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

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:

1. 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.
2. 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.
3. 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.
4. 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++ 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 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.

 # 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.

 <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>

3 4 7 3186

Time Complexity: O(N)
Auxiliary Space: O(1)