Modify a given array by replacing each element with the sum or product of their digits based on a given condition

Last Updated : 16 Jul, 2021

Given an array arr[] consisting of N integers, the task is to modify the array elements after performing only one of the following operations on each array elements:

If the count of even digits is greater than the count of odd digits in an array element, then update that element to the sum of all the digits of that element.
Otherwise, update that element to the product of all the digits of that element.

Examples:

Input: arr[] = {113, 141, 214, 3186}
Output: 3 4 7 3186
Explanation:
Following are the operation performed on each array elements:

For element arr[0](= 113): count of even and odd digits are 0 and 3. As count of even < count of odd digit, therefore update arr[0](= 113) to the product of each digit of the number 113 i.e., 1 * 1 * 3 = 3.
For element arr[1](= 141): count of even and odd digits are 1 and 2. As count of even < count of odd digit, therefore update arr[1](= 141) to the product of each digit of the number 141 i.e., 1 * 4 * 1 = 4.
For element arr[2]:(= 214) count of even and odd digits are 2 and 1. As count of even > count of odd digit, therefore update arr[2](= 214) to the sum of each digit of the number 214 i.e., 2 + 1 + 4 = 7.
For element arr[3](= 3186): count of even and odd digits are 2 and 2. As count of even is the same as the count of odd digit, then no operation is performed. Therefore, arr[3](= 3186) remains the same.

After the above operations, the array modifies to {3, 4, 7, 3186}.

Input: arr[] = {2, 7, 12, 22, 110}
Output: 2 7 12 4 0

Approach: The given problem can be solved by performing the given operations for each array element and print the result accordingly. Follow the steps below to solve the problem:

Traverse the given array arr[] and perform the following steps:
- Find the count of even and odd digits of the current element of the array.
- If the count of even and odd digits are the same, then no operation is needed to perform.
- If the count of even digits is greater than the count of odd digits in an array element, then update that element to the sum of all the digits of that element.
- Otherwise, update that element to the product of all the digits of that element.
After completing the above steps, print the array arr[] as the modified array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach  #include <bits/stdc++.h> using namespace std;  // Function to modify the given array // as per the given conditions void evenOdd(int arr[], int N) {     // Traverse the given array arr[]     for (int i = 0; i < N; i++) {          // Initialize the count of even         // and odd digits         int even_digits = 0;         int odd_digits = 0;          // Initialize temp with the         // current array element         int temp = arr[i];          // For count the number of         // even digits         while (temp) {              // Increment the odd count             if ((temp % 10) & 1)                 odd_digits++;              // Otherwise             else                 even_digits++;              // Divide temp by 10             temp /= 10;         }          // Performe addition         if (even_digits > odd_digits) {              int res = 0;             while (arr[i]) {                  res += arr[i] % 10;                 arr[i] /= 10;             }             cout << res << " ";         }          // Performe multiplication         else if (odd_digits > even_digits) {              int res = 1;             while (arr[i]) {                  res *= arr[i] % 10;                 arr[i] /= 10;             }             cout << res << " ";         }          // Otherwise         else             cout << arr[i] << " ";     } }  // Driver Code int main() {     int arr[] = { 113, 141, 214, 3186 };     int N = sizeof(arr) / sizeof(arr[0]);     evenOdd(arr, N);      return 0; }

Java

// Java program for the above approach  import java.io.*;  class GFG {      // Function to modify the given array     // as per the given conditions     static void evenOdd(int[] arr, int N)     {         // Traverse the given array arr[]         for (int i = 0; i < N; i++) {              // Initialize the count of even             // and odd digits             int even_digits = 0;             int odd_digits = 0;              // Initialize temp with the             // current array element             int temp = arr[i];              // For count the number of             // even digits             while (temp > 0) {                  // Increment the odd count                 if ((temp % 10) % 2 != 0)                     odd_digits++;                  // Otherwise                 else                     even_digits++;                  // Divide temp by 10                 temp /= 10;             }              // Performe addition             if (even_digits > odd_digits) {                  int res = 0;                 while (arr[i] > 0) {                      res += arr[i] % 10;                     arr[i] /= 10;                 }                 System.out.print(res + " ");             }              // Performe multiplication             else if (odd_digits > even_digits) {                  int res = 1;                 while (arr[i] > 0) {                      res *= arr[i] % 10;                     arr[i] /= 10;                 }                 System.out.print(res + " ");             }              // Otherwise             else                 System.out.print(arr[i] + " ");         }     }      // Driver Code     public static void main(String[] args)     {         int[] arr = { 113, 141, 214, 3186 };         int N = arr.length;         evenOdd(arr, N);     } }  // This code is contributed by rishavmahato348.

