# Lexicographically smallest rotated sequence | Set 2

Write code to find lexicographic minimum in a circular array, e.g. for the array BCABDADAB, the lexicographic minimum is ABBCABDAD
Input Constraint: 1 < n < 1000

Examples:

```Input:  GEEKSQUIZ
Output: EEKSQUIZG

Input:  GFG
Output: FGG

Input :  CAPABCQ
Output : ABCQCAP```

We have discussed a O(n2Logn) solution in Lexicographically minimum string rotation | Set 1. Here we need to find the starting index of minimum rotation and then print the rotation.

```1) Initially assume 0 to be current min
starting index.
2) Loop through i = 1 to n-1.
a) For each i compare sequence starting
at i with current min starting index
b) If sequence starting at i is lexicographically
smaller, update current min starting
index.```

Here is pseudo-code for algorithm

```function findIndexForSmallestSequence(S, n):
result = 0
for i = 1:n-1
if (sequence beginning at i <
sequence beginning at result)
result = i
end if
end for
return result```

Here is implementation of above algorithm.

## C++

 `// C++ program to find lexicographically ` `// smallest sequence with rotations. ` `#include ` `using` `namespace` `std; `   `// Function to compare lexicographically ` `// two sequence with different starting ` `// indexes. It returns true if sequence ` `// beginning with y is lexicographically ` `// greater. ` `bool` `compareSeq(``char` `S[], ``int` `x, ``int` `y, ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(S[x] < S[y]) ` `            ``return` `false``; ` `        ``else` `if` `(S[x] > S[y]) ` `            ``return` `true``; ` `        ``x = (x + 1) % n; ` `        ``y = (y + 1) % n; ` `    ``} ` `    ``return` `true``; ` `} `   `// Function to find starting index ` `// of lexicographically smallest sequence ` `int` `smallestSequence(``char` `S[], ``int` `n) ` `{ ` `    ``int` `index = 0; ` `    ``for` `(``int` `i = 1; i < n; i++) `   `        ``// if new sequence is smaller ` `        ``if` `(compareSeq(S, index, i, n)) `   `            ``// change index of current min ` `            ``index = i; `   `    ``return` `index; ` `} `   `// Function to print lexicographically ` `// smallest sequence ` `void` `printSmallestSequence(``char` `S[], ``int` `n) ` `{ ` `    ``int` `starting_index = smallestSequence(S, n); ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << S[(starting_index + i) % n]; ` `} `   `// driver code ` `int` `main() ` `{ ` `    ``char` `S[] = ``"DCACBCAA"``; ` `    ``int` `n = 8; ` `    ``printSmallestSequence(S, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find lexicographically` `// smallest sequence with rotations.` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`   `/* Name of the class */` `class` `LexoSmallest {` `    ``// Function to compare lexicographically` `    ``// two sequence with different starting` `    ``// indexes. It returns true if sequence` `    ``// beginning with y is lexicographically` `    ``// greater.` `    ``static` `boolean` `compareSeq(``char``[] S, ``int` `x, ``int` `y, ``int` `n)` `    ``{` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(S[x] < S[y])` `                ``return` `false``;` `            ``else` `if` `(S[x] > S[y])` `                ``return` `true``;` `            ``x = (x + ``1``) % n;` `            ``y = (y + ``1``) % n;` `        ``}` `        ``return` `true``;` `    ``}`   `    ``// Function to find starting index` `    ``// of lexicographically smallest sequence` `    ``static` `int` `smallestSequence(``char``[] S, ``int` `n)` `    ``{` `        ``int` `index = ``0``;` `        ``for` `(``int` `i = ``1``; i < n; i++)`   `            ``// if new sequence is smaller` `            ``if` `(compareSeq(S, index, i, n))`   `                ``// change index of current min` `                ``index = i;`   `        ``return` `index;` `    ``}`   `    ``// Function to print lexicographically` `    ``// smallest sequence` `    ``static` `void` `printSmallestSequence(String str, ``int` `n)` `    ``{` `        ``char``[] S = str.toCharArray();` `        ``int` `starting_index = smallestSequence(S, n);` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``System.out.print(S[(starting_index + i) % n]);` `    ``}`   `    ``// driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String S = ``"DCACBCAA"``;` `        ``int` `n = ``8``;` `        ``printSmallestSequence(S, n);` `    ``}` `}` `// This code is contributed by Mr Somesh Awasthi`

