DXC Coding Questions and Answers

February 4, 2025

DXC Coding Questions for 2025

DXC has updated its pattern for its 2025 drive. There are some new sections added in their recruitment drive, which will be testing students on their problem solving and coding capabilities. DXC Coding Questions and Answers in which there will be 2 problem statements, which you have to code in your desired coding languages.

Note: Automata Section now contains Coding Questions in placement pattern of 2025, and automata fix is not asked in DXC On-campus placement exam.

DXC Coding Questions

DXC has introduced this new section where they will be testing the students upon their logical thinking and problem solving techniques, there will be two problem statements, which may represent some day to day problem or some specific issue, for which you have to write a complete code in any of your preferred coding languages. Some of the languages that you can use to code are-:

C
C++
Java
Python

Rules for DXC Coding Round Questions Section:

There are two question for 45 minutes.
We must start our code from the scratch.
The coding platform is divided into two, one for writing the code and other for output.
The errors are clearly mentioned.
Atleast one Partial and One Complete Output is required for clearing the round.

DXC Coding Round details

Total number of Questions	2 Question
Total Time Duration	45 minutes
Type of Test	Non- Adaptive
Negative Marking	No

Below are some coding questions which will help you in preparing for DXC Coding Round.

Note We have observed that for some of the colleges this round is also asked in DXC drives along with Automata Fix section.

DXC Coding Questions and Answers

Question 1

Problem statement

In an auditorium, the lighting system of N bulbs is placed in a row. Due to some technical fault in the power supply, only some of the bulbs remain ON and the rest of them go OFF. Due to this flaw in the network, the bulbs which are OFF are unable to work. So till the technical team finds the actual cause of the fault, the technical head Akshay makes some temporary arrangements for the OFF bulbs at a minimum cost. Akshay decides to connect all the OFF bulbs to the nearby ON bulbs so that the length of the cable used to connect them is minimum. He calculates the distance of the systems from the first system.

Write an algorithm to help Akshay find the minimum length of the cable used to turn all the bulbs ON.

Input

The first line of the input consists of an integral num, representing the number of bulbs (N).
The second line consists of N space-separated integers representing the initial state of each bulb, ON(1) or OFF(0)
The last line consists of N space-separated integers representing the distance of the bulbs from the first bulb.

Output

Print an integer representing the minimum length of the cable used to turn all the bulbs ON.

Example

Input:
- 3
- 1 0 0
- 1 5 6

Output:
- 5

Explanation:

Length of the cable required to connect the 2nd bulb to the 1st bulb =4
Length of the cable required to connect the 3rd bulb to the 2nd bulb =1
The total length of the cable = 5(4+1)

So, the output is 5.

C++

Java

Python

#include <stdio.h>
#include <limits.h>

int main() {
    int N;
    scanf("%d", &N);

    int bulbs[N];
    for (int i = 0; i < N; i++) {
        scanf("%d", &bulbs[i]);
    }

    int distance[N];
    for (int i = 0; i < N; i++) {
        scanf("%d", &distance[i]);
    }

    int minCableLength = 0;
    int lastOnIndex = -1;

    for (int i = 0; i < N; i++) {
        if (bulbs[i] == 1) {
            if (lastOnIndex != -1) {
                minCableLength += distance[i] - distance[lastOnIndex];
            }
            lastOnIndex = i;
        }
    }

    printf("%d\n", minCableLength);

    return 0;
}

C++

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int N;
    cin >> N;

    vector state(N);
    vector distance(N);

    for (int i = 0; i < N; i++) {
        cin >> state[i];
    }

    for (int i = 0; i < N; i++) {
        cin >> distance[i];
    }

    int cableLength = 0;
    int lastOnIndex = -1;

    for (int i = 0; i < N; i++) {
        if (state[i] == 1) {
            if (lastOnIndex != -1) {
                cableLength += distance[i] - distance[lastOnIndex];
            }
            lastOnIndex = i;
        }
    }

    cout << cableLength << endl;

    return 0;
}
>

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();

        int[] state = new int[N];
        int[] distance = new int[N];

        for (int i = 0; i < N; i++) {
            state[i] = scanner.nextInt();
        }

        for (int i = 0; i < N; i++) {
            distance[i] = scanner.nextInt();
        }

        int cableLength = 0;
        int lastOnIndex = -1;

        for (int i = 0; i < N; i++) {
            if (state[i] == 1) {
                if (lastOnIndex != -1) {
                    cableLength += distance[i] - distance[lastOnIndex];
                }
                lastOnIndex = i;
            }
        }

