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Create Construct the Lexicographically Largest Valid Sequence

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Daily Question/2025/February
- Construct the Lexicographically Largest Valid Sequence

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`‎Daily Question/2025/February/Construct the Lexicographically Largest Valid Sequence`

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	`1`	`+想法:利用 Recursive Backtracking 的技巧,找過所有的可能序列,並輸出最大者。`
	`2`	`+由於題目要求要回傳字典序最大者;所以我們從數字大填到小,只要一找到解即可停止遞迴並回傳`
	`3`	`+`
	`4`	`+Time Complexity : O(n!) for there are n! possible sequence`
	`5`	`+(or you can think as the first digit has n candidates , second digit has n - 1 ... to the last digit , total complexity is`
	`6`	`+n * (n - 1) * (n - 2) ... * 1 = O(n!) )`
	`7`	`+Space Complexity : O(n! for the recursion (recursion depth = O(n))`
	`8`	`+`
	`9`	`+class Solution {`
	`10`	`+public:`
	`11`	`+ vector<int> ans , appear ;`
	`12`	`+ bool construct(int n , int index) {`
	`13`	`+ if (index == 2 * n - 1)`
	`14`	`+ return true ;`
	`15`	`+ if ( ans[index] != 0 )`
	`16`	`+ return construct(n , index + 1) ;`
	`17`	`+`
	`18`	`+ for(int num = n ; num > 0 ; num--) {`
	`19`	`+ if ( appear[num] == 0 && ans[index] == 0`
	`20`	`+ && (num == 1 \|\| (index + num < 2 * n - 1 && ans[index + num] == 0)) ) {`
	`21`	`+ appear[num] = 1 ;`
	`22`	`+ if (num == 1)`
	`23`	`+ ans[index] = 1 ;`
	`24`	`+ else`
	`25`	`+ ans[index] = ans[index + num] = num ;`
	`26`	`+ if ( construct(n , index + 1) )`
	`27`	`+ return true ;`
	`28`	`+`
	`29`	`+ appear[num] = 0;`
	`30`	`+ if (num == 1)`
	`31`	`+ ans[index] = 0 ;`
	`32`	`+ else`
	`33`	`+ ans[index] = ans[index + num] = 0;`
	`34`	`+ }`
	`35`	`+ }`
	`36`	`+ return false ;`
	`37`	`+ }`
	`38`	`+ vector<int> constructDistancedSequence(int n) {`
	`39`	`+ ans.resize(2 * n - 1) ;`
	`40`	`+ appear.resize(n + 1) ;`
	`41`	`+ construct(n , 0) ;`
	`42`	`+ return ans ;`
	`43`	`+ }`
	`44`	`+};`

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`‎Daily Question/2025/February/Construct the Lexicographically Largest Valid Sequence`

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