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

Last Updated : 30 Apr, 2024

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 : 0Array is already a palindrome. So wedo not need any merge operation.Input : arr[] = {1, 4, 5, 1}Output : 1We can make given array palindrome withminimum one merging (merging 4 and 5 tomake 9)Input : arr[] = {11, 14, 15, 99}Output : 3We need to merge all elements to makea 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.

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).
2. 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).
3. 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++ ```// 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 ```// 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)); } } ``` Python3 ```# 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# ```// 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 ``` Javascript ```<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 // 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. ?> ```

Output
```Count of minimum operations is 1

```

Time Complexity: O(n)
Auxiliary Space: O(1)