        System.out.println(cableLength);
    }
}

Python

n = int(input())
states = list(map(int, input().split()))
distances = list(map(int, input().split()))
distance = 0
for i in range(len(distances)-1):
    distance = distance + distances[i+1] - distances[i]
print (distance)

Question 2

Mayur is developing a game that has three levels in it. He wants to display the results after the successful completion of all the levels for this purpose he is using a function int giveResult(int s1, s2, s3), where s1, s2, and s3 are the score of every level. This function will return “Good” if the s1<s2, “Best” if s1>= s2 and s1<=s3 and “Can do better” if s1>s3. Can you help Mayur in developing that function

Input:
- s1 : 10
- s2 : 12
- s3 : 15

Output :
- Good

Explanation :
- Since 10 is less than 12, hence “Good” is returned.

C++

Java

Python

#include  <stdio.h>
int main()
{
    int s1, s2, s3;
    printf ("s1 : ");
    scanf ("%d", &s1);
    printf("s2 : ");
    scanf("%d",&s2);
    printf("s3 : ");
    scanf("%d",&s3);
    if(s1<s2)
    {
        printf("Good");
    }
    if(s1>s3)
    {
        printf("Can do better");
    }
    if((s1>=s2)&&(s1<=s3))
    {
        printf("Best");
    }
    return 0;
}

C++

#include <iostream>
#include <string>
using namespace std;

string giveResult(int s1, int s2, int s3) {
    if (s1 < s2) {
        return "Good";
    } else if (s1 >= s2 && s1 <= s3) {
        return "Best";
    } else {
        return "Can do better";
    }
}

int main() {
    int s1, s2, s3;
    cin >> s1 >> s2 >> s3;

    string result = giveResult(s1, s2, s3);
    cout << result << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static String giveResult(int s1, int s2, int s3) {
        if (s1 < s2) {
            return "Good";
        } else if (s1 >= s2 && s1 <= s3) {
            return "Best";
        } else {
            return "Can do better";
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int s1 = scanner.nextInt();
        int s2 = scanner.nextInt();
        int s3 = scanner.nextInt();

        String result = giveResult(s1, s2, s3);
        System.out.println(result);
    }
}

Python

def giveResult(s1, s2, s3):
    if s1 < s2:
        return "Good"
    elif s1 >= s2 and s1 <= s3:
        return "Best"
    else:
        return "Can do better"

# Input
s1 = int(input())
s2 = int(input())
s3 = int(input())

# Call the function and print the result
result = giveResult(s1, s2, s3)
print(result)

Question 3

Problem Statement:

There are 3 Horses with different speeds(in m/s) in a horse race. The length of the Racing track is given by m meters. You have to calculate the minimum time (in secs.) the fastest horse will win the race.

Input

The first line of input consists of an integer – Distance, representing the length of the racing track.
The second line of the input consists of 3 space-separated integers representing the speeds of horses running in the race.

Output

Print the minimum time required to complete the race.

Note:

Consider all arithmetic operations as integer operations
Computed values lie within the integer range.

Example

Input:
- distance: 72
- Speed of horse1: 8
- Speed of horse2: 4
- Speed of horse3: 6

Output:
- 9

Explanation :

Time is taken by three horses to finish the race are :

Time of first horse = 72/8 = 9
Time of second horse = 72/4 = 18
Time of third horse = 72/6 = 12

