Remove trailing zeros from the sum of two numbers ( Using Stack )
Last Updated :
23 Jul, 2025
Given two numbers A and B, the task is to remove the trailing zeros present in the sum of the two given numbers using a stack.
Examples:
Input: A = 124, B = 186
Output: 31
Explanation: Sum of A and B is 310. Removing the trailing zeros modifies the sum to 31.
Input: A=130246, B= 450164
Output : 58041
Approach: The given problem can be solved using the string and stack data structures. Follow the steps below to solve the problem:
- Calculate A + B and store it in a variable, say N.
- Initialize a stack < char >, say S, to store the digits of N.
- Convert the integer N to string and then push the characters into the stack S.
- Iterate while S is not empty(), If the top element of the stack is '0', then pop it out of the stack. Otherwise, break.
- Initialize a string, say res, to store the resultant string.
- Iterate while S is not empty(), and push all the characters in res and, then pop the top element.
- Reverse the string res and print the res as the answer.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to remove trailing
// zeros from the sum of two numbers
string removeTrailing(int A, int B)
{
// Stores the sum of A and B
int N = A + B;
// Stores the digits
stack<int> s;
// Stores the equivalent
// string of integer N
string strsum = to_string(N);
// Traverse the string
for (int i = 0; i < strsum.length(); i++) {
// Push the digit at i
// in the stack
s.push(strsum[i]);
}
// While top element is '0'
while (s.top() == '0')
// Pop the top element
s.pop();
// Stores the resultant number
// without trailing 0's
string res = "";
// While s is not empty
while (!s.empty()) {
// Append top element of S in res
res = res + char(s.top());
// Pop the top element of S
s.pop();
}
// Reverse the string res
reverse(res.begin(), res.end());
return res;
}
// Driver Code
int main()
{
// Input
int A = 130246, B = 450164;
// Function Call
cout << removeTrailing(A, B);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to remove trailing
// zeros from the sum of two numbers
static String removeTrailing(int A, int B)
{
// Stores the sum of A and B
int N = A + B;
// Stores the digits
Stack<Character> s = new Stack<Character>();
// Stores the equivalent
// string of integer N
String strsum = Integer.toString(N);
// Traverse the string
for(int i = 0; i < strsum.length(); i++)
{
// Push the digit at i
// in the stack
s.push(strsum.charAt(i));
}
// While top element is '0'
while (s.peek() == '0')
{
// Pop the top element
s.pop();
}
// Stores the resultant number
// without trailing 0's
String res = "";
// While s is not empty
while (s.empty() == false)
{
// Append top element of S in res
res = res + (char)s.peek();
// Pop the top element of S
s.pop();
}
StringBuilder str = new StringBuilder();
str.append(res);
// Reverse the string res
str.reverse();
return str.toString();
}
// Driver Code
public static void main (String[] args)
{
// Input
int A = 130246, B = 450164;
// Function Call
System.out.println(removeTrailing(A, B));
}
}
// This code is contributed by Dharanendra.L.V.
# Python 3 program for the above approach
# Function to remove trailing
# zeros from the sum of two numbers
def removeTrailing(A, B):
# Stores the sum of A and B
N = A + B
# Stores the digits
s = []
# Stores the equivalent
# string of integer N
strsum = str(N)
# Traverse the string
for i in range(len(strsum)):
# Push the digit at i
# in the stack
s.append(strsum[i])
# While top element is '0'
while (s[-1] == '0'):
# Pop the top element
s.pop()
# Stores the resultant number
# without trailing 0's
res = ""
# While s is not empty
while (len(s) != 0):
# Append top element of S in res
res = res + (s[-1])
# Pop the top element of S
s.pop()
# Reverse the string res
res = list(res)
res.reverse()
res = ''.join(res)
return res
# Driver Code
if __name__ == "__main__":
# Input
A = 130246
B = 450164
# Function Call
print(removeTrailing(A, B))
# This code is contributed by ukasp.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to remove trailing
// zeros from the sum of two numbers
static string removeTrailing(int A, int B)
{
// Stores the sum of A and B
int N = A + B;
// Stores the digits
Stack<char> s = new Stack<char>();
// Stores the equivalent
// string of integer N
string strsum = N.ToString();
// Traverse the string
for(int i = 0; i < strsum.Length; i++)
{
// Push the digit at i
// in the stack
s.Push(strsum[i]);
}
// While top element is '0'
while (s.Peek() == '0')
{
// Pop the top element
s.Pop();
}
// Stores the resultant number
// without trailing 0's
string res = "";
// While s is not empty
while (s.Count != 0)
{
// Append top element of S in res
res = res + (char)s.Peek();
// Pop the top element of S
s.Pop();
}
char[] str = res.ToCharArray();
Array.Reverse(str);
// Reverse the string res
return new string( str);
}
// Driver Code
static public void Main()
{
// Input
int A = 130246, B = 450164;
// Function Call
Console.WriteLine(removeTrailing(A, B));
}
}
// This code is contributed by avanitrachhadiya2155
<script>
// Javascript program for the above approach
// Function to remove trailing
// zeros from the sum of two numbers
function removeTrailing(A, B) {
// Stores the sum of A and B
let N = A + B;
// Stores the digits
let s = new Array();
// Stores the equivalent
// string of integer N
let strsum = N.toString();
// Traverse the string
for (let i = 0; i < strsum.length; i++) {
// Push the digit at i
// in the stack
s.push(strsum.charAt(i));
}
// While top element is '0'
while (s[s.length-1] === '0') {
// Pop the top element
s.pop();
}
// Stores the resultant number
// without trailing 0's
let res = "";
// While s is not empty
while (s.length != 0) {
// Append top element of S in res
res = res.concat(s[s.length-1]);
// Pop the top element of S
s.pop();
}
let str = "";
str = str.concat(res)
// Reverse the string res
str = str.split("").reverse().join("");
return str.toString();
}
// Driver Code
// Input
let A = 130246, B = 450164;
// Function Call
document.write(removeTrailing(A, B));
// This code is contributed by Hritik
</script>
Time Complexity: O(len(A + B))
Auxiliary Space: O(len(A + B))
