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.
Implementation:
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 codeint 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 seriesimport 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 seriesdef 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 codearr = [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 seriesusing 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 seriesfunction 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.?> |
Javascript
<script>// javascript program for divide and// sum the number of series function divideAndSum(arr , n) { var 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 += parseInt(arr[i] / arr[i - 1]); } return sum; } // Driver code var arr = [ 3, 7, 9, 10, 12, 18 ]; var n = arr.length; document.write( divideAndSum(arr, n));// This code contributed by umadevi9616 </script> |
Output
9
Time Complexity: O(N )
Auxiliary Space: O(1)
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