The minimum time is 9 secs. Thus, the output is 9

Sample Input

distance: 18
- Speed of horse1 : 3
- Speed of horse2: 9
- Speed of horse3: 6

Sample Output
- 2

C++

Java

Python

#include <stdio.h>

int main() {
    int distance;
    scanf("%d", &distance);

    int speed1, speed2, speed3;
    scanf("%d %d %d", &speed1, &speed2, &speed3);

    int time1 = distance / speed1;
    int time2 = distance / speed2;
    int time3 = distance / speed3;

    int minTime;

    if (time1 <= time2 && time1 <= time3) {
        minTime = time1;
    } else if (time2 <= time1 && time2 <= time3) {
        minTime = time2;
    } else {
        minTime = time3;
    }

    printf("%d\n", minTime);

    return 0;
}

C++

#include <iostream>
using namespace std;

int main() {
    int distance;
    cin >> distance;

    int speed1, speed2, speed3;
    cin >> speed1 >> speed2 >> speed3;

    int time1 = distance / speed1;
    int time2 = distance / speed2;
    int time3 = distance / speed3;

    int minTime = min(min(time1, time2), time3);

    cout << minTime << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int distance = scanner.nextInt();

        int speed1 = scanner.nextInt();
        int speed2 = scanner.nextInt();
        int speed3 = scanner.nextInt();

        int time1 = distance / speed1;
        int time2 = distance / speed2;
        int time3 = distance / speed3;

        int minTime = Math.min(Math.min(time1, time2), time3);

        System.out.println(minTime);
    }
}

Python

def main():
    distance = int(input())
    speed1, speed2, speed3 = map(int, input().split())

    time1 = distance // speed1
    time2 = distance // speed2
    time3 = distance // speed3

    min_time = min(time1, time2, time3)

    print(min_time)

if __name__ == "__main__":
    main()

Question 4

Problem Statement

Chinese Government wants to build a secret message encryption algorithm to encrypt important messages before transmission. The algorithm will have a message along with two keys which will be required necessarily for encryption as well as decryption of the secret message. The formula to encrypt the code is given below:

(((messagekey2 % 5)key1)%7000000001)

Write an algorithm to Encrypt the Code.

Input

The input consists of three space-separated integers – Message, Key1, and Key2, representing the secret code, the first key value, and the second key-value, respectively.

Output

Print an integer representing the code encrypted.

Constraints

1<= Message <= 109
0 <= Key1, Key2<= 109

Example

Input
- 4
- 5
- 1

Output
- 1024

C++

Java

Python

#include  <stdio.h>
#include <math.h>
int main()
{
    long long int mes, k1, k2, result1, result2, temp;
    scanf("%lld %lld %lld",&s,&n,&m);
    result1 = (pow(mes,k1));
    result1 = result1 % 10;
    result2 = pow(result1,k2);
    result2 = result2 % 1000000007;
    printf("%lld",result2);
    return 0;
}

C++

#include 
using namespace std;

int main() {
    int message, key1, key2;
    cin >> message >> key1 >> key2;

    long long encryptedCode = ((static_cast(message) * key2 % 5) * key1) % 7000000001;

    cout << encryptedCode << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int message = scanner.nextInt();
        int key1 = scanner.nextInt();
        int key2 = scanner.nextInt();

        long encryptedCode = (((long)message * key2 % 5) * key1) % 7000000001L;

        System.out.println(encryptedCode);
    }
}

Python

message = int(input())
key1 = int(input())
key2 = int(input())
print(pow((pow(message, key2)%5),key1)%7000000001)

Question 5

Problem Statement

A Monitor of the class wanted to take the attendance, so he circulated a blank page in the classroom asking students to write there student_Id. But some naughty students wrote their id multiple times to trouble the monitor. Monitor went to the teacher and reported the incident, the Teacher asked him to mark the absence of students whose id is repeated in the attendance sheet and give him the number of students with repeated id in the attendance sheet.

Input

The input consists of a string representing the ids of students.

Output

Print an integer representing the count of characters that are not repeated in the string.

Note: Student id is nothing but an alphabet from a-z.

