Replace every array element by multiplication of previous and next

1.5

Given an array of integers, update every element with multiplication of previous and next elements with following exceptions.
a) First element is replaced by multiplication of first and second.
b) Last element is replaced by multiplication of last and second last.

Input: arr[] = {2, 3, 4, 5, 6}
Output: arr[] = {6, 8, 15, 24, 30}

// We get the above output using following
// arr[] = {2*3, 2*4, 3*5, 4*6, 5*6} 

Source: Top 25 Interview Questions

A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.

An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Below is C++ implementation of this idea.

C

// C++ program to update every array element with
// multiplication of previous and next numbers in array
#include<iostream>
using namespace std;

void modify(int arr[], int n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
      return;

    // store current value of arr[0] and update it
    int prev = arr[0];
    arr[0] = arr[0] * arr[1];

    // Update rest of the array elements
    for (int i=1; i<n-1; i++)
    {
        // Store current value of next interation
        int curr = arr[i];

        // Update current value using previos value
        arr[i] = prev * arr[i+1];

        // Update previous value
        prev = curr;
    }

    // Update last array element
    arr[n-1] = prev * arr[n-1];
}

// Driver program
int main()
{
    int arr[] = {2, 3, 4, 5, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    modify(arr, n);
    for (int i=0; i<n; i++)
      cout << arr[i] << " ";
    return 0;
}

Java

// Java program to update every array element with
// multiplication of previous and next numbers in array
import java.io.*;
import java.util.*;
import java.lang.Math;

class Multipy
{
   static void modify(int arr[], int n)
    {
        // Nothing to do when array size is 1
        if (n <= 1)
            return;

        // store current value of arr[0] and update it
        int prev = arr[0];
        arr[0] = arr[0] * arr[1];

        // Update rest of the array elements
        for (int i=1; i<n-1; i++)
        {
            // Store current value of next interation
            int curr = arr[i];

            // Update current value using previos value
            arr[i] = prev * arr[i+1];

            // Update previous value
            prev = curr;
        }

        // Update last array element
        arr[n-1] = prev * arr[n-1];
    }

    // Driver program to test above function
    public static void main(String[] args)
    {
        int arr[] = {2, 3, 4, 5, 6};
        int n = arr.length;
        modify(arr, n);
        for (int i=0; i<n; i++)
         System.out.print(arr[i]+" ");
    }
}
/* This code is contributed by Devesh Agrawal */


Output:
6 8 15 24 30

This article is contributed by Ravi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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