How to Check whether a number is unique | Python, Java , C & C++ Program to Check whether the number is unique | How to check if the input number is unique or not in python, Java, C, C++

How to Check whether a number is unique | Python, Java , C & C++ Program to Check whether the number is unique | How to check if the input number is unique or not in python, Java, C, C++

Explanation of the Code:  Check whether a number is unique in Python, Java , C & C++

The problem asks us to determine if a given number has unique digits, i.e., no digit repeats in the number.

Input and Output: (Check whether a number is unique in Python, Java , C & C++)

• The program asks the user to input a number n.
• The program checks whether all the digits of the number n are unique. If they are, it outputs that the number is unique; otherwise, it outputs that the number is not unique.

Algorithm (Main Steps): (Check whether a number is unique in Python, Java , C & C++)

• The function isUnique(n) checks if the digits of the number n are unique.
• For each digit from 0 to 9, the function counts how many times that digit appears in n.
•  It does so by repeatedly extracting the last digit of n (using num % 10), comparing it with the current digit i, and reducing n by dividing it by 10.
• If any digit appears more than once (i.e., count > 1), the function returns false, indicating the number is not unique.
• If no digit is repeated, the function returns true.

Steps in Detail: (Check whether a number is unique in Python, Java , C & C++)

How to Check whether a number is unique | Python, Java , C & C++ Program to Check whether the number is unique | How to check if the input number is unique or not in python, Java, C, C++

• The outer loop runs through digits from 0 to 9.
• The inner loop processes the number n by extracting each digit and counting its occurrences.
• The isUnique function returns true if all digits in the number are unique (no repetition), and false otherwise.

Java Code

import java.util.Scanner;

public class uniqueNumber {

public static boolean isUnique(int n)

{

for(int i=0; i<9; i++) {

int c = 0, num = n;

while(num != 0) {

int d = num%10;

if(d == i)

c++;

num = num/10;

}

if(c > 1)

return false;

}

return true;

}

public static void main(String[] args) {

int n;

Scanner sc=new Scanner(System.in);

System.out.print("Enter a number: ");

n=sc.nextInt();

if(isUnique(n))

{

System.out.println(n+" is a Unique Number.");

}

else

{

System.out.println(n+" is not a Unique Number.");

}

sc.close();

}

}

// |VICTORY| The code ran successfully !

Python:

def is_unique(n):
    for i in range(10):
        count = 0
        num = n
        while num != 0:
            d = num % 10
            if d == i:
                count += 1
            num = num // 10
        if count > 1:
            return False
    return True

n = int(input("Enter a number: "))
if is_unique(n):
    print(f"{n} is a Unique Number.")
else:
    print(f"{n} is not a Unique Number.")

C:

#include <stdio.h>

int isUnique(int n) {
    for (int i = 0; i < 10; i++) {
        int count = 0, num = n;
        while (num != 0) {
            int d = num % 10;
            if (d == i)
                count++;
            num = num / 10;
        }
        if (count > 1)
            return 0;  // not a unique number
    }
    return 1;  // unique number
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", &n);

    if (isUnique(n)) {
        printf("%d is a Unique Number.\n", n);
    } else {
        printf("%d is not a Unique Number.\n", n);
    }

    return 0;
}

C++:

#include <iostream>
using namespace std;

bool isUnique(int n) {
    for (int i = 0; i < 10; i++) {
        int count = 0, num = n;
        while (num != 0) {
            int d = num % 10;
            if (d == i)
                count++;
            num = num / 10;
        }
        if (count > 1)
            return false;  // not a unique number
    }
    return true;  // unique number
}

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;

    if (isUnique(n)) {
        cout << n << " is a Unique Number." << endl;
    } else {
        cout << n << " is not a Unique Number." << endl;
    }

    return 0;
}

Time Complexity:

• Inner Loop Complexity:
  • For each digit from 0 to 9, the inner loop processes the number n by repeatedly dividing it by 10 (to extract digits). This loop runs for a maximum of O(d) iterations, where d is the number of digits in the number n.
  • In the worst case, the number n could have up to O(log n) digits (since a number n has roughly log n digits in base 10).
• Outer Loop Complexity:
  • The outer loop runs 10 times (for each digit from 0 to 9).

Thus, the total time complexity is:

• Time Complexity: O(10 * d) = O(d), where d is the number of digits in the number n. Since d is proportional to log n, the overall time complexity is O(log n).

Space Complexity:

• The function uses only a few integer variables to count occurrences and to manipulate the number n.

• No additional data structures (such as arrays or hash maps) are used to store values.

• Therefore, the space complexity is constant.

• Space Complexity: O(1).

Summary: 

• Time Complexity: O(log n) (proportional to the number of digits in the number n).
• Space Complexity: O(1) (constant space usage).

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