Program to find Minimum number of merge operations required to make an array palindrome in C++

October 13, 2022

Minimum number of merge operations required to make an array palindrome in C++

Here, in this page we will discuss the program to find Minimum number of merge operations required to make an array palindrome in C++ programming language. We are given with an array and we need to print an integer value denoting the number of operations required.

Minimum number of merge operations

Method 1 :

  • Declare a variable say, count=0, that will count the required merging operation.
  • Take two variables, i=0, and j=n-1.
  • Now, run a loop till i<j, inside loop check if (arr[i]==arr[j]), then increase the value of i by 1 and decrease the value of j by 1.
  • Else if (arr[i]>arr[j]), then set arr[j-1] = arr[j-1]+arr[j] and decrease the value of j and increase the value of count by 1.
  • Else set, arr[i+1] = arr[i]+arr[i+1] and increase the value of i and count by 1.
  • After the traversal print the value of count.

Time and Space Complexity :

  • Time-Complexity : O(n)
  • Space-Complexity : O(1)
Number of merge operations

Code in C++

Run 
#include<bits/stdc++.h>
using namespace std;

int main(){

int arr[] = {1, 4, 5, 9, 1};
int n = sizeof(arr)/sizeof(arr[0]), count = 0;

int i=0, j=n-1;

while(i< j){ 
    if(arr[i]==arr[j])
    { 
        i++; 
        j--; 
        
    } 
    else if(arr[i]>arr[j]){
        arr[j-1] = arr[j]+arr[j-1];
        j--;
        count++;
    }
else{
    arr[i+1] = arr[i]+arr[i+1];
    i++;
    count++;
    }
}

cout<< "Required Minimum Operations : "<< count;
}

Output :
Required Minimum Operations : 1

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Method 2 (Using Recursion) :

  • Create a recursive function say, fun() pass arr and two values 0 and n-1.
  • In the fun(), return 0, if(i==j or i>j)
  • Otherwise, check if (arr[i]==arr[j]), then return(1+fun(arr, i+1, j-1))
  • Else check if (arr[i]>arr[j]), then set arr[j-1] = arr[j-1]+arr[j] and return(1+fun(arr, i, j-1))
  • Else set, arr[i+1] = arr[i]+arr[i+1] and return(1+fun(arr, i+1, j)).

Time and Space Complexity :

  • Time-Complexity : O(n)
  • Space-Complexity : O(1)

Code in C++

Run 
#include<bits/stdc++.h>
using namespace std;

int fun(int arr[], int i, int j){

      if( i==j or i>j )
         return 0;

     if(arr[i]==arr[j])
         return fun(arr, i+1, j-1);

     else if(arr[i]>arr[j]){
          arr[j-1] = arr[j]+arr[j-1];
          return (1+fun( arr, i, j-1));
     }
     else{
          arr[i+1] = arr[i]+arr[i+1];
          return (1+fun( arr, i+1, j));
     }

}

int main(){

int arr[] = {1, 4, 5, 9, 1};
int n = sizeof(arr)/sizeof(arr[0]);

cout<<"Required Minimum Operations : "<< fun(arr, 0 , n-1);
}
Output :
Required Minimum Operations : 1

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