# Reverse digits of an integer with overflow handled

• Difficulty Level : Medium
• Last Updated : 24 Nov, 2022

Write a program to reverse an integer assuming that the input is a 32-bit integer. If the reversed integer overflows, print -1 as the output.
Let us see a simple approach to reverse digits of an integer

## C++

 `// A simple C program to reverse digits of` `// an integer.` `#include ` `using` `namespace` `std;`   `int` `reversDigits(``int` `num)` `{` `    ``int` `rev_num = 0;` `    ``while` `(num != 0) {` `        ``rev_num = rev_num * 10 + num % 10;` `        ``num = num / 10;` `    ``}` `    ``return` `rev_num;` `}`   `/* Driver program to test reversDigits */` `int` `main()` `{` `    ``int` `num = 5896;` `    ``cout << ``"Reverse of no. is "` `<< reversDigits(num);` `    ``return` `0;` `}`   `// This code is contributed by Sania Kumari Gupta (kriSania804)` `// This code is improved by Md. Owais Ashraf (professor_011)`

## C

 `// A simple C program to reverse digits of` `// an integer.` `#include `   `int` `reversDigits(``int` `num)` `{` `    ``int` `rev_num = 0;` `    ``while` `(num != 0) {` `        ``rev_num = rev_num * 10 + num % 10;` `        ``num = num / 10;` `    ``}` `    ``return` `rev_num;` `}`   `/* Driver program to test reversDigits */` `int` `main()` `{` `    ``int` `num = 5896;` `    ``printf``(``"Reverse of no. is %d"``, reversDigits(num));` `    ``return` `0;` `}`   `// This code is contributed by Sania Kumari Gupta (kriSania804)` `// This code is improved by Md. Owais Ashraf (professor_011)`

## Java

 `// Java program to reverse a number `   `class` `GFG` `{` `    ``/* Iterative function to reverse` `    ``digits of num*/` `    ``static` `int` `reversDigits(``int` `num)` `    ``{` `        ``int` `rev_num = ``0``;` `        ``while``(num!=``0``)` `        ``{` `            ``rev_num = rev_num * ``10` `+ num % ``10``;` `            ``num = num / ``10``;` `        ``}` `        ``return` `rev_num;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `num = ``5896``;` `        ``System.out.println(``"Reverse of no. is "` `                           ``+ reversDigits(num));` `    ``}` `}`   `// This code is contributed by Anant Agarwal.` `// This code is improved by Md. Owais Ashraf (professor_011)` `// This code is improved by Jeet Purohit  (jeetpurohit989)`

## Python3

 `# Python program to reverse a number `   `n ``=` `5896``;` `rev ``=` `0`   `while``(n !``=` `0``):` `    ``a ``=` `n ``%` `10` `    ``rev ``=` `rev ``*` `10` `+` `a` `    ``n ``=` `n ``/``/` `10` `    `  `print``(rev)`   `# This code is contributed by Shariq Raza ` `# This code is improved by Jeet Purohit  (jeetpurohit989)`

## C#

 `// C# program to reverse a number ` `using` `System;`   `class` `GFG` `{` `    ``// Iterative function to ` `    ``// reverse digits of num` `    ``static` `int` `reversDigits(``int` `num)` `    ``{` `        ``int` `rev_num = 0;` `        ``while``(num != 0)` `        ``{` `            ``rev_num = rev_num * 10 + num % 10;` `            ``num = num / 10;` `        ``}` `        ``return` `rev_num;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main() ` `    ``{` `        ``int` `num = 5896;` `        ``Console.Write(``"Reverse of no. is "` `                        ``+ reversDigits(num));` `    ``}` `}`   `// This code is contributed by Sam007` `// This code is improved by Md. Owais Ashraf (professor_011)`

## PHP

 ``

## Javascript

 ``

Output

`Reverse of no. is 6985`

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

However, if the number is large such that the reverse overflows, the output is some garbage value. If we run the code above with input as any large number say 1000000045, then the output is some garbage value like 1105032705 or any other garbage value. See this for the output.
How to handle overflow?
The idea is to store the previous value of the sum can be stored in a variable that can be checked every time to see if the reverse overflowed or not.

Below is the implementation to deal with such a situation.