Python3

 # Python program for the above approach  # Function to modify the given array # as per the given conditions def evenOdd(arr,N):        # Traverse the given array arr[]     for i in range(N):          # Initialize the count of even         # and odd digits         even_digits = 0;         odd_digits = 0;          # Initialize temp with the         # current array element         temp = arr[i];          # For count the number of         # even digits         while (temp):              # Increment the odd count             if ((temp % 10) & 1):                 odd_digits += 1;              # Otherwise             else:                 even_digits += 1;              # Divide temp by 10             temp = temp//10          # Performe addition         if (even_digits > odd_digits):              res = 0;             while (arr[i]):                  res += arr[i] % 10;                 arr[i] = arr[i]//10;                  print(res, end=" ");          # Performe multiplication         elif (odd_digits > even_digits):              res = 1;             while (arr[i]):                  res *= arr[i] % 10;                 arr[i] = arr[i]//10                          print(res, end=" ");                  # Otherwise         else:             print(arr[i], end=" ");      # Driver Code arr = [113, 141, 214, 3186 ]; N = len(arr); evenOdd(arr, N);      # This code is contributed by _saurabh_jaiswal

// C# program for the above approach  using System;  class GFG {      // Function to modify the given array     // as per the given conditions     static void evenOdd(int[] arr, int N)     {         // Traverse the given array arr[]         for (int i = 0; i < N; i++) {              // Initialize the count of even             // and odd digits             int even_digits = 0;             int odd_digits = 0;              // Initialize temp with the             // current array element             int temp = arr[i];              // For count the number of             // even digits             while (temp > 0) {                  // Increment the odd count                 if ((temp % 10) % 2 != 0)                     odd_digits++;                  // Otherwise                 else                     even_digits++;                  // Divide temp by 10                 temp /= 10;             }              // Performe addition             if (even_digits > odd_digits) {                  int res = 0;                 while (arr[i] > 0) {                      res += arr[i] % 10;                     arr[i] /= 10;                 }                 Console.Write(res + " ");             }              // Performe multiplication             else if (odd_digits > even_digits) {                  int res = 1;                 while (arr[i] > 0) {                      res *= arr[i] % 10;                     arr[i] /= 10;                 }                 Console.Write(res + " ");             }              // Otherwise             else                 Console.Write(arr[i] + " ");         }     }      // Driver Code     public static void Main()     {         int[] arr = { 113, 141, 214, 3186 };         int N = arr.Length;         evenOdd(arr, N);     } }  // This code is contributed by subham348.

JavaScript

 <script>    // JavaScript program for the above approach  // Function to modify the given array // as per the given conditions function evenOdd(arr,N) {     // Traverse the given array arr[]     for (let i = 0; i < N; i++) {          // Initialize the count of even         // and odd digits         let even_digits = 0;         let odd_digits = 0;          // Initialize temp with the         // current array element         let temp = arr[i];          // For count the number of         // even digits         while (temp) {              // Increment the odd count             if ((temp % 10) & 1)                 odd_digits++;              // Otherwise             else                 even_digits++;              // Divide temp by 10             temp = parseInt(temp/10)         }          // Performe addition         if (even_digits > odd_digits) {              let res = 0;             while (arr[i]) {                  res += arr[i] % 10;                 arr[i] = parseInt(arr[i]/10);             }             document.write(res+" ");         }          // Performe multiplication         else if (odd_digits > even_digits) {              let res = 1;             while (arr[i]) {                  res *= arr[i] % 10;                 arr[i] = parseInt(arr[i]/10)             }             document.write(res+" ");         }          // Otherwise         else         document.write(arr[i]+" ");     } }  // Driver Code      let  arr = [113, 141, 214, 3186 ];     let N = arr.length;     evenOdd(arr, N);          // This code is contributed by Potta Lokesh    </script>