# Arrange numbers to form a valid sequence

Given an array arr[] with N distinct numbers and another array arr1[] with N-1 operators (either < or >), the task is to organize the numbers to form a valid sequence which obeys relational operator rules with respect to provided operators.

Examples:

Input: arr[] = {3, 12, 7, 8, 5}; arr1= {‘<‘, ‘>’, ‘>’, ‘<‘}
Output: {3, 12, 8, 5, 7}
Explanation:
3 < 12 > 8 > 5 < 7
There can be more such combinations. The task is to return one of the combinations.

Input: arr[] = {8, 2, 7, 1, 5, 9}; arr1[] = {‘>’, ‘>’, ‘<‘, ‘>’, ‘<‘}
Output:{9, 8, 1, 7, 2, 5}
Explanation:
9 > 8 > 1 < 7 > 2 < 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:
A naive approach is to try all different possible arrangement of numbers and check if the sequence is valid.

Time Complexity: O(2N).

Efficient Approach: The idea is to first sort the given array of numbers in ascending order. Then solve the problem using two pointers technique: one pointing at the front and other pointing at the end.

1. Take one resultant array of size same as given array.
2. If the current operator is ‘<‘ then include the element which the top pointer is pointing to in resultant array and increment it by 1.
3. If the current operator is ‘>’ then include the element which the last pointer is pointing to in resultant array and decrement it by 1.

Below is the implementation of the above approach.

 `// C++ implemenattion of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to organize the given numbers ` `// to form a valid sequence. ` `vector<``int``> orgazineInOrder(vector<``int``> vec, ` `                            ``vector<``int``> op, ``int` `n) ` `{ ` `    ``vector<``int``> result(n); ` `    ``// Sorting the array ` `    ``sort(vec.begin(), vec.end()); ` ` `  `    ``int` `i = 0, j = n - 1, k = 0; ` `    ``while` `(i <= j && k <= n - 2) { ` `        ``// Two pointer technique ` `        ``// to organize the numbers ` `        ``if` `(op[k] == ``'<'``) { ` `            ``result[k] = vec[i++]; ` `        ``} ` `        ``else` `{ ` `            ``result[k] = vec[j--]; ` `        ``} ` `        ``k++; ` `    ``} ` `    ``result[n - 1] = vec[i]; ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> vec({ 8, 2, 7, 1, 5, 9 }); ` ` `  `    ``vector<``int``> op({ ``'>'``, ``'>'``, ``'<'``, ` `                     ``'>'``, ``'<'` `}); ` ` `  `    ``vector<``int``> result ` `        ``= orgazineInOrder(vec, ` `                          ``op, vec.size()); ` ` `  `    ``for` `(``int` `i = 0; i < result.size(); i++) { ` `        ``cout << result[i] << ``" "``; ` `    ``} ` `    ``return` `0; ` `} `

 `// Java implemenattion of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to organize the given numbers ` `// to form a valid sequence. ` `static` `int``[] orgazineInOrder(``int` `[]vec,``int``[] op, ``int` `n) ` `{ ` `    ``int` `[]result = ``new` `int``[n]; ` `     `  `    ``// Sorting the array ` `    ``Arrays.sort(vec); ` ` `  `    ``int` `i = ``0``, j = n - ``1``, k = ``0``; ` `    ``while` `(i <= j && k <= n - ``2``)  ` `    ``{ ` `        ``// Two pointer technique ` `        ``// to organize the numbers ` `        ``if` `(op[k] == ``'<'``)  ` `        ``{ ` `            ``result[k] = vec[i++]; ` `        ``} ` `        ``else` `        ``{ ` `            ``result[k] = vec[j--]; ` `        ``} ` `        ``k++; ` `    ``} ` `    ``result[n - ``1``] = vec[i]; ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `[]vec ={ ``8``, ``2``, ``7``, ``1``, ``5``, ``9` `}; ` ` `  `    ``int``[] op ={ ``'>'``, ``'>'``, ``'<'``, ` `                    ``'>'``, ``'<'` `}; ` ` `  `    ``int` `[]result = orgazineInOrder(vec, ` `                        ``op, vec.length); ` ` `  `    ``for` `(``int` `i = ``0``; i < result.length; i++)  ` `    ``{ ` `        ``System.out.print(result[i]+ ``" "``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 `# Python3 implemenattion of the above approach  ` ` `  `# Function to organize the given numbers  ` `# to form a valid sequence.  ` `def` `orgazineInOrder(vec, op, n) :  ` ` `  `    ``result ``=` `[``0``] ``*` `n; ` `     `  `    ``# Sorting the array ` `    ``vec.sort(); ` `    ``i ``=` `0``; ` `    ``j ``=` `n ``-` `1``; ` `    ``k ``=` `0``;  ` `     `  `    ``while` `(i <``=` `j ``and` `k <``=` `n ``-` `2``) : ` `         `  `        ``# Two pointer technique  ` `        ``# to organize the numbers  ` `        ``if` `(op[k] ``=``=` `'<'``) : ` `            ``result[k] ``=` `vec[i];  ` `            ``i ``+``=` `1``; ` `         `  `        ``else` `: ` `            ``result[k] ``=` `vec[j];  ` `            ``j ``-``=` `1``; ` `         `  `        ``k ``+``=` `1``;  ` ` `  `    ``result[n ``-` `1``] ``=` `vec[i];  ` ` `  `    ``return` `result;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``vec ``=` `[ ``8``, ``2``, ``7``, ``1``, ``5``, ``9` `]; ` `    ``op ``=` `[ ``'>'``, ``'>'``, ``'<'``, ``'>'``, ``'<'` `];  ` ` `  `    ``result ``=` `orgazineInOrder(vec, op, ``len``(vec));  ` ` `  `    ``for` `i ``in` `range``(``len``(result)) : ` `        ``print``(result[i], end ``=` `" "``);  ` ` `  `# This code is contributed by AnkitRai01 `

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to organize the given numbers  ` `    ``// to form a valid sequence.  ` `    ``static` `int``[] orgazineInOrder(``int` `[]vec,``int``[] op, ``int` `n)  ` `    ``{  ` `        ``int` `[]result = ``new` `int``[n];  ` `         `  `        ``// Sorting the array  ` `        ``Array.Sort(vec);  ` `     `  `        ``int` `i = 0, j = n - 1, k = 0;  ` `        ``while` `(i <= j && k <= n - 2)  ` `        ``{  ` `            ``// Two pointer technique  ` `            ``// to organize the numbers  ` `            ``if` `(op[k] == ``'<'``)  ` `            ``{  ` `                ``result[k] = vec[i++];  ` `            ``}  ` `            ``else` `            ``{  ` `                ``result[k] = vec[j--];  ` `            ``}  ` `            ``k++;  ` `        ``}  ` `        ``result[n - 1] = vec[i];  ` `     `  `        ``return` `result;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]vec ={ 8, 2, 7, 1, 5, 9 };  ` `     `  `        ``int``[] op ={ ``'>'``, ``'>'``, ``'<'``,  ` `                        ``'>'``, ``'<'` `};  ` `     `  `        ``int` `[]result = orgazineInOrder(vec,  ` `                            ``op, vec.Length);  ` `     `  `        ``for` `(``int` `i = 0; i < result.Length; i++)  ` `        ``{  ` `            ``Console.Write(result[i] + ``" "``);  ` `        ``}  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output:
```9 8 1 7 2 5
```

Time Complexity: O(NlogN)

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Improved By : 29AjayKumar, AnkitRai01