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There are different kind of Array. 1鈥檚, 2鈥檚, 3^rd dimensional and multi-dimensional array鈥檚. In the above question there will be sorted array technique through we can combine single sorted array in k/n elements. To check that see below detail related to combine theory of array.

Consider the any array of integers x [1...n] can be sorted in O (n + M) time where,

$\mathit{M}\mathbf{=}\begin{array}{c}{\mathbf{maxx}}_{\mathbf{i}}\mathbf{-}{\mathbf{minx}}_{\mathbf{i}}\\ \mathbf{i}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{i}\end{array}$

Counting sorting takes O(n) times to find maxX_i and minX_i , and then the size can be established as,

$\begin{array}{rcl}S& =& \max \left(x_i\right)-\min \left(x_i\right)+1\\ & =& M+1\\ & & \end{array}$

The array for counting is $C\left[\min \left(x_i\right),鈥�,\max \left(x_i\right)\right]$. Travers the array x [1...n] and for each occurrence of x [i] , the corresponding $C\left[x\left[i\right]\right]$ is incremented by 1. If C [i] is greater than zero, then the element i is the output of the array. The total time complexity is $O\left(n+S\right)=O\left(n+M\right)$.

Since counting sort is not based on comparison, it is not subject to lower bound $惟\left(n\log n\right)$limit.

Therefore,The $惟\left(n\log n\right)$lower bound does not apply because this is not a comparison-based sort, and also because the information beforehand about the range of the elements in the array is given.