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Sum of array elements after reversing each element
• Difficulty Level : Easy
• Last Updated : 09 Apr, 2021

Given an array arr[] consisting of N positive integers, the task is to find the sum of all array elements after reversing digits of every array element.

Examples:

Input: arr[] = {7, 234, 58100}
Output: 18939
Explanation:
Modified array after reversing each array elements = {7, 432, 18500}.
Therefore, the sum of the modified array = 7 + 432 + 18500 = 18939.

Input: arr[] = {0, 100, 220}
Output: 320
Explanation:
Modified array after reversing each array elements = {0, 100, 220}.
Therefore, the sum of the modified array = 0 + 100 + 220 = 320.

Approach: The idea is to reverse each number of the given array as per the given conditions and find sum of all array elements formed after reversing. Below steps to solve the problem:

1. Initialize a variable, say sum, to store the required sum.
2. Initialize variable count as 0 and f as false to store count of ending 0s of arr[i] and flag to avoid all non-ending 0.
3. Initialize rev as 0 to store reversal of each array element.
4. Traverse the given array and for each array element do the following operation:
• Increment count and divide arr[i] by 10 until all the zeros at the end are traversed.
• Reset f with true that means all ending 0 digits have been considered.
• Now, reverse arr[i] by updating rev = rev*10 + arr[i] % 10 and  arr[i] = arr[i]/10.
• After traversing each digit of arr[i], update rev = rev * Math.pow(10, count) to add all ending 0’s to the end of reversed number.
5. For each reverse number in the above step, add that value to the resultant sum.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the sum of elements``// after reversing each element in arr[]``void` `totalSum(``int` `arr[], ``int` `n)``{``    ` `    ``// Stores the final sum``    ``int` `sum = 0;` `    ``// Traverse the given array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Stores count of ending 0s``        ``int` `count = 0;` `        ``int` `rev = 0, num = arr[i];` `        ``// Flag to avoid count of 0s``        ``// that doesn't ends with 0s``        ``bool` `f = ``false``;` `        ``while` `(num > 0)``        ``{``            ` `            ``// Count of ending 0s``            ``while` `(num > 0 && !f &&``                   ``num % 10 == 0)``            ``{``                ``count++;``                ``num = num / 10;``            ``}` `            ``// Update flag with true``            ``f = ``true``;` `            ``// Reversing the num``            ``if` `(num > 0)``            ``{``                ``rev = rev * 10 +``                      ``num % 10;` `                ``num = num / 10;``            ``}``        ``}` `        ``// Add all ending 0s to``        ``// end of rev``        ``if` `(count > 0)``            ``rev = rev * ``pow``(10, count);` `        ``// Update sum``        ``sum = sum + rev;``    ``}` `    ``// Print total sum``    ``cout << sum;``}` `// Driver Code``int` `main()``{``    ` `    ``// Given arr[]``    ``int` `arr[] = { 7, 234, 58100 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``  ` `    ``// Function call``    ``totalSum(arr, n);``  ` `    ``return` `0;``}` `// This code is contributed by akhilsaini`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG {` `    ``// Function to find the sum of elements``    ``// after reversing each element in arr[]``    ``static` `void` `totalSum(``int``[] arr)``    ``{``        ``// Stores the final sum``        ``int` `sum = ``0``;` `        ``// Traverse the given array``        ``for` `(``int` `i = ``0``;``            ``i < arr.length; i++) {` `            ``// Stores count of ending 0s``            ``int` `count = ``0``;` `            ``int` `rev = ``0``, num = arr[i];` `            ``// Flag to avoid count of 0s``            ``// that doesn't ends with 0s``            ``boolean` `f = ``false``;` `            ``while` `(num > ``0``) {` `                ``// Count of ending 0s``                ``while` `(num > ``0` `&& !f``                    ``&& num % ``10` `== ``0``) {``                    ``count++;``                    ``num = num / ``10``;``                ``}` `                ``// Update flag with true``                ``f = ``true``;` `                ``// Reversing the num``                ``if` `(num > ``0``) {``                    ``rev = rev * ``10``                        ``+ num % ``10``;` `                    ``num = num / ``10``;``                ``}``            ``}` `            ``// Add all ending 0s to``            ``// end of rev``            ``if` `(count > ``0``)``                ``rev = rev``                    ``* (``int``)Math.pow(``10``,``                                    ``count);` `            ``// Update sum``            ``sum = sum + rev;``        ``}` `        ``// Print total sum``        ``System.out.print(sum);``    ``}` `    ``// Driver Code``    ``public` `static` `void``        ``main(String[] args)``    ``{``        ``// Given arr[]``        ``int``[] arr = { ``7``, ``234``, ``58100` `};` `        ``// Function Call``        ``totalSum(arr);``    ``}``}`

