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Smallest array that can be obtained by replacing adjacent pairs with their products
  • Last Updated : 06 Oct, 2020

Given an array arr[] of size N, the task is to print the least possible size the given array can be reduced to by performing the following operations: 

  • Remove any two adjacent elements, say arr[i] and arr[i+1] and insert a single element arr[i] * arr[i+1] at that position in the array.
  • If all array elements become equal, then print the size of the array.

Examples: 

Input: arr[] = {1, 7, 7, 1, 7, 1} 
Output:
Explanation: 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {7, 7, 1, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 1, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343} 

Input: arr[] = {2, 2, 2, 2} 
Output:

 

Approach: The approach is based on the idea that if all array elements are same, then the given operations can’t be performed on the given array. Otherwise, in every case, the array size can be reduced to 1. Below are the steps:



  1. Iterate over the given array arr[].
  2. If all elements of the array are same, then print N as the required answer.
  3. Otherwise, always pick adjacent elements which gives the maximum product to reduce the size of arr[] to 1. Therefore, the minimum possible size will be 1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to minimize the size of
// array by performing given operations
int minLength(int arr[], int N)
{
 
    for (int i = 1; i < N; i++) {
 
        // If all array elements
        // are not same
        if (arr[0] != arr[i]) {
            return 1;
        }
    }
 
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 3, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << minLength(arr, N);
 
    return 0;
}

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Java

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// Java implementation of the above approach
import java.io.*;
 
class GFG{
     
// Function to minimize the size of
// array by performing given operations
static int minLength(int arr[], int N)
{
    for(int i = 1; i < N; i++)
    {
         
        // If all array elements
        // are not same
        if (arr[0] != arr[i])
        {
            return 1;
        }
    }
     
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 2, 1, 3, 1 };
    int N = arr.length;
 
    // Function call
    System.out.print(minLength(arr, N));
}
}
 
// This code is contributed by akhilsaini

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Python3

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# Python3 implementation of the above approach
 
# Function to minimize the size of
# array by performing given operations
def minLength(arr, N):
     
    for i in range(1, N):
         
        # If all array elements
        # are not same
        if (arr[0] != arr[i]):
            return 1
             
    # If all array elements
    # are same
    return N
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 2, 1, 3, 1 ]
    N = len(arr)
     
    # Function call
    print(minLength(arr, N))
 
# This code is contributed by akhilsaini

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C#

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// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to minimize the size of
// array by performing given operations
static int minLength(int[] arr, int N)
{
    for(int i = 1; i < N; i++)
    {
         
        // If all array elements
        // are not same
        if (arr[0] != arr[i])
        {
            return 1;
        }
    }
 
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 2, 1, 3, 1 };
    int N = arr.Length;
 
    // Function call
    Console.Write(minLength(arr, N));
}
}
 
// This code is contributed by akhilsaini

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Output: 

1



 

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

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