Given an array of positive integers. We need to make the given array a ‘Palindrome’. The only allowed operation is”merging” (of two adjacent elements). Merging two adjacent elements means replacing them with their sum. The task is to find the minimum number of merge operations required to make the given array a ‘Palindrome’.
To make any array a palindrome, we can simply apply merge operation n-1 times where n is the size of the array (because a single-element array is always palindromic, similar to single-character string). In that case, the size of array will be reduced to 1. But in this problem, we are asked to do it in the minimum number of operations.
Example :
Input : arr[] = {15, 4, 15}
Output : 0
Array is already a palindrome. So we
do not need any merge operation.
Input : arr[] = {1, 4, 5, 1}
Output : 1
We can make given array palindrome with
minimum one merging (merging 4 and 5 to
make 9)
Input : arr[] = {11, 14, 15, 99}
Output : 3
We need to merge all elements to make
a palindrome.
The expected time complexity is O(n).
Let f(i, j) be minimum merging operations to make subarray arr[i..j] a palindrome. If i == j answer is 0. We start i from 0 and j from n-1.
- If arr[i] == arr[j], then there is no need to do any merging operations at index i or index j. Our answer in this case will be f(i+1, j-1).
-
Else, we need to do merging operations. Following cases arise.
- If arr[i] > arr[j], then we should do merging operation at index j. We merge index j-1 and j, and update arr[j-1] = arr[j-1] + arr[j]. Our answer in this case will be 1 + f(i, j-1).
- For the case when arr[i] < arr[j], update arr[i+1] = arr[i+1] + arr[i]. Our answer in this case will be 1 + f(i+1, j).
- Our answer will be f(0, n-1), where n is the size of array arr[].
Therefore this problem can be solved iteratively using two pointers (first pointer pointing to start of the array and second pointer pointing to the last element of the array) method and keeping count of total merging operations done till now.
Below is an implementation of the above idea.
// C++ program to find number of operations // to make an array palindrome #include <bits/stdc++.h> using namespace std;
// Returns minimum number of count operations // required to make arr[] palindrome int findMinOps( int arr[], int n)
{ int ans = 0; // Initialize result
// Start from two corners
for ( int i=0,j=n-1; i<=j;)
{
// If corner elements are same,
// problem reduces arr[i+1..j-1]
if (arr[i] == arr[j])
{
i++;
j--;
}
// If left element is greater, then
// we merge right two elements
else if (arr[i] > arr[j])
{
// need to merge from tail.
j--;
arr[j] += arr[j+1] ;
ans++;
}
// Else we merge left two elements
else
{
i++;
arr[i] += arr[i-1];
ans++;
}
}
return ans;
} // Driver program to test above int main()
{ int arr[] = {1, 4, 5, 9, 1};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << "Count of minimum operations is "
<< findMinOps(arr, n) << endl;
return 0;
} |
// Java program to find number of operations // to make an array palindrome class GFG
{ // Returns minimum number of count operations
// required to make arr[] palindrome
static int findMinOps( int [] arr, int n)
{
int ans = 0 ; // Initialize result
// Start from two corners
for ( int i= 0 ,j=n- 1 ; i<=j;)
{
// If corner elements are same,
// problem reduces arr[i+1..j-1]
if (arr[i] == arr[j])
{
i++;
j--;
}
// If left element is greater, then
// we merge right two elements
else if (arr[i] > arr[j])
{
// need to merge from tail.
j--;
arr[j] += arr[j+ 1 ] ;
ans++;
}
// Else we merge left two elements
else
{
i++;
arr[i] += arr[i- 1 ];
ans++;
}
}
return ans;
}
// Driver method to test the above function
public static void main(String[] args)
{
int arr[] = new int []{ 1 , 4 , 5 , 9 , 1 } ;
System.out.println( "Count of minimum operations is " +
findMinOps(arr, arr.length));
}
} |
# Python program to find number of operations # to make an array palindrome # Returns minimum number of count operations # required to make arr[] palindrome def findMinOps(arr, n):
ans = 0 # Initialize result
# Start from two corners
i,j = 0 ,n - 1
while i< = j:
# If corner elements are same,
# problem reduces arr[i+1..j-1]
if arr[i] = = arr[j]:
i + = 1
j - = 1
# If left element is greater, then
# we merge right two elements
elif arr[i] > arr[j]:
# need to merge from tail.
j - = 1
arr[j] + = arr[j + 1 ]
ans + = 1
# Else we merge left two elements
else :
i + = 1
arr[i] + = arr[i - 1 ]
ans + = 1
return ans
# Driver program to test above arr = [ 1 , 4 , 5 , 9 , 1 ]
n = len (arr)
print ( "Count of minimum operations is " + str (findMinOps(arr, n)))
# This code is contributed by Pratik Chhajer |
// C# program to find number of operations // to make an array palindrome using System;
class GFG
{ // Returns minimum number of count operations
// required to make arr[] palindrome
static int findMinOps( int []arr, int n)
{
int ans = 0; // Initialize result
// Start from two corners
for ( int i = 0, j = n - 1; i <= j;)
{
// If corner elements are same,
// problem reduces arr[i+1..j-1]
if (arr[i] == arr[j])
{
i++;
j--;
}
// If left element is greater, then
// we merge right two elements
else if (arr[i] > arr[j])
{
// need to merge from tail.
j--;
arr[j] += arr[j + 1] ;
ans++;
}
// Else we merge left two elements
else
{
i++;
arr[i] += arr[i-1];
ans++;
}
}
return ans;
}
// Driver Code
public static void Main()
{
int []arr = new int []{1, 4, 5, 9, 1} ;
Console.Write( "Count of minimum operations is " +
findMinOps(arr, arr.Length));
}
} // This code is contributed by nitin mittal |
<script> // JavaScript program to find number of operations // to make an array palindrome // Returns minimum number of count operations
// required to make arr[] palindrome
function findMinOps(arr, n)
{
let ans = 0; // Initialize result
// Start from two corners
for (let i=0,j=n-1; i<=j;)
{
// If corner elements are same,
// problem reduces arr[i+1..j-1]
if (arr[i] == arr[j])
{
i++;
j--;
}
// If left element is greater, then
// we merge right two elements
else if (arr[i] > arr[j])
{
// need to merge from tail.
j--;
arr[j] += arr[j+1] ;
ans++;
}
// Else we merge left two elements
else
{
i++;
arr[i] += arr[i-1];
ans++;
}
}
return ans;
}
// Driver Code let arr = [1, 4, 5, 9, 1];
document.write( "Count of minimum operations is " +
findMinOps(arr, arr.length));
</script> |
<?php // PHP program to find number // of operations to make an // array palindrome // Returns minimum number of // count operations required // to make arr[] palindrome function findMinOps( $arr , $n )
{ // Initialize result
$ans = 1;
// Start from two corners
for ( $i = 0, $j = $n - 1; $i <= $j 😉
{
// If corner elements are same,
// problem reduces arr[i+1..j-1]
if ( $arr [ $i ] == $arr [ $j ])
{
$i ++;
$j --;
}
// If left element is greater, then
// we merge right two elements
else if ( $arr [ $i ] > $arr [ $j ])
{
// need to merge from tail.
$j --;
$arr [ $j ] += $arr [ $j + 1] ;
$ans ++;
}
// Else we merge
// left two elements
else
{
$i ++;
$arr [ $i ] += $arr [ $i - 1];
$ans ++;
}
}
return $ans ;
} // Driver Code $arr [] = array (1, 4, 5, 9, 1);
$n = sizeof( $arr );
echo "Count of minimum operations is " ,
findMinOps( $arr , $n ) ;
// This code is contributed by nitin mittal. ?> |
Count of minimum operations is 1
Time Complexity: O(n)
Auxiliary Space: O(1)
Approach#2: Using dynamic programming
Create a 2D array dp of size n x n where n is the length of the input array arr. Initialize all elements of dp to 0. For each gap value from 1 to n – 1, loop over all i values from 0 to n – gap – 1. Compute the corresponding j value as i + gap. If arr[i] is equal to arr[j], then set dp[i][j] to dp[i + 1][j – 1].Otherwise, set dp[i][j] to min(dp[i][j – 1], dp[i + 1][j]) + 1. The minimum number of merge operations required to make arr a palindrome is dp[0][n – 1].
Algorithm
1. Initialize count variable to 0.
2. Initialize two pointers i and j to the start and end of the array respectively.
3. While i is less than j,
a. If arr[i] is equal to arr[j], increment i and decrement j.
b. Else, if arr[i] is less than arr[j], increment i by 1 and add arr[i-1] to arr[i]. Increment count by 1.
c. Else, decrement j by 1 and add arr[j+1] to arr[j]. Increment count by 1.
