What is the key pattern for solving Range Sum Query - Immutable?

The key pattern is using a prefix sum array to optimize range sum queries. This allows each query to be answered in constant time after an O(n) preprocessing step.

How can I improve my solution to Range Sum Query - Immutable?

Focus on the initialization phase and make sure to build the prefix sum array efficiently. Also, ensure that you handle queries in constant time after the preprocessing.

What are the main challenges in solving Range Sum Query - Immutable?

The main challenge is designing the solution to handle multiple queries efficiently, balancing the space used by the prefix sum array with the time required for queries.

Is Range Sum Query - Immutable a common pattern in coding interviews?

Yes, this problem is commonly used to assess understanding of arrays and optimization techniques such as prefix sum for efficient querying.

How can GhostInterview assist with Range Sum Query - Immutable?

GhostInterview provides insights into the prefix sum technique and helps you avoid common pitfalls, guiding you to optimize for both space and time complexity.

#303

Easy

auto_awesome前缀和

LeetCode 题解工作台

区域和检索 - 数组不可变

给定一个整数数组 nums ，处理以下类型的多个查询: 计算索引 left 和 right （包含 left 和 right ）之间的 nums 元素的和，其中 left 实现 NumArray 类： NumArray(int[] nums) 使用数组 nums 初始化对象 int sumRan…

数组设计前缀和

题目描述

给定一个整数数组 nums，处理以下类型的多个查询:

计算索引 left 和 right （包含 left 和 right）之间的 nums 元素的和，其中 left <= right

实现 NumArray 类：

NumArray(int[] nums) 使用数组 nums 初始化对象
int sumRange(int left, int right) 返回数组 nums 中索引 left 和 right 之间的元素的总和，包含 left 和 right 两点（也就是 nums[left] + nums[left + 1] + ... + nums[right] )

示例 1：

输入：
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出：
[null, 1, -1, -3]

解释：
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

提示：

1 <= nums.length <= 10⁴
-10⁵ <= nums[i] <= 10⁵
0 <= left <= right < nums.length
最多调用 10⁴ 次 sumRange 方法

lightbulb

解题思路

方法一：前缀和

我们创建一个长度为 $n + 1$ 的前缀和数组 $s$ ，其中 $s[i]$ 表示前 $i$ 个元素的前缀和，即 $s[i] = \sum_{j=0}^{i-1} nums[j]$ ，那么索引 $[left, right]$ 之间的元素的和就可以表示为 $s[right + 1] - s[left]$ 。

初始化前缀和数组 $s$ 的时间复杂度为 $O(n)$ ，查询的时间复杂度为 $O(1)$ 。空间复杂度 $O(n)$ 。

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class NumArray:
    def __init__(self, nums: List[int]):
        self.s = list(accumulate(nums, initial=0))

    def sumRange(self, left: int, right: int) -> int:
        return self.s[right + 1] - self.s[left]


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(left,right)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Understanding the trade-off between preprocessing and query time is crucial.
question_mark
Candidates should demonstrate an ability to optimize for multiple queries.
question_mark
Pay attention to how the solution scales with large inputs.

warning

常见陷阱

外企场景

error
Not optimizing for multiple queries by recalculating sums from scratch.
error
Incorrectly calculating the range sum due to off-by-one errors with indices.
error
Failing to properly initialize the prefix sum array, leading to incorrect query results.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Query sum over a segment of the array after each update.
arrow_right_alt
Handle range queries with modifications to the array in between.
arrow_right_alt
Designing a solution for a multi-dimensional range sum query.

help

常见问题

外企场景

继续练习

#304 二维区域和检索 - 矩阵不可变 #731 我的日程安排表 II #307 区域和检索 - 数组可修改