【leetcode】1208. Get Equal Substrings Within Budget
题目如下:
You are given two strings
sandtof the same length. You want to changestot. Changing thei-th character ofstoi-th character oftcosts|s[i] - t[i]|that is, the absolute difference between the ASCII values of the characters.You are also given an integer
maxCost.Return the maximum length of a substring of
sthat can be changed to be the same as the corresponding substring oftwith a cost less than or equal tomaxCost.If there is no substring from
sthat can be changed to its corresponding substring fromt, return0.Example 1:
Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.Example 2:
Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to charactor int, so the maximum length is 1.Example 3:
Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.Constraints:
1 <= s.length, t.length <= 10^50 <= maxCost <= 10^6sandtonly contain lower case English letters.
解题思路:本题包了一层壳,去掉外表后题目是给定一个正整数组成的数组,求出最长的一段子数组的长度,要求子数组的和不大于cost。解题方法也不难,记per_cost[i]为abs(s[i] - t[i])的值,cost[i]为sum(per_cost[0:i])的值。对于任意一个下标i,很容易通过二分查找的方法找出cost中另外一个下标j,使得cost[i:j] <= cost。
代码如下:
class Solution(object):
def equalSubstring(self, s, t, maxCost):
"""
:type s: str
:type t: str
:type maxCost: int
:rtype: int
"""
cost = []
amount = 0
per_cost = []
for cs,ct in zip(s,t):
amount += abs(ord(cs) - ord(ct))
cost.append(amount)
per_cost.append(abs(ord(cs) - ord(ct)))
#cost.sort()
#print cost
#print per_cost
import bisect
res = -float('inf')
for i in range(len(cost)):
inx = bisect.bisect_right(cost,cost[i] + maxCost - per_cost[i])
res = max(res,inx - i)
return res
【leetcode】1208. Get Equal Substrings Within Budget的更多相关文章
- 【LeetCode】1208. 尽可能使字符串相等 Get Equal Substrings Within Budget (Python)
作者: 负雪明烛 id: fuxuemingzhu 公众号:每日算法题 本文关键词:LeetCode,力扣,算法,算法题,字符串,并查集,刷题群 目录 题目描述 示例 解题思路 滑动窗口 代码 刷题心 ...
- 【LeetCode】696. Count Binary Substrings 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:暴力解法(TLE) 方法二:连续子串计算 日 ...
- 【LeetCode】416. Partition Equal Subset Sum 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 动态规划 日期 题目地址:https://l ...
- 【leetcode】927. Three Equal Parts
题目如下: Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these ...
- 【leetcode】1224. Maximum Equal Frequency
题目如下: Given an array nums of positive integers, return the longest possible length of an array prefi ...
- 【leetcode】416. Partition Equal Subset Sum
题目如下: 解题思路:对于这种判断是否的题目,首先看看动态规划能不能解决.本题可以看成是从nums中任选i个元素,判断其和是否为sum(nums)/2,很显然从nums中任选i个元素的和的取值范围是[ ...
- 【LeetCode】647. Palindromic Substrings 解题报告(Python)
[LeetCode]647. Palindromic Substrings 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/p ...
- 【leetcode】698. Partition to K Equal Sum Subsets
题目如下: 解题思路:本题是[leetcode]473. Matchsticks to Square的姊妹篇,唯一的区别是[leetcode]473. Matchsticks to Square指定了 ...
- 【leetcode】486. Predict the Winner
题目如下: Given an array of scores that are non-negative integers. Player 1 picks one of the numbers fro ...
随机推荐
- CTF—攻防练习之HTTP—PUT上传漏洞
主机:192.168.32.152 靶机:192.168.32.159 中间件PUT漏洞 包括apache,tomcat,IIS等中间件设置支持的HTTP方法(get,post,head,delete ...
- Unity3D 旋转
Unity有两种设置物体旋转的方式,一种时用Rotate()函数来旋转,另一种时直接构造目标Quaternion来直接赋予rotation. 好吧,不知到写什么,各种旋转和unity2D差不多.在国内 ...
- docker搭建环境的时候常用的命令有哪些
1.docker搭建环境的时候常用的命令有哪些 docker如果要删除镜像,现在停止container docker ps 查询正在运行的镜像docker stop +containerid停止后再删 ...
- centos8飞行驾驶舱和docker安装
零.先解决cenos8的网络(systemctl restart network.service已被废弃) 1.# vim /etc/sysconfig/network-scripts/ifcfg-e ...
- nginx配置本地域名反向代理实现本地多域名80访问
什么是反向代理? 代理:通过客户机的配置,实现让一台服务器代理客户机,客户的所有请求都交给代理服务器处理. 反向代理:用一台服务器,代理真实服务器,用户访问时,不再是访问真实服务器,而是代理服务器. ...
- Flask-wtf导入Regexp规则库验证手机号码合法性(测试通过)
手机号码在项目有着很重要的地位,保证用户输入的号码准确无误就显得很关键. 废话不多说,现在页面中引入Regexp规则库: from wtforms.validators import Regexp 验 ...
- logging模块及日志框架
logging模块及日志框架 logging模块 一.导入方式 import logging 二.作用 写日志 三.模块功能 3.1 经常使用 # V1 import logging logging ...
- Django 前端通过json 取出后端数据
Django 前端通过json 取出后端数据 前端通过json 取出后端数据 步骤1:后台数据通过 JSON 序列化成字符串a 注意:1.json是1个字符串 2.通过json.dumps('xx ...
- C - 卿学姐与诡异村庄(并查集+One face meng bi)
卿学姐与诡异村庄 Time Limit: 4500/1500MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others) Submit ...
- 12、MA图的计算过程
为了简化问题,假设有3张芯片,每组数有9个探针: Data: 2,4,6,7,9,10,4,7,8,3 9,5,3,2,5,7,9,10,3,12 6,4,3,2,7,8,1,2,6,9 一.给3组数 ...