4. Return count.
#include <iostream> #include <vector> using namespace std;
int minMergeOperations(vector< int > arr)
{ int n = arr.size();
vector<vector< int > > dp(n, vector< int >(n, 0));
// Fill the dp table
for ( int gap = 1; gap < n; gap++) {
for ( int i = 0; i < n - gap; i++) {
int j = i + gap;
if (arr[i] == arr[j]) {
dp[i][j] = dp[i + 1][j - 1];
}
else {
dp[i][j]
= min(dp[i][j - 1], dp[i + 1][j]) + 1;
}
}
}
return dp[0][n - 1];
} int main()
{ vector< int > arr = { 11, 14, 15, 99 };
cout << minMergeOperations(arr) << endl;
return 0;
} |
import java.util.Arrays;
import java.util.Vector;
public class MinMergeOperations {
// Function to calculate the minimum number of merge operations
// to make the array palindrome
static int minMergeOperations(Vector<Integer> arr) {
int n = arr.size();
// Create a 2D array for dynamic programming
int [][] dp = new int [n][n];
// Fill the dp table
for ( int gap = 1 ; gap < n; gap++) {
for ( int i = 0 ; i < n - gap; i++) {
int j = i + gap;
// If the elements at the ends are equal,
// no additional merge operation is needed
if (arr.get(i).equals(arr.get(j))) {
dp[i][j] = dp[i + 1 ][j - 1 ];
}
// If the elements are different, choose the minimum
// of the two adjacent elements and add 1 for the merge
else {
dp[i][j] = Math.min(dp[i][j - 1 ], dp[i + 1 ][j]) + 1 ;
}
}
}
// The result is stored at the top-right corner of the dp table
return dp[ 0 ][n - 1 ];
}
public static void main(String[] args) {
// Example array
Vector<Integer> arr = new Vector<>(Arrays.asList( 11 , 14 , 15 , 99 ));
// Print the minimum merge operations needed
System.out.println(minMergeOperations(arr));
}
} // This code is contributed by shivamgupta0987654321 |
def min_merge_operations(arr):
n = len (arr)
dp = [[ 0 for j in range (n)] for i in range (n)]
for gap in range ( 1 , n):
for i in range (n - gap):
j = i + gap
if arr[i] = = arr[j]:
dp[i][j] = dp[i + 1 ][j - 1 ]
else :
dp[i][j] = min (dp[i][j - 1 ], dp[i + 1 ][j]) + 1
return dp[ 0 ][n - 1 ]
arr = [ 11 , 14 , 15 , 99 ]
print (min_merge_operations(arr))
|
using System;
using System.Collections.Generic;
class Program {
// Function to find the minimum merge operations
static int MinMergeOperations(List< int > arr)
{
int n = arr.Count;
int [][] dp = new int [n][];
for ( int i = 0; i < n; i++) {
dp[i] = new int [n];
}
// Fill the dp table using dynamic programming
for ( int gap = 1; gap < n; gap++) {
for ( int i = 0; i < n - gap; i++) {
int j = i + gap;
// If the elements at the ends are equal, no
// merge is needed.
if (arr[i] == arr[j]) {
dp[i][j] = dp[i + 1][j - 1];
}
else {
// If the elements at the ends are
// different, we take the minimum of
// merging from the left and merging
// from the right, and add 1.
dp[i][j] = Math.Min(dp[i][j - 1],
dp[i + 1][j])
+ 1;
}
}
}
// The result is stored at the top-right corner of
// the dp table.
return dp[0][n - 1];
}
static void Main()
{
List< int > arr = new List< int >{ 11, 14, 15, 99 };
int minMergeOps = MinMergeOperations(arr);
Console.WriteLine( "Minimum merge operations: "
+ minMergeOps);
}
} |
function min_merge_operations(arr) {
let n = arr.length;
let dp = Array.from(Array(n), () => Array(n).fill(0));
for (let gap = 1; gap < n; gap++) {
for (let i = 0; i < n - gap; i++) {
let j = i + gap;
if (arr[i] == arr[j]) {
dp[i][j] = dp[i + 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i][j - 1], dp[i + 1][j]) + 1;
}
}
}
return dp[0][n - 1];
} let arr = [11, 14, 15, 99]; console.log(min_merge_operations(arr)); |
3
Time Complexity: O(n), where n is length of array
Auxiliary Space: O(1)