## Python 3

 `# Python 3 program to find lexicographically` `# smallest sequence with rotations.`   `# Function to compare lexicographically` `# two sequence with different starting` `# indexes. It returns true if sequence` `# beginning with y is lexicographically` `# greater.` `import` `copy`     `def` `printSmallestSequence(s):` `    ``m ``=` `copy.copy(s)` `    ``for` `i ``in` `range``(``len``(s) ``-` `1``):`   `        ``if` `m > s[i:] ``+` `s[:i]:` `            ``m ``=` `s[i:] ``+` `s[:i]`   `    ``return` `m`   `#Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``st ``=` `'DCACBCAA'` `    ``print``(printSmallestSequence(st))`     `# This code is contributed by Koushik Reddy B`

## C#

 `// C# program to find lexicographically` `// smallest sequence with rotations.` `using` `System;`   `class` `LexoSmallest {` `    `  `    ``// Function to compare lexicographically` `    ``// two sequence with different starting` `    ``// indexes. It returns true if sequence` `    ``// beginning with y is lexicographically` `    ``// greater.` `    ``static` `bool` `compareSeq(``string` `S, ``int` `x, ``int` `y, ``int` `n)` `    ``{` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(S[x] < S[y])` `                ``return` `false``;` `            ``else` `if` `(S[x] > S[y])` `                ``return` `true``;` `            ``x = (x + 1) % n;` `            ``y = (y + 1) % n;` `        ``}` `        ``return` `true``;` `    ``}`   `    ``// Function to find starting index` `    ``// of lexicographically smallest sequence` `    ``static` `int` `smallestSequence(``string` `S, ``int` `n)` `    ``{` `        ``int` `index = 0;` `        ``for` `(``int` `i = 1; i < n; i++)`   `            ``// if new sequence is smaller` `            ``if` `(compareSeq(S, index, i, n))`   `                ``// change index of current min` `                ``index = i;`   `        ``return` `index;` `    ``}`   `    ``// Function to print lexicographically` `    ``// smallest sequence` `    ``static` `void` `printSmallestSequence(``string` `str, ``int` `n)` `    ``{` `        ``// char[] S=str.toCharArray();` `        ``int` `starting_index = smallestSequence(str, n);` `        ``for` `(``int` `i = 0; i < n; i++)` `        ``Console.Write(str[(starting_index + i) % n]);` `    ``}`   `    ``// driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `S = ``"DCACBCAA"``;` `        ``int` `n = 8;` `        ``printSmallestSequence(S, n);` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 ` ``\$S``[``\$y``])` `            ``return` `true;` `        ``\$x` `= (``\$x` `+ 1) % ``\$n``;` `        ``\$y` `= (``\$y` `+ 1) % ``\$n``;` `    ``}` `    ``return` `true;` `}`   `// Function to find starting index` `// of lexicographically smallest` `// sequence` `function` `smallestSequence(``\$S``, ``\$n``)` `{` `    ``\$index` `= 0;` `    ``for` `( ``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++)`   `        ``// if new sequence is smaller` `        ``if` `(compareSeq(``\$S``, ``\$index``, ``\$i``, ``\$n``))`   `            ``// change index of current min` `            ``\$index` `= ``\$i``;`   `    ``return` `\$index``;` `}`   `// Function to print lexicographically` `// smallest sequence` `function` `printSmallestSequence(``\$S``, ``\$n``)` `{` `    ``\$starting_index` `= smallestSequence(``\$S``, ``\$n``);` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)` `        ``echo` `\$S``[(``\$starting_index` `+ ``\$i``) % ``\$n``];` `}`   `    ``// Driver Code` `    ``\$S``= ``"DCACBCAA"``;` `    ``\$n` `= 8;` `    ``printSmallestSequence(``\$S``, ``\$n``);`   `// This code is contributed by Ajit.` `?>`

## Javascript

 ``

Output

`AADCACBC`

Time Complexity : O(n^2)
Auxiliary Space : O(1)

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