In the Reverse Integer problem, you are given an integer that can lie from [-(2^31), (2^31 – 1)]. You have to reverse its digits and return this new number. If the new integer does not lie between [-(2^31), (2^31 – 1)] return 0.<\/strong><\/p>\n\n\n\n Here is an example:<\/p>\n\n\n\n Input<\/strong>: -213<\/p>\n\n\n\n Output<\/strong>: -312<\/p>\n\n\n\n Explanation<\/strong>: After reversing 213 we got 312 but as the input is negative return -312<\/p>\n\n\n\n Take this another example:<\/p>\n\n\n\n Input<\/strong>: 1534236469<\/p>\n\n\n\n Output <\/strong>: 0<\/p>\n\n\n\n Explanation<\/strong>: After reversing the digits we got 9646324351 which is greater than (2^31)-1 so we return 0.<\/p>\n\n\n\n The basic approach is to store all the digits of the given integer into an array and after that reverse that array and make a new integer out of it. Here is the process:<\/p>\n\n\n\n The time complexity for the above approach is O(32) ~ O(1). As we keep on dividing the given integer by 10. And maximum number of digits a number can have is 31 as our value lies between [-(2^31), (2^31 – 1)]. The space complexity for the above code is O(31) as we are storing our digits in an array.\u00a0<\/p>\n\n\n\n Another way to reverse an integer is to first convert the given integer into a string and then reverse that string through an inbuilt reverse string function.<\/strong> Once the string is reversed again convert the string into the integer. Check if the resultant integer lies in between the range or not. If it does not lie between the range return 0. Otherwise, return the integer.<\/p>\n\n\n\n Let\u2019s examine step by step approach:<\/p>\n\n\n\n The time complexity for the above code is O(32) as we are using the inbuilt function reverse whose time complexity is O(n) where n is the size of the string. As in our case the size will never be more than 32 so the time complexity is O(32). The space complexity for the above code is O(32) as we are using a reverse function which took O(n) space complexity.<\/p>\n\n\n\n This approach is the same as the brute force approach. Inside of storing the values in the array we are directly updating our resultant array. Because of this, our space complexity will be reduced from O(32) to O(1). <\/p>\n\n\n\n Let\u2019s examine how we can implement it:<\/p>\n\n\n\n Here is the C++ program to solve the Reverse Integer problem:<\/p>\n\n\n\n Below is how to code it with Java:<\/p>\n\n\n\n You can do the same with Python also:<\/p>\n\n\n\n Output:<\/strong><\/p>\n\n\n\n The time complexity for the above code is O(1). As we are just iterating over the integer whose maximum size is 31 so time complexity will be O(32) ~ O(1). The space complexity is also reduced to O(1).<\/p>\n\n\n\n Let’s compare all the approaches to solve the Reverse Integer problem in the end:<\/p>\n\n\n\n![\"Example](\"https:\/\/lh7-us.googleusercontent.com\/Z-HJDFOqYcLDwTfkJJUS0GRP6ZcVFEYuofIDPYsCtFS1KslgKMsIhBoO5G1vbwRFOBIsWhvZmwjBEP5tqegJhpVv8jkCg0zVqflooaR9UH31D7U8GKXN_oZqZm6cDSme27Rc_VB0Jqo1Xl17hP_KLgE\")
Brute Force Approach<\/strong><\/h2>\n\n\n\n
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Inbuilt Function Approach<\/strong><\/h2>\n\n\n\n
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Optimized Approach<\/strong><\/h2>\n\n\n\n
![\"Given](\"https:\/\/lh7-us.googleusercontent.com\/8Cqvn9L-K99lesvBtKXs6oB6XPLeQUIPNSqBVBp8P_kGswrdm1AC-ivce1eh-Hr6wHZxFCa3wX5QEyPdkF7aLjdgRzMZ39GDbPWUarvrgzoMcO5hkALdUgrUYhEhuy6IuwfLlpW47-GKNDp_qz3Q60c\")
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![\"Result\"\/](\"https:\/\/lh7-us.googleusercontent.com\/sRyfNGMz2b7wmZ7-P_97NC24o0he-wNWnVbKB1skmjz7zUdO2fi6V-RXFDhESsYRTUZI3Ohw_N00RWCSjnOm6dUJi6dg2Y0O59xs-gZ0j2cELN5E-eTlBfDOnQHNgLT_UEeShf_75zIK62L9K_LiewA\")
C++ Code<\/strong><\/h3>\n\n\n\n
#include<bits\/stdc++.h>\nusing namespace std;\n\nint main(){\n long long int num = -241;\n\n bool flag = false;\n if(num < 0) {\n flag = true;\n num = -num;\n }\n long long int result = 0;\n\n while(num>0){\n result = (result*10) + (num%10);\n num = num\/10;\n }\n\n if(flag) result = - result;\n\n if((result > INT_MAX) || (result < INT_MIN)) {\n cout<<0<<endl;\n return 0;\n }\n\n \/\/ result = -142\n cout<<result<<endl;\n \n return 0;\n}<\/pre><\/div>\n\n\n\nJava Code<\/strong><\/h3>\n\n\n\n
class ReverseInteger {\n public static void main(String[] args) {\n long num = -241;\n\n boolean flag = false;\n if (num < 0) {\n flag = true;\n num = -num;\n }\n long result = 0;\n\n while (num > 0) {\n result = (result * 10) + (num % 10);\n num = num \/ 10;\n }\n\n if (flag) result = -result;\n\n if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {\n System.out.println(0);\n return;\n }\n\n \/\/ result = -142\n System.out.println(result);\n }\n}<\/pre><\/div>\n\n\n\nPython Code<\/strong><\/h3>\n\n\n\n
num = -241\n\nflag = False\nif num < 0:\n flag = True\n num = -num\nresult = 0\n\nwhile num > 0:\n result = (result * 10) + (num % 10)\n num = num \/\/ 10\n\nif flag:\n result = -result\n\nif result > 2**31 - 1 or result < -2**31:\n print(0)\nelse:\n # result = -142\n print(result)<\/pre><\/div>\n\n\n\n
-142<\/code><\/pre>\n\n\n\n