Example:

Input:
- ghygbvghyghnhjuhjumnj

Output:
- 3
Explanation:
- As there are only 3 students whose ids are not repeating, i.e, d f g v c x and the students with ids repeating are a and s

C++

Java

Python

#include 

int main() {
    char ids[1001];
    scanf("%s", ids);

    int count[26] = {0}; // Initialize count for each character to 0

    // Count the occurrences of each character
    for (int i = 0; ids[i] != '\0'; i++) {
        count[ids[i] - 'a']++;
    }

    int notRepeatedCount = 0;
    
    // Count characters that are not repeated
    for (int i = 0; i < 26; i++) {
        if (count[i] == 1) {
            notRepeatedCount++;
        }
    }

    printf("%d\n", notRepeatedCount);

    return 0;
}

C++

#include 
using namespace std;

int main() {
    string ids;
    cin >> ids;

    int count[26] = {0}; // Initialize count for each character to 0

    // Count the occurrences of each character
    for (char ch : ids) {
        count[ch - 'a']++;
    }

    int notRepeatedCount = 0;
    
    // Count characters that are not repeated
    for (int i = 0; i < 26; i++) {
        if (count[i] == 1) {
            notRepeatedCount++;
        }
    }

    cout << notRepeatedCount << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String ids = scanner.next();

        int[] count = new int[26]; // Initialize count for each character to 0

        // Count the occurrences of each character
        for (char ch : ids.toCharArray()) {
            count[ch - 'a']++;
        }

        int notRepeatedCount = 0;

        // Count characters that are not repeated
        for (int i = 0; i < 26; i++) {
            if (count[i] == 1) {
                notRepeatedCount++;
            }
        }

        System.out.println(notRepeatedCount);
    }
}

Python

attendance = input()
count = 0
for i in attendance:
    if attendance.count(i) < 2 :
        count+=1
print(count)

Question 6

Problem Statement

You have to select an integer number “check” from the range 1 to max and then you have to print the following output after checking that number.

If the check is a multiple of both 2 and 7, print “PrepInsta”
If the check is a multiple of 2(but not 7), print “Prep”
If the check is a multiple of 7(but not 2), print “Insta”
If i is not a multiple of 2 and 7, print “Prepster”

C++

Java

Python

#include  <stdio.h>
int main()
{
    int max, check;
    scanf(" %d",&max);
    for(check=1; check<=max; check++)
    {
        if((check%2==0)&&(check%7==0))
        {
            printf("PrepInsta\n");
        }
        else if(check%2==0)
        {
            printf("Prep\n");
        }
        else if(check%7==0)
        {
            printf("Insta\n");
        }
        else
        {
           printf("Prepster\n");
        }
    }
    return 0;
}

C++

#include <iostream>
using namespace std;

int main() {
    int max;
    cin >> max;

    for (int check = 1; check <= max; check++) {
        if (check % 2 == 0 && check % 7 == 0) {
            cout << "PrepInsta" << endl;
        } else if (check % 2 == 0) {
            cout << "Prep" << endl;
        } else if (check % 7 == 0) {
            cout << "Insta" << endl;
        } else {
            cout << "Prepster" << endl;
        }
    }

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int max = scanner.nextInt();

        for (int check = 1; check <= max; check++) {
            if (check % 2 == 0 && check % 7 == 0) {
                System.out.println("PrepInsta");
            } else if (check % 2 == 0) {
                System.out.println("Prep");
            } else if (check % 7 == 0) {
                System.out.println("Insta");
            } else {
                System.out.println("Prepster");
            }
        }
    }
}

Python

check = int(input())
if check % 2 == 0:
	print("Prep",end="")
	if check % 7 == 0 :
    		print("Insta")
elif check % 7 == 0 :
	print("Insta")
else:
	print("Prepster")

Question 7

A company wishes to bucketize their item IDs for better search operations. The bucket for the item ID is chosen on the basis of the maximum value of the digit in the item ID. Write an algorithm to find the bucket to which the item ID will be assigned.

Input:
The input consists of an integer itemID, representing the identity number of the item.
Output:
Print an integer representing the bucket to which the itemID will be assigned.

Example:
723894
Output: 4

C++

Java

Python

#include <stdio.h>

int findMaxDigit(int n) {
    int maxDigit = 0;
    
    while (n > 0) {
        int digit = n % 10;
        maxDigit = (digit > maxDigit) ? digit : maxDigit;
        n /= 10;
    }
    
    return maxDigit;
}

int main() {
    int itemID;
    scanf("%d", &itemID);

    int bucket = findMaxDigit(itemID);
    printf("%d\n", bucket);

    return 0;
}

C++

#include <iostream>
using namespace std;

int findMaxDigit(int n) {
    int maxDigit = 0;
    
    while (n > 0) {
        int digit = n % 10;
        maxDigit = max(maxDigit, digit);
        n /= 10;
    }
    
