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:
- Initialize a variable, say sum, to store the required sum.
- 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.
- Initialize rev as 0 to store reversal of each array element.
- 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.
- 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 <bits/stdc++.h>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 Codeint main(){ // Given arr[] int arr[] = { 7, 234, 58100 }; int n = sizeof(arr) / sizeof(arr[0]); // 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 Codeif __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 approachusing 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 Codestatic public void Main(){ // Given arr[] int[] arr = { 7, 234, 58100 }; // Function call totalSum(arr);}}// This code is contributed by akhilsaini |
Javascript
<script>// Javascript program for the above approach// Function to find the sum of elements// after reversing each element in arr[]function totalSum(arr, n){ // Stores the final sum let sum = 0; // Traverse the given array for(let i = 0; i < n; i++) { // Stores count of ending 0s let count = 0; let rev = 0, num = arr[i]; // Flag to avoid count of 0s // that doesn't ends with 0s let f = false; while (num > 0) { // Count of ending 0s while (num > 0 && !f && num % 10 == 0) { count++; num = Math.floor(num / 10); } // Update flag with true f = true; // Reversing the num if (num > 0) { rev = rev * 10 + num % 10; num = Math.floor(num / 10); } } // Add all ending 0s to // end of rev if (count > 0) rev = rev * Math.pow(10, count); // Update sum sum = sum + rev; } // Print total sum document.write(sum);}// Driver Code // Given arr[] let arr = [ 7, 234, 58100 ]; let n = arr.length; // Function call totalSum(arr, n);// This code is contributed by Mayank Tyagi</script> |
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
