Consider Algorithm 1 and the input array \(A\) below. \(A\)…
Consider Algorithm 1 and the input array \(A\) below. \(A\) is a square matrix. \[\begin{array}{l}\textbf{Algorithm 1} \,\, processSquareMatrix(A, n): \\\quad s = \sqrt n\\\quad \textbf{for} \, i = 1\, \textbf{to} \,s \, \textbf{do}\\\quad \quad \textbf{for} \, j = 1\, \textbf{to} \, \lfloor s/2 \rfloor \, \textbf{do}\\\quad \quad \quad temp = A[i][j]\\\quad \quad \quad A[i][j] = A[i][s-j+1]\\\quad \quad \quad A[i][s-j+1] = temp\\\end{array}\] 22 17 9 76 55 61 29 83 2 45 90 22 23 42 44 3 \[ \lfloor x \rfloor \, \text{is the floor of} \, x . n \text{ is the number of items in } A \] Enter the values of the first row of \(A\) after running the algorithm and passing \(A\) as the first parameter, and \(n\) as the second parameter: Briefly explain what this algorithm does in simple terms: Using the proper asymptotic notation, the running time of Algorithm 1 is )
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