# Algorithms / Merge two sorted arrays with extra space

Posted On 04.02.2022

The problem

Given two non-decreasing arrays a of length m, and b of length n. Merge the two arrays into array a. Since it’s an in-place merge, a has some extra spaces to fit the result.

The solution

Iterating from the end of both input arrays, at each iteration, pick the larger of the last two inputs and insert it into the end of the target array. Repeat this process until we reach the beginning of either input.

For example:

Given two arrays a and b of length m and n as follow:

\begin{align} a &= [ 1, 3, 5, 0, 0, 0 ] &m = 3 \\ b &= [ 2, 4, 7 ] &n = 3 \\ \end{align}

Let k be the last index of the merged array:

$$k = m + n - 1$$

Compare the two numbers $a[m-1]$ and $b[n-1]$ and insert the larger one into the end of the merged array (at position $a[k]$). After that, decrease the index $k$ and the value of $m$ or $n$ depending on which value is selected.

Repeat the process until either $m$ or $n$ reach the beginning of the array.

Here’s how the algorithm running. From the beginning:

\begin{align} a &= [ 1, 3, \overset{\overset{m}\downarrow}{5}, 0, 0, \overset{\overset{k}\downarrow}{0} ] \\ b &= [ 2, 4, \underset{\underset{n}\uparrow}{7} ] \\ \end{align}

We have $a[m] = 5$, which is smaller than $b[n] = 7$. Move $b[n]$ into the position $a[k]$, decrease the value of both $n$ and $k$ by 1:

\begin{align} a &= [ 1, 3, \overset{\overset{m}\downarrow}{5}, 0, \overset{\overset{k}\downarrow}{0}, 7 ] \\ b &= [ 2, \underset{\underset{n}\uparrow}{4}, \emptyset ] \\ \end{align}

Now, $a[m] = 5$ is larger than $b[n] = 4$. Move $a[m]$ into $a[k]$ and decrease both $m$ and $k$ by 1:

\begin{align} a &= [ 1, \overset{\overset{m}\downarrow}{3}, \emptyset, \overset{\overset{k}\downarrow}{0}, 5, 7 ] \\ b &= [ 2, \underset{\underset{n}\uparrow}{4}, \emptyset ] \\ \end{align}
Next, $a[m] = 3$, less than $b[n] = 4$. Move $b[n]$ into $a[k]$ and decrease both $n$ and $k$:

\begin{align} a &= [ 1, \overset{\overset{m}\downarrow}{3}, \overset{\overset{k}\downarrow}{\emptyset}, 4, 5, 7 ] \\ b &= [ \underset{\underset{n}\uparrow}{2}, \emptyset, \emptyset ] \\ \end{align}

Next, $a[m] = 3$, larger than $b[n] = 2$. Move $a[m]$ into $a[k]$ and decrease $m$ and $k$:

\begin{align} a &= [ \overset{\overset{m}\downarrow}{1}, \overset{\overset{k}\downarrow}{\emptyset}, 3, 4, 5, 7 ] \\ b &= [ \underset{\underset{n}\uparrow}{2}, \emptyset, \emptyset ] \\ \end{align}
Finally, move $b[n]$ into $a[k]$ since $a[m] = 1$ is less than $b[n] = 2$, decreasing $n$ and $k$:

\begin{align} a &= [ \overset{\overset{m,k}\downarrow}{1}, 2, 3, 4, 5, 7 ] \\ b &= \underset{\underset{n}\uparrow}{[} \emptyset, \emptyset, \emptyset ] \\ \end{align}

At this point, we reach the end of the algorithm, the merged result is:

$$a = [1, 2, 3, 4, 5, 7]$$