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feat: add solutions to lc problem: No.2744~2746 (doocs#2207)

* No.2744.Find Maximum Number of String Pairs * No.2745.Construct the Longest New String * No.2746.Decremental String Concatenation

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`@@ -58,6 +58,16 @@ It can be proven that 1 is the maximum number of pairs that can be formed.`
`58`	`58`
`59`	`59`	`## Solutions`
`60`	`60`
	`61`	`+Solution 1: Hash Table`
	`62`	`+`
	`63`	`+We can use a hash table $cnt$ to store the number of occurrences of each string's reversed string in the array $words$.`
	`64`	`+`
	`65`	`+Traverse the array $words,ドル for each string $w,ドル we directly add the value of $cnt[w]$ to the answer, then increase the occurrence count of $w$'s reversed string by 1ドル$.`
	`66`	`+`
	`67`	`+After the traversal ends, we can get the maximum number of matches.`
	`68`	`+`
	`69`	`+The time complexity is $O(L),ドル and the space complexity is $O(L),ドル where $L$ is the sum of the lengths of the strings in the array $words$.`
	`70`	`+`
`61`	`71`	`<!-- tabs:start -->`
`62`	`72`
`63`	`73`	`### Python3`

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`@@ -40,6 +40,16 @@ That string has length 14, and we can show that it is impossible to construct a`
`40`	`40`
`41`	`41`	`## Solutions`
`42`	`42`
	`43`	`+Solution 1: Case Discussion`
	`44`	`+`
	`45`	`+We observe that the string 'AA' can only be followed by 'BB', and the string 'AB' can be placed at the beginning or end of the string. Therefore:`
	`46`	`+`
	`47`	`+- If $x < y,ドル we can first alternately place 'BBAABBAA..BB', placing a total of $x$ 'AA' and $x+1$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x \times 2 + z + 1) \times 2$;`
	`48`	`+- If $x > y,ドル we can first alternately place 'AABBAABB..AA', placing a total of $y$ 'BB' and $y+1$ 'AA', then place the remaining $z$ 'AB', with a total length of $(y \times 2 + z + 1) \times 2$;`
	`49`	`+- If $x = y,ドル we only need to alternately place 'AABB', placing a total of $x$ 'AA' and $y$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x + y + z) \times 2$.`
	`50`	`+`
	`51`	`+The time complexity is $O(1),ドル and the space complexity is $O(1)$.`
	`52`	`+`
`43`	`53`	`<!-- tabs:start -->`
`44`	`54`
`45`	`55`	`### Python3`

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`@@ -68,6 +68,16 @@ It can be shown that the minimum possible length of str<sub>2</sub> is 6.`
`68`	`68`
`69`	`69`	`## Solutions`
`70`	`70`
	`71`	`+Solution 1: Case Discussion`
	`72`	`+`
	`73`	`+We observe that the string 'AA' can only be followed by 'BB', and the string 'AB' can be placed at the beginning or end of the string. Therefore:`
	`74`	`+`
	`75`	`+- If $x < y,ドル we can first alternately place 'BBAABBAA..BB', placing a total of $x$ 'AA' and $x+1$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x \times 2 + z + 1) \times 2$;`
	`76`	`+- If $x > y,ドル we can first alternately place 'AABBAABB..AA', placing a total of $y$ 'BB' and $y+1$ 'AA', then place the remaining $z$ 'AB', with a total length of $(y \times 2 + z + 1) \times 2$;`
	`77`	`+- If $x = y,ドル we only need to alternately place 'AABB', placing a total of $x$ 'AA' and $y$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x + y + z) \times 2$.`
	`78`	`+`
	`79`	`+The time complexity is $O(1),ドル and the space complexity is $O(1)$.`
	`80`	`+`
`71`	`81`	`<!-- tabs:start -->`
`72`	`82`
`73`	`83`	`### Python3`

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