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Array sum after dividing numbers from previous
  • Difficulty Level : Basic
  • Last Updated : 02 Feb, 2021

Find the sum of number of series after dividing the element of array from previous element.
Examples: 

Input : 3 7 9 10 12 18
Explanation:  3 + 7/3 + 9/7 + 10/9 + 
12/10 + 18/12 = 9 (taking only integer 
part)   
Output : 9

Input : 1 12 24 30 60
Output : 18 

Approach: We take elements in an array and divide the element from previous element. We do this process for all the elements of an array except very first element. Add the result after division and very first element.
Note: If any element is zero in an array then it fails to do the task and return minus one.  

C++




// C++  program for divide and
// sum the number of series
#include <bits/stdc++.h>
using namespace std;
 
int divideAndSum(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++) {
         
        // checking whether element
        // is zero or not
        if (arr[i] == 0)
            return -1;
         
        if (i == 0)
            sum += arr[i];
        else
 
            // divide element from
            // previous element
            sum += arr[i] / arr[i - 1];
    }
     
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 7, 9, 10, 12, 18 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << divideAndSum(arr, n);
    return 0;
}

Java




// java program for divide and
// sum the number of series
import java.io.*;
 
class GFG
{
     
    static int divideAndSum(int arr[], int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
             
            // checking whether element
            // is zero or not
            if (arr[i] == 0)
                return -1;
             
            if (i == 0)
                sum += arr[i];
            else
     
                // divide element from
                // previous element
                sum += arr[i] / arr[i - 1];
        }
         
        return sum;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 3, 7, 9, 10, 12, 18 };
        int n = arr.length;
        System.out.println( divideAndSum(arr, n));
 
    }
}
 
// This code is contributed by vt_m

Python3




# Python 3 program for divide and
# sum the number of series
 
def divideAndSum(arr, n):
  
    sum = 0
    for i in range(0,n): 
         
         # checking whether element
         # is zero or not
        if (arr[i] == 0):
            return -1
         
        if (i == 0):
            sum += arr[i]
        else:
 
                     # divide element from
             # previous element
             sum += int(arr[i] / arr[i - 1])
      
     
    return int(sum)
  
 
# Driver code
arr =   [3, 7, 9, 10, 12, 18]  
 
n = len(arr)
print(divideAndSum(arr, n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#




// C# program for divide and
// sum the number of series
using System;
 
class GFG
{
     
    static int divideAndSum(int []arr, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
             
            // checking whether element
            // is zero or not
            if (arr[i] == 0)
                return -1;
             
            if (i == 0)
                sum += arr[i];
            else
     
                // divide element from
                // previous element
                sum += arr[i] / arr[i - 1];
        }
         
        return sum;
    }
     
    // Driver code
    public static void Main ()
    {
        int []arr = { 3, 7, 9, 10, 12, 18 };
        int n = arr.Length;
        Console.WriteLine( divideAndSum(arr, n));
 
    }
}
 
// This code is contributed by vt_m

PHP




<?php
// php program for divide and
// sum the number of series
 
function divideAndSum($arr, $n)
{
    $sum = 0;
    for ($i = 0; $i < $n; $i++) {
         
        // checking whether element
        // is zero or not
        if ($arr[$i] == 0)
            return -1;
         
        if ($i == 0)
            $sum += $arr[$i];
        else
 
            // divide element from
            // previous element
            $sum += floor($arr[$i] /
                    $arr[$i - 1]);
    }
     
    return $sum;
}
 
// Driver code
{
    $arr = array(3, 7, 9, 10, 12, 18);
    $n = sizeof($arr) / sizeof($arr[0]);
    echo divideAndSum($arr, $n);
    return 0;
}
 
// This code is contributed by nitin mittal.
?>

Output: 

9

Time Complexity: O(N )

Auxiliary Space: O(N)




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