Check if a + b = c is valid after removing all zeroes from a, b and c

Given two integers a and b. Now, c can be found as a + b = c. The task is to check if the equation is still valid after removing all zeroes from a, b and c. If valid then print Yes else print No.

Examples:

Input: a = 101, b = 102 
Output: Yes 
Current equation is 101 + 102 = 203 
After removing 0s, 11 + 12 = 23 (which is still correct)

Input: a = 105, b = 106 
Output: No 
105 + 106 = 211 
15 + 16 = 211 (Incorrect) 

Approach:

• Calculate c.
• Remove all 0s from a, b and c.
• Check if the new values form a correct equation.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
#include<string>
#include<iostream>
#include<sstream>

using namespace std;

// Function to remove zeroes from a number
int remove(int x)
{
    // Converting x into a string

    string y = to_string(x);

    // To store the new integer without 0s
    string num;
    int i;
    for(i = 0; i < y.length(); i++)
    {
        // Skip if current character is 0
        if(y[i] == 0)
            continue;
        num += y[i];
    }

    // Return the integer after removing 0s
    return stoi(num);
}

// Function that returns true if
// the given condition is satisfied
bool check(int a, int b)
{
    // Calculate c
    int c = a + b;

    // Remove 0s from a, b and c
    a = remove(a);
    b = remove(b);
    c = remove(c);

    // Check if the equation is still correct
    if((a + b) == c)
        return true;
    else
        return false;
}

// Driver code
int main()
{
    int a = 101;
    int b = 102;

    if(check(a, b))
        cout << "Yes";
    else
        cout << "No";
}

// This code is contributed by ita_c

Java

// Java implementation of the approach
import java.util.*;

class GFG{

// Function to remove zeroes from a number
public static int remove(int x)
{

    // Converting x into a string
    String y = String.valueOf(x);

    // To store the new integer without 0s
    String num = "";
    int i;

    for(i = 0; i < y.length(); i++)
    {

       // Skip if current character is 0
       if(y.charAt(i) == 0)
          continue;
       num += y.charAt(i);
    }

    // Return the integer after
    // removing 0s
    return Integer.parseInt(num);
}


// Function that returns true if
// the given condition is satisfied
public static boolean check(int a, int b)
{

    // Calculate c
    int c = a + b;

    // Remove 0s from a, b and c
    a = remove(a);
    b = remove(b);
    c = remove(c);

    // Check if the equation is
    // still correct
    if((a + b) == c)
        return true;
    else
        return false;
}

// Driver code
public static void main(String[] args)
{
    int a = 101;
    int b = 102;

    if(check(a, b))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by grandmaster

Python3

# Python3 implementation of the approach

# Function that returns true if
# the given condition is satisfied
def check(a, b):

    # Calculate c
    c = a + b

    # Remove 0s from a, b and c
    a = remove(a)
    b = remove(b)
    c = remove(c)

    # Check if the equation is still correct
    if((a + b) == c):
        return True
    else:
        return False

# Function to remove zeroes from a number
def remove(x):

    # Converting x into a string
    y = str(x)

    # To store the new integer without 0s
    num = ""
    for i in range(len(y)):

        # Skip if current character is 0
        if(y[i] == "0"):
            continue
        num += y[i]

    # Return the integer after removing 0s
    return int(num)

# Driver code
a = 101
b = 102

if(check(a, b)):
    print("Yes")
else:
    print("No")

C#

// C# implementation of the approach
using System;

class GFG{

// Function to remove zeroes from a number
public static int remove(int x)
{

    // Converting x into a string
    string y = x.ToString();

    // To store the new integer without 0s
    string num = "";
    int i;

    for(i = 0; i < y.Length; i++)
    {

        // Skip if current character is 0
        if (y[i] == 0)
            continue;

        num += y[i];
    }

    // Return the integer after
    // removing 0s
    return Int32.Parse(num);
}

// Function that returns true if
// the given condition is satisfied
public static bool check(int a, int b)
{

    // Calculate c
    int c = a + b;

    // Remove 0s from a, b and c
    a = remove(a);
    b = remove(b);
    c = remove(c);

    // Check if the equation is
    // still correct
    if ((a + b) == c)
        return true;
    else
        return false;
}

// Driver code
public static void Main(string[] args)
{
    int a = 101;
    int b = 102;

    if (check(a, b))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by rutvik_56

