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When it comes to coding interview prep for software developers or engineers, sorting algorithms is a topic you cannot afford to miss. Problems based on sorting algorithms regularly feature in tech interviews at FAANG and other tier-1 tech companies. In this article, we’ll help you review the iterative merge sort. Here’s what we will cover:

- What Is Iterative Merge Sort?
- How Does Iterative Merge Sort Work?
- Iterative Merge Sort Algorithm
- Recursive Merge Sort Code
- Iterative Merge Sort Code
- Iterative Merge Sort Complexities
- Advantages of Iterative Merge Sort
- Disadvantages of Iterative Merge Sort
- FAQs on Iterative Merge Sort

In Iterative merge sort, we implement merge sort in a bottom-up manner. This is how it works:

- We start by sorting all sub-arrays of length 1
- Then, we sort all sub-arrays of length 2 by merging length-1 sub-arrays
- Then, we sort all sub-arrays of length 4 by merging length-2 sub-arrays
- We repeat the above step for sub-arrays of lengths 8, 16, 32, and so on until the whole array is sorted

Let’s assume that the array Arr[] = {3, 2, 1, 9, 5, 4, 10, 11} of size N = 8 is to be sorted.

Arrays of length 1 are trivially sorted. First, we take sub_size = 1 and merge all pairs of sub-arrays of size 1.

Then, we multiply sub_size by 2, and sub_size becomes 2. Now, we merge all pairs of sub-arrays of size 2.

Again, we multiply sub_size by 2, and sub_size becomes 4, and we merge all pairs of sub-arrays of size 4.

Now, we stop, as sub_size is >= N and the array is sorted.

Consider an array Arr[] of size N that we want to sort:

**Step 1: **Initialize sub_size with 1 and multiply it by 2 as long as it is less than N. And for each sub_size, do the following:

**Step 2:** Initialize L with 0 and add 2*sub_size as long as it is less than N. Calculate Mid as min(L + sub_size - 1, N-1) and R as min(L + (2* sub_size) -1, N-1) and do the following:

**Step 3:** Copy sub-array [L, Mid-1] in list A and sub-array [Mid, R] in list B and merge these sorted lists to make a sorted list C using the following method:

**Step 3.1: **Compare the first elements of lists A and B and remove the first element from the list whose first element is smaller and append it to C. Repeat this until either list A or B becomes empty.

**Step 3.2:** Copy the list(A or B), which is not empty, to C.

**Step 4:** Copy list C to Arr[] from index L to R.

Recursive Merge Sort Implementation

Here’s the implementation of recursive merge sort algorithm in C++:

#include<bits/stdc++.h>

using namespace std;

void merge(int Arr[], int l, int m, int r) {

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = Arr[l + i];

for (j = 0; j < n2; j++)

R[j] = Arr[m + 1 + j];

i = 0, j = 0, k = l;

while (i < n1 && j < n2) {

if (L[i] <= R[j]) {

Arr[k] = L[i];

i++;

} else {

Arr[k] = R[j];

j++;

}

k++;

}

while (i < n1) {

Arr[k] = L[i];

i++;

k++;

}

while (j < n2) {

Arr[k] = R[j];

j++;

k++;

}

}

void merge_sort(int L, int R, int Arr[]){

if(L==R)

return ;

int Mid= (L+R)/2;

// Dividing sub-array from L to R into

// two parts and recursively solving

merge_sort(L, Mid, Arr);

merge_sort(Mid+1, R, Arr);

// merging two sorted sub-arrays

merge(Arr,L, Mid, R);

}

int main()

{

int i;

int N = 8;

int Arr[N] = {3, 2, 1, 9, 5, 4, 10, 11};

cout<<"Unsorted Array: ";

for(i=0;i<N;i++)

cout<<Arr[i]<<" ";

cout<<endl;

merge_sort(0, N-1, Arr);

cout<<"Sorted Array: ";

for(i=0;i<N;i++)

cout<<Arr[i]<<" ";

return 0;

}

Unsorted Array: 3 2 1 9 5 4 10 11

Sorted Array: 1 2 3 4 5 9 10 11

And this is how iterative merge sort can be implemented in C++:

#include<bits/stdc++.h>

using namespace std;

void merge(int Arr[], int l, int m, int r) {

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = Arr[l + i];

for (j = 0; j < n2; j++)

R[j] = Arr[m + 1+ j];

i = 0, j = 0, k = l;

while (i < n1 && j < n2) {

if (L[i] <= R[j]) {

Arr[k] = L[i];

i++;

} else {

Arr[k] = R[j];

j++;

}

k++;

}

while (i < n1) {

Arr[k] = L[i];

i++;

k++;

}

while (j < n2) {

Arr[k] = R[j];

j++;

k++;

}

}

void merge_sort(int Arr[], int N){

for(int sub_size=1;sub_size<N;sub_size*=2)

{

for(int L=0; L<N; L+=(2*sub_size))

{

int Mid=min(L+sub_size-1,N-1);

int R=min(L+2*sub_size-1,N-1);

// function to merge two sub-arrays of

// size sub_size starting from L and Mid

merge(Arr, L,Mid,R);

}

}

}

int main()

{

int i;

int N = 8;

int Arr[N] = {3, 2, 1, 9, 5, 4, 10, 11};

cout<<"Unsorted Array: ";

for(i=0;i<N;i++)

cout<<Arr[i]<<" ";

cout<<endl;

merge_sort(Arr, N);

cout<<"Sorted Array: ";

for(i=0;i<N;i++)

cout<<Arr[i]<<" ";

return 0;

}

Unsorted Array: 3 2 1 9 5 4 10 11

Sorted Array: 1 2 3 4 5 9 10 11

Given: N = 8, Arr[] = {3, 2, 1, 9, 5, 4, 10, 11}

- When sub_size = 1

Arr[] = {2, 3, 1, 9, 4, 5, 10, 11}

- When sub_size = 2

Arr[] = {1, 2, 3, 9, 4, 5, 10, 11}

- When sub_size = 4

Arr[] = {1, 2, 3, 4, 5, 9, 10, 11}

- Worst-case time complexity:
**O(N*logN).** - Average-case time complexity:
**O(N*logN).** - Best-case time complexity:
**O(N*logN).**

- Worst-case, best-case, and average-case time complexity of merge sort are O(N*logN), making it very efficient.
- Iterative merge sort is slightly faster than recursive merge sort.

- Space complexity of iterative merge sort is O(N), whereas quicksort has O(1) space complexity.
- Recursive merge sort is easier to implement than iterative merge sort.
- Iterative merge sort is marginally slower than quicksort in practice.

*To know more about the merge sort vs. quicksort, read:*

**Question 1: Is iterative merge sort stable?**

Answer: Yes, iterative merge sort is an example of a stable sorting algorithm, as it does not change the relative order of elements of the same value in the input.

**Question 2. Is iterative merge sort an in-place sorting algorithm?**

Answer: No, iterative merge sort is not an in-place sorting algorithm. In in-place sorting algorithms, only a small/constant auxiliary space is used; in iterative merge sort, we use auxiliary lists to merge to sub-arrays.

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*Article contributed Abhinav Tiwari*