# Find minimum number of merge operations to make an array palindrome in C++

C++Server Side ProgrammingProgramming

#### C in Depth: The Complete C Programming Guide for Beginners

45 Lectures 4.5 hours

#### Practical C++: Learn C++ Basics Step by Step

Most Popular

50 Lectures 4.5 hours

#### Master C and Embedded C Programming- Learn as you go

66 Lectures 5.5 hours

In this problem, we are given an array arr[] consisting of n positive numbers. Our task is to Find minimum number of merge operations to make an array palindrome.

Palindrome array are similar to palindrome strings, the elements at index i and n-i should be the same.

## Example

{5, 1, 7, 2, 7, 1, 5}

Problem Description − We need to make the array palindrome by performing operations on it. And the only operation which is valid on the array is the merge operation in which we will add elements at index i and i+1.

We need to return the minimum number of such operations required to make the given array palindrome.

Let’s take an example to understand the problem,

## Input

arr[] = {4, 1, 7, 6, 1, 5}

## Output

2

## Explanation

We need two merge operations,

Merging elements at index 0 and 1, makes the array {5, 7, 6, 1, 5}.

After this merging elements at index 2 and 3, makes the array {5, 7, 7, 5}.

## Solution Approach

A simple solution to the problem is by finding the number of operations to make the string palindrome. This is done by using two pointers start and end. If both the points meet i.e. start == end the array is a palindrome. So, we will loop for start and end pointer and perform the operation based on these conditions −

• If arr[start] == arr[end], this means for the current index, the array satisfies palindrome condition. For will move them to the next index. I.e. start++ and end--.

• If arr[start] > arr[end], in this case we need to perform the merge operation for the end index and increment the mergeCount by 1.

• If arr[start] < arr[end], in this case we need to perform the merge operation for the start index and increment the mergeCount by 1.

We will return the merge count when the start == end.

Program to illustrate the working of our solution,

## Example

Live Demo

#include <iostream>
using namespace std;
int findMergeCount(int arr[], int n){
int mergeCount = 0;
int start = 0;
int end = n-1;
while(start <= end){
if (arr[start] == arr[end]){
start++;
end--;
}
else if (arr[start] > arr[end]){
end--;
arr[end] += arr[end+1] ;
mergeCount++;
} else {
start++;
arr[start] += arr[start-1];
mergeCount++;
}
}
return mergeCount;
}
int main(){
int arr[] = {4, 1, 7, 6, 1, 5};
int n = sizeof(arr)/sizeof(arr);
cout<<"The minimum number of merge operations required to make the array palindrome is "<<findMergeCount(arr, n);
return 0;
}

## Output

The minimum number of merge operations required to make the array palindrome is 2