walmart的面试中，考生展现出了在高压力环境下快速解决问题的能力。面试过程中所涉及的算法题目覆盖了动态规划、数组操作以及哈希表的运用，每一道题目都要求考生在有限的时间内写出高效且正确的代码。这不仅考验了考生的技术能力，更考验了他们在面试场景中的应变和适应能力。

如果你也想在面试中展现如此优异的表现，我们的代面服务可以帮助你更好地准备和应对各种编程挑战。我们提供的面试代面、OA代做、面试辅导等服务，能够为你提供针对性的指导，确保你在面试中脱颖而出，成功拿到心仪的offer。

想要了解更多或获取我们的服务，欢迎添加微信 leetcode-king。

---

在这次Walmart的面试过程中，主要聚焦在编程挑战与算法问题的解决。面试分为多个环节，每个环节对应不同类型的算法问题，要求考生在规定的时间内编写出正确且高效的代码。

环节一：Ways to Sum 问题（编程技巧，动态规划，递归）

Problem Description:

An automated packaging system is responsible for packing boxes. A box is certified to hold a certain weight. Given an integer total, calculate the number of possible ways to achieve total as a sum of the weights of items weighing integer weights from 1 to k, inclusive.

Example:

total = 8
k = 2

To reach a weight of 8, there are 5 different ways that items with weights between 1 and 2 can be combined:

[1, 1, 1, 1, 1, 1, 1, 1]
[1, 1, 1, 1, 1, 1, 2]
[1, 1, 1, 1, 2, 2]
[1, 2, 2, 2, 2]
[2, 2, 2, 2]

Function Description:

Complete the function ways in the editor below.

ways has the following parameter(s):

• int total: the value to sum to
• int k: the maximum of the range of integers to consider when summing to total

Code:

def ways_helper(total, k, min_k):
    if total == 0:
        return 1
    if total < 0:
        return 0
    possible_way = 0
    for candidate_k in range(min_k, k+1):
        possible_way += ways_helper(total - candidate_k, k, candidate_k)
    return possible_way

def ways(total, k):
    if total == 0:
        return 1
    return ways_helper(total, k, 1)

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')
    total = int(input().strip())
    k = int(input().strip())
    result = ways(total, k)
    fptr.write(str(result) + '\n')
    fptr.close()

环节二：Remove Duplicates 问题（数组操作，双指针技巧）

Problem Description:

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.

Example:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

Code:

def removeDuplicate(nums):
    i = 0
    for j in range(len(nums)):
        if j > 0 and nums[j] == nums[j-1]:
            continue
        else:
            nums[i] = nums[j]
            i+=1
    return i

print(removeDuplicate([1,1,2]))
print(removeDuplicate([1,1,1,1,1,2,3,3,4]))

环节三：Degree of Array 问题（哈希表，子数组分析）

Problem Description:

Given an array of integers, its degree is defined as the number of occurrences of the element that occurs most frequently in the array. Given a list of integers, determine two properties:

1. the degree of the array
2. the length of the shortest sub-array that shares that degree

Example:

arr = [1, 2, 1, 3, 2]

The array has a degree of 2 based on the occurrences of values 1 and 2. The subarray of degree 2 based on the 1's is [1, 2, 1] and based on 2's is [2, 1, 3, 2]. Their lengths are 3 and 4, so the shortest is length 3. Return the shortest length.

Code:

def degreeOfArray(arr):
    index = {}
    for i, ele in enumerate(arr):
        if ele not in index:
            index[ele] = []
        index[ele].append(i)
    
    degree = 0
    min_length = len(arr)
    
    for k, v in index.items():
        degree = max(degree, len(v))
        min_length = min(min_length, v[-1] - v[0] + 1)
    
    return min_length

if __name__ == '__main__':
    arr_count = int(input().strip())
    arr = list(map(int, input().rstrip().split()))
    result = degreeOfArray(arr)
    print(result)