    return maxDigit;
}

int main() {
    int itemID;
    cin >> itemID;

    int bucket = findMaxDigit(itemID);
    cout << bucket << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static int findMaxDigit(int n) {
        int maxDigit = 0;
        
        while (n > 0) {
            int digit = n % 10;
            maxDigit = Math.max(maxDigit, digit);
            n /= 10;
        }
        
        return maxDigit;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int itemID = scanner.nextInt();

        int bucket = findMaxDigit(itemID);
        System.out.println(bucket);
    }
}

Python

def findMaxDigit(n):
    maxDigit = 0
    
    while n > 0:
        digit = n % 10
        maxDigit = max(digit, maxDigit)
        n //= 10
    
    return maxDigit

itemID = int(input())
bucket = findMaxDigit(itemID)
print(bucket)

Question 8

A retail company is planning a promotional sale where they will offer a discount on all items in their store. The discount will not be a standard percentage or fixed amount but will be calculated based on the price of each individual item. The company has decided to calculate the discount as the sum of all prime digits in the item’s price. Write an algorithm to find the amount of discount the company will offer.
Input:
The input consists of an integer priceItem, representing the price of the item.
Output:
Print an integer representing the amount of discount.

Example:
84302
Output: 5

C++

Java

Python

#include <stdio.h>

int isPrime(int n) {
    if (n <= 1) return 0;
    if (n <= 3) return 1;

    if (n % 2 == 0 || n % 3 == 0) return 0;

    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) return 0;
    }

    return 1;
}

int calculateDiscount(int priceItem) {
    int discount = 0;

    while (priceItem > 0) {
        int digit = priceItem % 10;
        if (isPrime(digit)) {
            discount += digit;
        }
        priceItem /= 10;
    }

    return discount;
}

int main() {
    int priceItem;
    scanf("%d", &priceItem);

    int discount = calculateDiscount(priceItem);
    printf("%d\n", discount);

    return 0;
}

C++

#include <iostream>
using namespace std;

bool isPrime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;

    if (n % 2 == 0 || n % 3 == 0) return false;

    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) return false;
    }

    return true;
}

int calculateDiscount(int priceItem) {
    int discount = 0;

    while (priceItem > 0) {
        int digit = priceItem % 10;
        if (isPrime(digit)) {
            discount += digit;
        }
        priceItem /= 10;
    }

    return discount;
}

int main() {
    int priceItem;
    cin >> priceItem;

    int discount = calculateDiscount(priceItem);
    cout << discount << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        if (n <= 3) return true;

        if (n % 2 == 0 || n % 3 == 0) return false;

        for (int i = 5; i * i <= n; i += 6) {
            if (n % i == 0 || n % (i + 2) == 0) return false;
        }

        return true;
    }

    public static int calculateDiscount(int priceItem) {
        int discount = 0;

        while (priceItem > 0) {
            int digit = priceItem % 10;
            if (isPrime(digit)) {
                discount += digit;
            }
            priceItem /= 10;
        }

        return discount;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int priceItem = scanner.nextInt();

        int discount = calculateDiscount(priceItem);
        System.out.println(discount);
    }
}

Python

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True

    if n % 2 == 0 or n % 3 == 0:
        return False

    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6

    return True

def calculate_discount(price_item):
    discount = 0

    while price_item > 0:
        digit = price_item % 10
        if is_prime(digit):
            discount += digit
        price_item //= 10

    return discount

price_item = int(input())
discount = calculate_discount(price_item)
print(discount)

Question 9

You’ve just been hired as a software engineer for a tech company. Your first task is to work on an array processing algorithm.
You are given an array of integers, which could be both positive or negative. The company wants the array to be sorted in a specific way for their processing algorithm.
They need the array to be rearranged so that all the even numbers come before all the odd numbers. However, the relative order of the even and odd numbers should not be changed. In other words, they want to preserve the relative ordering of numbers of the same parity.
You are required to write a program that takes an array of integers as input and returns the array sorted according to these rules.
Input:
The input will consist of a single line containing n space-separated integers a[i] (-10^9 ≤ a[i] ≤ 10^9), where n is the size of the array.
Output:
Your program should output a single line containing n space-separated integers, representing the sorted array.
Example:
8
3 2 5 8 1 6 9 4
Output:
2 8 6 4 3 5 1 9