PHP

<?php
// PHP implementation of the approach

// Function that returns true if
// the given condition is satisfied
function check($a, $b)
{

    // Calculate c
    $c = $a + $b ;

    // Remove 0s from a, b and c
    $a = remove($a);
    $b = remove($b);
    $c = remove($c);

    // Check if the equation is
    // still correct
    if(($a + $b) == $c)
        return true;
    else
        return false;
}

// Function to remove zeroes
// from a number
function remove($x)
{

    // Converting x into a string
    $y = (string)$x;

    // To store the new integer without 0s
    $num = "";
    for ($i = 0; $i < strlen($y); $i++)
    {

        // Skip if current character is 0
        if($y[$i] == "0")
            continue ;
        $num .= $y[$i];
    }

    // Return the integer after removing 0s
    return (int)$num;
}

// Driver code
$a = 101;
$b = 102;

if(check($a, $b))
    echo "Yes";
else
    echo "No";

// This code is contributed by Ryuga
?>

Javascript

<script>

// JavaScript implementation of the approach

// Function to remove zeroes from a number
function remove(x)
{
    // Converting x into a string

    var y = x.toString();

    // To store the new integer without 0s
    var num = "";
    var i;
    for(i = 0; i < y.length; i++)
    {
        // Skip if current character is 0
        if(y[i] == 0)
            continue;
        num += y[i];
    }

    // Return the integer after removing 0s
    return parseInt(num);
}

// Function that returns true if
// the given condition is satisfied
function check(a, b)
{
    // Calculate c
    var c = a + b;

    // Remove 0s from a, b and c
    a = remove(a);
    b = remove(b);
    c = remove(c);

    // Check if the equation is still correct
    if((a + b) == c)
        return true;
    else
        return false;
}

// Driver code
var a = 101;
var b = 102;

if(check(a, b))
    document.write( "Yes");
else
    document.write( "No");

</script>

Output

Yes

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

One approach is to use the str.replace() method to remove all zeroes from a, b and c, and then check if the resulting values form a correct equation.

Here is an example of how this can be done:

C++

#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

int main() {
    int a = 101;
    int b = 102;
    int c = a + b;

    string str_a = to_string(a);
    string str_b = to_string(b);
    string str_c = to_string(c);

    // Remove all occurrences of '0' in a
    str_a.erase(remove(str_a.begin(), str_a.end(), '0'), str_a.end());

    // Remove all occurrences of '0' in b
    str_b.erase(remove(str_b.begin(), str_b.end(), '0'), str_b.end());

    // Remove all occurrences of '0' in c
    str_c.erase(remove(str_c.begin(), str_c.end(), '0'), str_c.end());

    // Check if the sum of a and b is equal to c
    if (stoi(str_a) + stoi(str_b) == stoi(str_c)) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }

    return 0;
}

Java

// Java code for the above approach
import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int a = 101;
        int b = 102;
        int c = a + b;

        String aString
            = Integer.toString(a).replace("0", "");
        String bString
            = Integer.toString(b).replace("0", "");
        String cString
            = Integer.toString(c).replace("0", "");

        if (Integer.parseInt(aString)
                + Integer.parseInt(bString)
            == Integer.parseInt(cString)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

// This code is contributed by lokeshpotta20.

Python3

a = 101
b = 102
c = a + b

a = str(a).replace("0", "")
b = str(b).replace("0", "")
c = str(c).replace("0", "")

if int(a) + int(b) == int(c):
    print("Yes")
else:
    print("No")
#This code is contributed by Edula Vinay Kumar Reddy

C#

using System;

namespace ConsoleApp1
{
    class GFG
    {
        static void Main(string[] args)
        {
            int a = 101;
            int b = 102;
            int c = a + b;

            a = int.Parse(a.ToString().Replace("0", ""));
            b = int.Parse(b.ToString().Replace("0", ""));
            c = int.Parse(c.ToString().Replace("0", ""));

            if (a + b == c)
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }
    }
}

Javascript

let a = "101";
let b = "102";
let x = parseInt(a) + parseInt(b);
let c=x.toString();

a = a.replace("0", "");
b = b.replace("0", "");
c = c.replace("0", "");

if (parseInt(a) + parseInt(b) == parseInt(c))
    console.log("Yes");
else
    console.log("No");

Output

Yes

Time Complexity: O(1) (Given is integer, even largest integer contains constant digits which is constant time to replace “0”s )
Auxiliary Space: O(1)