LeetCode 974. Subarray Sums Divisible by K--Python解法--数学题--取模求余

LeetCode 974. Subarray Sums Divisible by K–Python解法–数学题–取模求余

LeetCode题解专栏：LeetCode题解
LeetCode 所有题目总结：LeetCode 所有题目总结
大部分题目C++，Python，Java的解法都有。

题目地址：Subarray Sums Divisible by K - LeetCode

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by K = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Note:

1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000

这道题目最容易想到穷举的方法，肯定会超时。
Python解法如下：

class Solution:def subarraysDivByK(self, A: List[int], K: int) -> int:length = len(A)count = 0for i in range(0, length):for j in range(i, length):if sum(A[i:j+1]) % K == 0:count += 1return count

进行少量优化后还是超时：

class Solution:def subarraysDivByK(self, A: List[int], K: int) -> int:length = len(A)count = 0for i in range(0, length):now = 0for j in range(i, length):now += A[j]if now % K == 0:count += 1return count

但是用动态规划不能解决。

最后看了解法后发现是数学题，无语了。

详细解法参考：Count all sub-arrays having sum divisible by k - GeeksforGeeks

class Solution:def subarraysDivByK(self, A: List[int], K: int) -> int:length = len(A)# create auxiliary hash# array to count frequency# of remaindersmod = [0]*K# Traverse original array# and compute cumulative# sum take remainder of this# current cumulative# sum and increase count by# 1 for this remainder# in mod[] arraycumSum = 0for i in range(length):cumSum = cumSum + A[i]# as the sum can be negative,# taking modulo twicemod[cumSum % K] = mod[cumSum % K] + 1result = 0  # Initialize result# Traverse mod[]for i in range(K):# If there are more than# one prefix subarrays# with a particular mod value.if mod[i] > 1:result = result + (mod[i]*(mod[i]-1))//2# add the elements which# are divisible by k itself# i.e., the elements whose sum = 0return result + mod[0]

LeetCode 974. Subarray Sums Divisible by K--Python解法--数学题--取模求余相关推荐

leetcode 974. Subarray Sums Divisible by K的解法（统计共同余数）
题目大意:给定一个整数数组 A,返回其中元素之和可被 K 整除的(连续.非空)子数组的数目大致思路:比如数组A = [1,2,1], K=2,那么1%2 =1,(1+2)%2=1,所以 {2}是符合 ...
leetcode 974 Subarray Sums Divisible by K
leetcode 974 Subarray Sums Divisible by K 1.题目描述 2.解题思路 3.Python代码 1.题目描述给定一个整数数组 A,返回其中元素之和可被 K 整除 ...
leetcode 974. Subarray Sums Divisible by K
974. Subarray Sums Divisible by K 题意:给你一个数组A和一个数K,求改数组有多少连续子序列的和能被K整除. 思路:简单DP.假设sum[i]表示[0-i]的求和,那么 ...
LeetCode：974. Subarray Sums Divisible by K - Python
974. 和可被 K 整除的子数组问题描述: 给定一个整数数组 A,返回其中元素之和可被K整除的(连续.非空)子数组的数目. 示例: 输入:A = [4,5,0,-2,-3,1], K = 5 输出 ...
【leetcode】974. Subarray Sums Divisible by K
题目如下: Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have ...
974. Subarray Sums Divisible by K
Title 给定一个整数数组 A,返回其中元素之和可被 K 整除的(连续.非空)子数组的数目. 示例: 输入:A = [4,5,0,-2,-3,1], K = 5 输出:7 解释: 有 7 个子数组满 ...
974. Subarray Sums Divisible by K [Medium]
用了prefix sum的思想,之前没用过,感觉很难 /*** Runtime: 18 ms, faster than 56.05%* Memory Usage: 42.9 MB, less than ...
取模是什么意思python_原来Python中的取模运算方法竟然是这样的！
今天小编就为大家分享一篇Python中的取模运算方法,具有很好的参考价值,希望对大家有所帮助.一起跟随小编过来看看吧所谓取模运算,就是计算两个数相除之后的余数,符号是%.如a % b就是计算a除以b ...
python怎么取模,Python运算符之取模%
为了可以成为测试,最近在学Python.菜鸟教程中叫%运算为取余,而之前我在学习Java时记得书上是叫%运算为取模.因此想知道这两者到底有啥区别. 一.定义在matlab中,关于取余和取模是这么定义 ...

LeetCode 974. Subarray Sums Divisible by K--Python解法--数学题--取模求余

LeetCode 974. Subarray Sums Divisible by K--Python解法--数学题--取模求余相关推荐

最新文章

热门文章