## Python3

 `# Python3 program for the above approach` `# Function to find the sum of elements``# after reversing each element in arr[]``def` `totalSum(arr, n):``    ` `    ``# Stores the final sum``    ``sums ``=` `0` `    ``# Traverse the given array``    ``for` `i ``in` `range``(``0``, n):``        ` `        ``# Stores count of ending 0s``        ``count ``=` `0` `        ``rev ``=` `0``        ``num ``=` `arr[i]` `        ``# Flag to avoid count of 0s``        ``# that doesn't ends with 0s``        ``f ``=` `False` `        ``while` `num > ``0``:` `            ``# Count of ending 0s``            ``while` `(num > ``0` `and` `f ``=``=` `False` `and``                   ``num ``%` `10` `=``=` `0``):``                ``count ``=` `count ``+` `1``                ``num ``=` `num ``/``/` `10` `            ``# Update flag with true``            ``f ``=` `True` `            ``# Reversing the num``            ``if` `num > ``0``:``                ``rev ``=` `rev ``*` `10` `+` `num ``%` `10``                ``num ``=` `num ``/``/` `10` `        ``# Add all ending 0s to``        ``# end of rev``        ``if` `(count > ``0``):``            ``rev ``=` `rev ``*` `pow``(``10``, count)` `            ``# Update sum``        ``sums ``=` `sums ``+` `rev` `    ``# Print total sum``    ``print``(sums)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``# Given arr[]``    ``arr ``=` `[ ``7``, ``234``, ``58100` `]` `    ``n ``=` `len``(arr)` `    ``# Function call``    ``totalSum(arr, n)` `# This code is contributed by akhilsaini`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the sum of elements``// after reversing each element in arr[]``static` `void` `totalSum(``int``[] arr)``{``    ` `    ``// Stores the final sum``    ``int` `sum = 0;` `    ``// Traverse the given array``    ``for``(``int` `i = 0; i < arr.Length; i++)``    ``{``        ` `        ``// Stores count of ending 0s``        ``int` `count = 0;` `        ``int` `rev = 0, num = arr[i];` `        ``// Flag to avoid count of 0s``        ``// that doesn't ends with 0s``        ``bool` `f = ``false``;` `        ``while` `(num > 0)``        ``{``            ` `            ``// Count of ending 0s``            ``while` `(num > 0 && !f &&``                   ``num % 10 == 0)``            ``{``                ``count++;``                ``num = num / 10;``            ``}` `            ``// Update flag with true``            ``f = ``true``;` `            ``// Reversing the num``            ``if` `(num > 0)``            ``{``                ``rev = rev * 10 +``                      ``num % 10;` `                ``num = num / 10;``            ``}``        ``}` `        ``// Add all ending 0s to``        ``// end of rev``        ``if` `(count > 0)``            ``rev = rev * (``int``)Math.Pow(10, count);` `        ``// Update sum``        ``sum = sum + rev;``    ``}` `    ``// Print total sum``    ``Console.Write(sum);``}` `// Driver Code``static` `public` `void` `Main()``{` `    ``// Given arr[]``    ``int``[] arr = { 7, 234, 58100 };` `    ``// Function call``    ``totalSum(arr);``}``}` `// This code is contributed by akhilsaini`

## Javascript

 ``
Output:
`18939`

Time Complexity: O(N*log10M), where N denotes the length of the array and M denotes maximum array element.
Auxiliary Space: O(1)

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