C++

Java

Python

#include <stdio.h>

void rearrangeArray(int arr[], int n) {
    int temp[n], evenIndex = 0, oddIndex = n;

    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            temp[evenIndex++] = arr[i];
        } else {
            temp[--oddIndex] = arr[i];
        }
    }

    for (int i = 0; i < n; i++) {
        arr[i] = temp[i];
    }
}

int main() {
    int n;
    scanf("%d", &n);

    int arr[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }

    rearrangeArray(arr, n);

    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");

    return 0;
}

C++

#include <iostream>
using namespace std;

void rearrangeArray(int arr[], int n) {
    int temp[n], evenIndex = 0, oddIndex = n;

    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            temp[evenIndex++] = arr[i];
        } else {
            temp[--oddIndex] = arr[i];
        }
    }

    for (int i = 0; i < n; i++) {
        arr[i] = temp[i];
    }
}

int main() {
    int n;
    cin >> n;

    int arr[n];
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }

    rearrangeArray(arr, n);

    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    cout << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static void rearrangeArray(int[] arr) {
        int n = arr.length;
        int[] temp = new int[n];
        int evenIndex = 0, oddIndex = n;

        for (int i = 0; i < n; i++) {
            if (arr[i] % 2 == 0) {
                temp[evenIndex++] = arr[i];
            } else {
                temp[--oddIndex] = arr[i];
            }
        }

        System.arraycopy(temp, 0, arr, 0, n);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scanner.nextInt();
        }

        rearrangeArray(arr);

        for (int num : arr) {
            System.out.print(num + " ");
        }
        System.out.println();
    }
}

Python

def rearrange_array(arr):
    n = len(arr)
    temp = [0] * n
    even_index, odd_index = 0, n

    for num in arr:
        if num % 2 == 0:
            temp[even_index] = num
            even_index += 1
        else:
            odd_index -= 1
            temp[odd_index] = num

    arr[:] = temp

n = int(input())
arr = list(map(int, input().split()))

rearrange_array(arr)

print(*arr)

Question 10

Ravi has given n number of elements which contains elements from 0 to n having 1 number is missing , his task is to return the missing number from the given elements

Constraints :
1<=n<= 10^3
0<= nums[i] <= 10^3

Input :
4
1 2 0 4

Output :
3

C++

Java

Python

#include <stdio.h>

int findMissingNumber(int nums[], int n) {
    int totalSum = n * (n + 1) / 2;
    int arraySum = 0;

    for (int i = 0; i < n; i++) {
        arraySum += nums[i];
    }

    return totalSum - arraySum;
}

int main() {
    int n;
    scanf("%d", &n);

    int nums[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
    }

    int missingNumber = findMissingNumber(nums, n);
    printf("%d\n", missingNumber);

    return 0;
}

C++

#include <iostream>
using namespace std;

int findMissingNumber(int nums[], int n) {
    int totalSum = n * (n + 1) / 2;
    int arraySum = 0;

    for (int i = 0; i < n; i++) {
        arraySum += nums[i];
    }

    return totalSum - arraySum;
}

int main() {
    int n;
    cin >> n;

    int nums[n];
    for (int i = 0; i < n; i++) {
        cin >> nums[i];
    }

    int missingNumber = findMissingNumber(nums, n);
    cout << missingNumber << endl;

    return 0;
}

Java

import java.util.Scanner;

public class Main {
    public static int findMissingNumber(int[] nums, int n) {
        int totalSum = n * (n + 1) / 2;
        int arraySum = 0;

        for (int i = 0; i < n; i++) {
            arraySum += nums[i];
        }

        return totalSum - arraySum;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = scanner.nextInt();
        }

        int missingNumber = findMissingNumber(nums, n);
        System.out.println(missingNumber);
    }
}

Python

def find_missing_number(nums):
    n = len(nums)
    total_sum = n * (n + 1) // 2
    array_sum = sum(nums)

    return total_sum - array_sum

n = int(input())
nums = list(map(int, input().split()))

missing_number = find_missing_number(nums)
print(missing_number)

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