How to scale algorithm to fit in one frame and centered in the frame at same time?

Simply invert minipage with \scalebox, i.e. embed the minipage inside \scalebox.

MWE:

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}
\begin{frame}{Hello world}
  \begin{center}
    \scalebox{0.75}{
    \begin{minipage}{0.7\linewidth}
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}%
    }
  \end{center}

\end{frame}
\end{document} 

Output:

For content that contains text, choosing an appropriate font size is to be preferred over scaling the content, see Why not scale elements that contain text for some explanation.

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}


\begin{frame}{Hello world}
  \begin{center}
    \begin{minipage}{0.8\linewidth}
        \scriptsize
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}
  \end{center}
\end{frame}

\end{document}

---

However if you don't mind a sub-optimal result, you could use one of the most evil beamer options: shrink:

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}

% This solution is evil!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
% use it at your own risk!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\begin{frame}[shrink]{Hello world}
  \begin{center}
    \begin{minipage}{0.8\linewidth}
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}
  \end{center}
\end{frame}



\end{document}