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GitBook: [greyireland#7] No subject

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`29`	`29`
`30`	`30`	`## 面试公司`
`31`	`31`
`32`		`-* [FaceBook](mian-shi-gong-si/facebook.md)`
	`32`	`+* [Facebook](mian-shi-gong-si/facebook.md)`
	`33`	`+* [Amazon](mian-shi-gong-si/amazon.md)`
	`34`	`+* [Google](mian-shi-gong-si/google.md)`
	`35`	`+* [Linkedin](mian-shi-gong-si/linkedin.md)`

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	`1`	`+# Amazon`
	`2`	`+`
	`3`	`+### OA questions.`
	`4`	`+`
	`5`	`+#### Amazon OA password strength`
	`6`	`+`
	`7`	`+Given a string password, find the strength of that password. The strength of a password, consisting only lowercase english letters only, is calculated as the sum of the number of distinct characters present in all possible substrings of that password. Example:- password = "good" possible sub string and count of distinct characters are g = 1 o = 1 o = 1 d = 1 go = 2 oo = 1 od = 2 goo = 2 ood = 2 good = 3 1+たす1+たす1+たす1+たす2+たす1+たす2+たす2+たす2+たす3 =わ 16`
	`8`	`+`
	`9`	+```cpp
	`10`	`+// Some code`
	`11`	`+// A 1`
	`12`	`+// A B prev + (i -(-1)) = 3;`
	`13`	`+// A B A`
	`14`	`+int passwdStrengh(string s)`
	`15`	`+{`
	`16`	`+ vector<int> pos(26,-1);`
	`17`	`+ int cur =0;`
	`18`	`+ int res =0;`
	`19`	`+ for(int i=0; i< s.size(); i++)`
	`20`	`+ {`
	`21`	`+ int index = s[i]-'a';`
	`22`	`+ cur += (i - pos[index]);`
	`23`	`+ res += cur;`
	`24`	`+ }`
	`25`	`+ return res;`
	`26`	`+}`
	`27`	+```
	`28`	`+`
	`29`	`+#### aaa `
	`30`	`+`
	`31`	`+****`
	`32`	`+`
	`33`	`+####`

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