Q:

Given an array A of size N. Find the minimum number of operations needed to convert the given array to Palindromic Array

0

Palindromic Array

Given an array A of size NFind the minimum number of operations needed to convert the given array to 'Palindromic Array'.

The only allowed operation is that you can merge two adjacent elements in the array and replace them with their sum.

Example:

Input array:
5 3 3 3
No of minimum operation: 3 (don't worry, you will know the reason soon!!)
Input array:
5 3 3 5
No of minimum operation: 0 (It's palindromic array!!)

All Answers

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What is palindromic array?

An array is known to palindromic if it forms a palindrome with all its elements.

What are the operations allowed to form a palindromic array from any input array?

The only operation allowed is to merge two adjacent elements and replacing them with their sum.

Can we convert any input array to a palindromic array?

Yah, we can convert any input array to a palindromic array since an array with a single element is always a palindromic array. Thus, any input array can ultimately lead to a palindromic array.

Algorithm:

For solving the above problem a recursive algorithm can be constructed based on the following conditions.

Let array [i, j] be the subarray to be checked where I be the staring index & j be the end index

So there can be three cases,

1)  array[i]==array[j]
2)  array[i]>array[j]
3)  array[i]<array[j]

Based on these three conditions we have constructed the recursive function:

findNoOfOperation(input array, starting index, ending index)
Let,
i=start index
j=end index

FUNCTION findNoOfOperation(input array, i, j)
BASE CASE
IF(i==j)
    return 0; //no more operations needed
    END IF
IF (i<=j)
IF(a[i]==a[j])
No of operations needed for array(i, j) is same as array(i+1, j-1)
return findNoOfOperation (array,i+1,j-1);//recursive call
ELSEIF(a[i]>a[j])
    We need to merge the two element at the end 
    to convert it into palindrome array
    array [j-1]=array[j]+array[j-1]; //merge
	No of operations needed for array(i, j) is one more than array(i, j-1)
   return findNoOfOperation(array,i,j-1)+1;
ELSE
	We need to merge the two element at the start 
    to convert it into palindrome array
    array[i+1]=array[i]+array[i+1];
	No of operations needed for array(i, j) is one more than array(i+1, j)
returnfindNoOfOperation(array,i+1,j)+1;
END IF-ELSE
END IF
END FUNCTION

 

Example with explanation:

For input array:
5 3 3 3
Array size: 4
In the main function we call findNoOfOperation(array, 0, 3) to check the whole array
Start=0
End=3 //n-1
Array to be considered
5	3	3	3

Start referred as i & end referred as j from now
findNoOfOperation(array, 0, 3)
i=0
j=3
i !=j
i<j
array[i]=5 //array[0]=5
array[j]=3 //array[3]=3
array[i]>array[j]
merge at the end
array[3-1]=array[3-1]+array[3]
array[2]=6
now array is: 
(we are not deleting any element, rather omitting out from our consideration)
5	3	6

It returns findNoOfOperation(array, 0, 3-1) +1
Call to findNoOfOperation(array, 0, 2)
------------------------------------------------------------------------
		
findNoOfOperation(array, 0, 2)
i=0
j=2
i !=j
i<j
array[i]=5 //array[0]=5
array[j]=6 //array[2]=6
array[i]<array[j]
merge at the start
array[0+1]=array[0]+array[0+1]
array[1]=8
now array is: 
(we are not deleting any element, rather omitting out from our consideration)
8	6

It returns findNoOfOperation(array, 0+1, 2) +1
Call to findNoOfOperation(array, 1, 2)
------------------------------------------------------------------------

findNoOfOperation(array, 1, 2)
i=1
j=2
i !=j
i<j
array[i]=8 //array[1]=8
array[j]=6 //array[2]=6
array[i]>array[j]
merge at the end
array[2-1]=array[2]+array[2-1]
array[1]=14
now array is: 
(we are not deleting any element, rather omitting out from our consideration)
14

It returns findNoOfOperation(array, 1, 2-1) +1
Call to findNoOfOperation(array, 1, 1)
------------------------------------------------------------------------

findNoOfOperation(array, 1, 1)
i=1
j=1
i==j
return 0;
------------------------------------------------------------------------

At findNoOfOperation(array, 1, 2)
It returns findNoOfOperation(array, 1, 1) +1 =1
------------------------------------------------------------------------

At findNoOfOperation(array, 0, 2)
It returns findNoOfOperation(array, 1, 2) +1 =1+1=2
------------------------------------------------------------------------

At findNoOfOperation(array, 0, 3)
It returns findNoOfOperation(array, 0, 2) +1 =2+1=3
------------------------------------------------------------------------

At main function we called findNoOfOperation(array, 0, 3)
Hence the no of operation we needed is : 3
And the palindromic array converted to is 14 (single element array)
 

C++ implementation:

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

//to print
void print(vector<int> a,int n){
for(int i=0;i<n;i++)
	cout<<a[i]<<" ";
cout<<endl;
}


int findNoOfOperation(vector<int> a, int i,int j){
	//base case
	if(i==j)
		return 0;

	if(i<=j){
		//process according to three conditions 
		//deiscussed previously
		if(a[i]==a[j])
			return findNoOfOperation(a,i+1,j-1);

		else if(a[i]>a[j]){
			a[j-1]=a[j]+a[j-1];
			return findNoOfOperation(a,i,j-1)+1;
		}   
		else{
			a[i+1]=a[i]+a[i+1];
			return findNoOfOperation(a,i+1,j)+1;
		}
	}
	return 0;
}


int main(){
	int n,item;
	
	cout<<"enter array size: ";
	scanf("%d",&n);
	
	vector<int> a;
	
	cout<<"input the array to be converted of size: "<<n<<endl;
	for(int j=0;j<n;j++){
		scanf("%d",&item);
		a.push_back(item);
	}

	cout<<"your input array is:\n";
	print(a,n);
	
	cout<<"No of operation needed to convert this array";
	cout<<" to a palindromic array is:";
	cout<<findNoOfOperation(a,0,n-1)<<endl;
	
	return 0;
}

Output

First run:
enter array size: 4
input the array to be converted of size: 4
5 3 3 3
your input array is:
5 3 3 3
No of operation needed to convert this array to a palindromic array is:3   

Second run:
enter array size: 6
input the array to be converted of size: 6
5 2 3 1 4 5
your input array is:
5 2 3 1 4 5
No of operation needed to convert this array to a palindromic array is:2

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interview C++ coding problems/challenges | Recursion

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