## C++

 `// C++ program to reverse digits ` `// of a number` `#include `   `using` `namespace` `std;`   `/* Iterative function to reverse ` `digits of num*/` `int` `reversDigits(``int` `num)` `{` `    ``// Handling negative numbers` `    ``bool` `negativeFlag = ``false``;` `    ``if` `(num < 0)` `    ``{` `        ``negativeFlag = ``true``;` `        ``num = -num ;` `    ``}`   `    ``int` `prev_rev_num = 0, rev_num = 0;` `    ``while` `(num != 0)` `    ``{` `        ``int` `curr_digit = num % 10;`   `        ``rev_num = (rev_num * 10) + curr_digit;`   `        ``// checking if the reverse overflowed or not.` `        ``// The values of (rev_num - curr_digit)/10 and` `        ``// prev_rev_num must be same if there was no` `        ``// problem.` `        ``if` `((rev_num - curr_digit) / ` `               ``10 != prev_rev_num)` `        ``{` `            ``cout << ``"WARNING OVERFLOWED!!!"` `                 ``<< endl;` `            ``return` `0;` `        ``}`   `        ``prev_rev_num = rev_num;` `        ``num = num / 10;` `    ``}`   `    ``return` `(negativeFlag == ``true``) ?` `                         ``-rev_num : rev_num;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `num = 12345;` `    ``cout << ``"Reverse of no. is "` `         ``<< reversDigits(num) << endl;`   `    ``num = 1000000045;` `    ``cout << ``"Reverse of no. is "` `         ``<< reversDigits(num) << endl;`   `    ``return` `0;` `}`   `// This code is contributed ` `// by Akanksha Rai(Abby_akku)`

## C

 `// C program to reverse digits of a number` `#include `   `/* Iterative function to reverse digits of num*/` `int` `reversDigits(``int` `num)` `{` `    ``// Handling negative numbers` `    ``bool` `negativeFlag = ``false``;` `    ``if` `(num < 0)` `    ``{` `        ``negativeFlag = ``true``;` `        ``num = -num ;` `    ``}`   `    ``int` `prev_rev_num = 0, rev_num = 0;` `    ``while` `(num != 0)` `    ``{` `        ``int` `curr_digit = num%10;`   `        ``rev_num = (rev_num*10) + curr_digit;`   `        ``// checking if the reverse overflowed or not.` `        ``// The values of (rev_num - curr_digit)/10 and` `        ``// prev_rev_num must be same if there was no` `        ``// problem.` `        ``if` `((rev_num - curr_digit)/10 != prev_rev_num)` `        ``{` `            ``printf``(``"WARNING OVERFLOWED!!!\n"``);` `            ``return` `0;` `        ``}`   `        ``prev_rev_num = rev_num;` `        ``num = num/10;` `    ``}`   `    ``return` `(negativeFlag == ``true``)? -rev_num : rev_num;` `}`   `/* Driver program to test reverse Digits */` `int` `main()` `{` `    ``int` `num = 12345;` `    ``printf``(``"Reverse of no. is %d\n"``, reversDigits(num));`   `    ``num = 1000000045;` `    ``printf``(``"Reverse of no. is %d\n"``, reversDigits(num));`   `    ``return` `0;` `}`

## Java

 `// Java program to reverse digits of a number`   `class` `ReverseDigits` `{` `    ``/* Iterative function to reverse digits of num*/` `    ``static` `int` `reversDigits(``int` `num)` `    ``{` `        ``// Handling negative numbers` `        ``boolean` `negativeFlag = ``false``;` `        ``if` `(num < ``0``)` `        ``{` `            ``negativeFlag = ``true``;` `            ``num = -num ;` `        ``}` `     `  `        ``int` `prev_rev_num = ``0``, rev_num = ``0``;` `        ``while` `(num != ``0``)` `        ``{` `            ``int` `curr_digit = num%``10``;` `     `  `            ``rev_num = (rev_num*``10``) + curr_digit;` `     `  `            ``// checking if the reverse overflowed or not.` `            ``// The values of (rev_num - curr_digit)/10 and` `            ``// prev_rev_num must be same if there was no` `            ``// problem.` `            ``if` `((rev_num - curr_digit)/``10` `!= prev_rev_num)` `            ``{` `                ``System.out.println(``"WARNING OVERFLOWED!!!"``);` `                ``return` `0``;` `            ``}` `     `  `            ``prev_rev_num = rev_num;` `            ``num = num/``10``;` `        ``}` `     `  `        ``return` `(negativeFlag == ``true``)? -rev_num : rev_num;` `    ``}` `    `  `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `num = ``12345``;` `        ``System.out.println(``"Reverse of no. is "` `+ reversDigits(num));` `     `  `        ``num = ``1000000045``;` `        ``System.out.println(``"Reverse of no. is "` `+ reversDigits(num));` `    ``}` `}`

## Python3

 `# Python program to reverse digits ` `# of a number `   `""" Iterative function to reverse ` `digits of num"""` `def` `reversDigits(num): `   `    ``# Handling negative numbers ` `    ``negativeFlag ``=` `False` `    ``if` `(num < ``0``):` `    `  `        ``negativeFlag ``=` `True` `        ``num ``=` `-``num ` `    `    `    ``prev_rev_num ``=` `0` `    ``rev_num ``=` `0` `    ``while` `(num !``=` `0``): ` `    `  `        ``curr_digit ``=` `num ``%` `10`   `        ``rev_num ``=` `(rev_num ``*` `10``) ``+` `curr_digit `   `        ``# checking if the reverse overflowed or not. ` `        ``# The values of (rev_num - curr_digit)/10 and ` `        ``# prev_rev_num must be same if there was no ` `        ``# problem.` `        ``if` `(rev_num >``=` `2147483647` `or` `            ``rev_num <``=` `-``2147483648``):` `            ``rev_num ``=` `0` `        ``if` `((rev_num ``-` `curr_digit) ``/``/` `10` `!``=` `prev_rev_num):` `        `  `            ``print``(``"WARNING OVERFLOWED!!!"``) ` `            ``return` `0` `        `    `        ``prev_rev_num ``=` `rev_num ` `        ``num ``=` `num ``/``/``10` `    `    `    ``return` `-``rev_num ``if` `(negativeFlag ``=``=` `True``) ``else` `rev_num `         `# Driver code ` `if` `__name__ ``=``=``"__main__"``:` `    ``num ``=` `12345` `    ``print``(``"Reverse of no. is "``,reversDigits(num)) `   `    ``num ``=` `1000000045` `    ``print``(``"Reverse of no. is "``,reversDigits(num)) ` `        `  `    `  `# This code is contributed` `# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `// C# program to reverse digits` `// of a number` `using` `System;`   `class` `GFG` `{`   `/* Iterative function to reverse` `   ``digits of num*/` `static` `int` `reversDigits(``int` `num) ` `{ ` `    ``// Handling negative numbers ` `    ``bool` `negativeFlag = ``false``; ` `    ``if` `(num < 0) ` `    ``{ ` `        ``negativeFlag = ``true``; ` `        ``num = -num ; ` `    ``} `   `    ``int` `prev_rev_num = 0, rev_num = 0; ` `    ``while` `(num != 0) ` `    ``{ ` `        ``int` `curr_digit = num % 10; `   `        ``rev_num = (rev_num * 10) + ` `                   ``curr_digit; `   `        ``// checking if the reverse overflowed ` `        ``// or not. The values of (rev_num - ` `        ``// curr_digit)/10 and prev_rev_num must ` `        ``// be same if there was no problem. ` `        ``if` `((rev_num - curr_digit) / 10 != prev_rev_num) ` `        ``{ ` `            ``Console.WriteLine(``"WARNING OVERFLOWED!!!"``); ` `            ``return` `0; ` `        ``} `   `        ``prev_rev_num = rev_num; ` `        ``num = num / 10; ` `    ``} `   `    ``return` `(negativeFlag == ``true``) ?` `                         ``-rev_num : rev_num; ` `} `   `// Driver Code` `static` `public` `void` `Main ()` `{` `    ``int` `num = 12345; ` `    ``Console.WriteLine(``"Reverse of no. is "` `+ ` `                         ``reversDigits(num)); `   `    ``num = 1000000045; ` `    ``Console.WriteLine(``"Reverse of no. is "` `+ ` `                         ``reversDigits(num)); ` `}` `}`   `// This code is contributed by ajit`

## Javascript

 ``

Output

```Reverse of no. is 54321
Reverse of no. is 1105032705```

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

Efficient Approach :

The above approach won’t work if we are given a signed 32-bit integer x, and return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 – 1], then return 0. So we cannot multiply the number*10 and then check if the number overflows or not.

We must check the overflow condition before multiplying by 10 by using the following logic :
You are checking the boundary case before you do the operation. (reversed >INT_MAX ) wouldn’t work because reversed will overflow and become negative if it goes past MAX_VALUE.  Dividing MAX_VALUE by 10 lets you check the condition without overflowing INT_MAX is equal 2147483647. INT_MIN is equal  -2147483648.  The last digits are 7 and 8. This is the reason why we also  check  rem > 7 and rem < -8 conditions

## C++

 `// C++ program to reverse digits ` `// of a number` `#include `   `using` `namespace` `std;` `int` `reversDigits(``int` `num) {` `    `  `    ``int` `rev = 0  ;` `    `  `    ``while``(num != 0){         ` `        ``int` `rem = num % 10 ;` `        ``num /= 10 ;` `        `  `        ``if``(rev > INT_MAX/10 || rev == INT_MAX/10 && rem > 7){` `            ``return` `0 ;` `        ``}` `        `  `        ``if``(rev < INT_MIN/10 || rev == INT_MIN/10 && rem < -8){` `            ``return` `0 ;` `        ``}` `        `  `        ``rev = rev*10 + rem ;` `    ``}` `    `  `    ``return` `rev ;` `    `  `}`   `// Driver Code` `int` `main()` `{` `    ``int` `num = 12345;` `    ``cout << ``"Reverse of no. is "` `         ``<< reversDigits(num) << endl;`   `    ``num = 1000000045;` `    ``cout << ``"Reverse of no. is "` `         ``<< reversDigits(num) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java program to reverse digits` `// of a number` `public` `class` `GFG` `{`   `  ``static` `int` `reversDigits(``int` `num)` `  ``{` `    ``int` `rev = ``0`  `;`   `    ``while``(num != ``0``){        ` `      ``int` `rem = num % ``10` `;` `      ``num /= ``10` `;`   `      ``if``(rev > Integer.MAX_VALUE/``10` `|| rev == Integer.MAX_VALUE/``10` `&& rem > ``7``){` `        ``return` `0` `;` `      ``}`   `      ``if``(rev < Integer.MIN_VALUE/``10` `|| rev == Integer.MIN_VALUE/``10` `&& rem < -``8``){` `        ``return` `0` `;` `      ``}`   `      ``rev = rev*``10` `+ rem ;` `    ``}`   `    ``return` `rev ;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `main (String[] args)` `  ``{` `    ``int` `num = ``12345``;` `    ``System.out.println(``"Reverse of no. is "` `+ reversDigits(num) );`   `    ``num = ``1000000045``;` `    ``System.out.println(``"Reverse of no. is "` `+ reversDigits(num) );` `  ``}` `}`   `// This code is contributed by jana_sayantan.`

## Python3

 `# Python program for the above approach` `INT_MAX ``=` `2147483647` `INT_MIN ``=` `-``2147483648`   `def` `reversDigits(num):` `    `  `    ``rev ``=` `0` `    `  `    ``while``(num > ``0``):    ` `        ``rem ``=` `num ``%` `10` `        ``num ``=` `(num``/``/``10``) ` `        `  `        ``if``(rev > INT_MAX``/``10` `or` `rev ``=``=` `INT_MAX``/``10` `and` `rem > ``7``):` `            ``return` `0` `        `  `        ``if``(rev < INT_MIN``/``10` `or` `rev ``=``=` `INT_MIN``/``10` `and` `rem < ``-``8``):` `            ``return` `0` `        `  `        ``rev ``=` `rev``*``10` `+` `rem ` `    `  `    ``return` `rev `   `# Driver Code` `num ``=` `12345` `print``(f``"Reverse of no. is {reversDigits(num)}"``)`   `num ``=` `1000000045` `print``(f``"Reverse of no. is {reversDigits(num)}"``)`   `# This code is contributed by shinjanpatra`

## C#

 `// C# program to reverse digits` `// of a number` `using` `System;`   `public` `class` `GFG` `{`   `  ``static` `int` `reversDigits(``int` `num)` `  ``{` `    ``int` `rev = 0  ;`   `    ``while``(num != 0){        ` `      ``int` `rem = num % 10 ;` `      ``num /= 10 ;`   `      ``if``(rev > Int32.MaxValue/10 || rev == Int32.MaxValue/10 && rem > 7){` `        ``return` `0 ;` `      ``}`   `      ``if``(rev < Int32.MinValue/10 || rev == Int32.MinValue/10 && rem < -8){` `        ``return` `0 ;` `      ``}`   `      ``rev = rev*10 + rem ;` `    ``}`   `    ``return` `rev ;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main (``string``[] args)` `  ``{` `    ``int` `num = 12345;` `    ``Console.WriteLine(``"Reverse of no. is "` `+ reversDigits(num) );`   `    ``num = 1000000045;` `    ``Console.WriteLine(``"Reverse of no. is "` `+ reversDigits(num) );` `  ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 ``

Output

```Reverse of no. is 54321
Reverse of no. is 0```

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

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