Wipro Milestone 1 Coding Questions & Answers 2024 (Java/C++/Python)

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QUESTIONS) Count set bits in an integer

Input

13

Solution:

PYTHON PROGRAM

SAMPLE 01

#include <bits/stdc++.h>
using namespace std;

int countSetBits(int n)
{
    // base case
    if (n == 0)
        return 0;
    else
        // if last bit set add 1 else add 0
        return (n & 1) + countSetBits(n >> 1);
}

// driver code
int main()
{
    int n = 13;
    // function calling
    cout << countSetBits(n);
    return 0;
}
 
SAMPLE 02

#include <bits/stdc++.h>
using namespace std;

int countSetBits(int n)
{
    // base case
    if (n == 0)
        return 0;
    else
        // if last bit set add 1 else add 0
        return (n & 1) + countSetBits(n >> 1);
}

// driver code
int main()
{
    int n = 13;
    // function calling
    cout << countSetBits(n);
    return 0;
}

C++ PROGRAM

# Python3 implementation of recursive
# approach to find the number of set
# bits in binary representation of
# positive integer n

def countSetBits( n):
    
    # base case
    if (n == 0):
        return 0

    else:

        # if last bit set add 1 else
        # add 0
        return (n & 1) + countSetBits(n >> 1)
        
# Get value from user
n = 13

# Function calling
print( countSetBits(n))

Output

3

QUESTIONS)  Find Majority Element in an array which appears more than n/2 times.(n is size of array)

Input

{6,4,7,6,6,6,6,9}

Solution:

PYTHON PROGRAM

#include <bits/stdc++.h>
using namespace std;


void findMajority(int arr[], int n)
{
    int maxCount = 0;
    int index = -1; // sentinels
    for (int i = 0; i < n; i++) {
        int count = 0;
        for (int j = 0; j < n; j++) {
            if (arr[i] == arr[j])
                count++;
        }

    
        if (count > maxCount) {
            maxCount = count;
            index = i;
        }
    }

    
    if (maxCount > n / 2)
        cout << arr[index] << endl;

    else
        cout << "No Majority Element" << endl;
}


int main()
{
    int arr[] = { 6,4,7,6,6,6,6,9};
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function calling
    findMajority(arr, n);

    return 0;
}

C++ PROGRAM

def findMajority(arr, n):

    maxCount = 0
    index = -1 
    for i in range(n):

        count = 1

        for j in range(i+1, n):

            if(arr[i] == arr[j]):
                count += 1

        # update maxCount if count of
        # current element is greater
        if(count > maxCount):

            maxCount = count
            index = i

    # if maxCount is greater than n/2
    # return the corresponding element
    if (maxCount > n//2):
        print(arr[index])

    else:
        print("No Majority Element")


# Driver code
if __name__ == "__main__":
    arr = [6,4,7,6,6,6,6,9]
    n = len(arr)

    # Function calling
    findMajority(arr, n)

Output

6

QUESTIONS)  Program to validate an IP address

Input

ip1 = "222.111.111.111"
ip2 = "5555..555"
ip3 = "0000.0000.0000.0000"
ip4 = "1.1.1.1"

Solution:

PYTHON PROGRAM

// Program to check if a given
// string is valid IPv4 address or not
#include <bits/stdc++.h>
using namespace std;
#define DELIM "."

/* function to check whether the
string passed is valid or not */
bool valid_part(char* s)
{
    int n = strlen(s);
    
    // if length of passed string is
    // more than 3 then it is not valid
    if (n > 3)
        return false;
    
    // check if the string only contains digits
    // if not then return false
    for (int i = 0; i < n; i++)
        if ((s[i] >= '0' && s[i] <= '9') == false)
            return false;
    string str(s);
    
    // if the string is "00" or "001" or
    // "05" etc then it is not valid
    if (str.find('0') == 0 && n > 1)
        return false;
    stringstream geek(str);
    int x;
    geek >> x;
    
    // the string is valid if the number
    // generated is between 0 to 255
    return (x >= 0 && x <= 255);
}

/* return 1 if IP string is
valid, else return 0 */
int is_valid_ip(char* ip_str)
{
    // if empty string then return false
    if (ip_str == NULL)
        return 0;
    int i, num, dots = 0;
    int len = strlen(ip_str);
    int count = 0;
    
    // the number dots in the original
    // string should be 3
    // for it to be valid
    for (int i = 0; i < len; i++)
        if (ip_str[i] == '.')
            count++;
    if (count != 3)
        return false;
    
    // See following link for strtok()

    char *ptr = strtok(ip_str, DELIM);
    if (ptr == NULL)
        return 0;

    while (ptr) {

        /* after parsing string, it must be valid */
        if (valid_part(ptr))
        {
            /* parse remaining string */
            ptr = strtok(NULL, ".");
            if (ptr != NULL)
                ++dots;
        }
        else
            return 0;
    }

    
    if (dots != 3)
        return 0;
    return 1;
}

// Driver code
int main()
{

    
            
    char ip1[] = "222.111.111.111";
    char ip2[] = "5555..555";
    char ip3[] = "0000.0000.0000.0000";
    char ip4[] = "1.1.1.1";

    is_valid_ip(ip1) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip2) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip3) ? cout<<"Valid\n" : cout<<"Not valid\n";
    is_valid_ip(ip4) ? cout<<"Valid\n" : cout<<"Not valid\n";
    return 0;
}

C++ PROGRAM

# your code goes here


def in_range(n): #check if every split is in range 0-255
    if n >= 0 and n<=255:
        return True
    return False
    
def has_leading_zero(n): # check if every split has leading zero or not.
    if len(n)>1:
        if n[0] == "0":
            return True
    return False
def isValid(s):
    
    s = s.split(".")
    if len(s) != 4: #if number of splitting element is not 4 it is not a valid ip address
        return 0
    for n in s:
        
        if has_leading_zero(n):
            return 0
        if len(n) == 0:
            return 0
        try: #if int(n) is not an integer it raises an error
            n = int(n)

            if not in_range(n):
                return 0
        except:
            return 0
    return 1
        

if __name__=="__main__":
    
    
    ip1 = "222.111.111.111"
    ip2 = "5555..555"
    ip3 = "0000.0000.0000.0000"
    ip4 = "1.1.1.1"
    print(isValid(ip1))
    print(isValid(ip2))
    print(isValid(ip3))
    print(isValid(ip4))

Output

Valid
Not valid
Not valid
Valid

QUESTIONS)  Maximum profit by buying and selling a share at most k times

Input

3

[100, 30, 15, 10, 8, 25, 80]

Solution:

PYTHON PROGRAM

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


int maxProfit(int price[], int n, int k)
{

    int profit[k + 1][n + 1];

    // For day 0, you can't earn money
    // irrespective of how many times you trade
    for (int i = 0; i <= k; i++)
        profit[i][0] = 0;

    // profit is 0 if we don't do any transaction
    // (i.e. k =0)
    for (int j = 0; j <= n; j++)
        profit[0][j] = 0;

    // fill the table in bottom-up fashion
    for (int i = 1; i <= k; i++) {
        for (int j = 1; j < n; j++) {
            int max_so_far = INT_MIN;

            for (int m = 0; m < j; m++)
                max_so_far = max(max_so_far,
                                price[j] - price[m] + profit[i - 1][m]);

            profit[i][j] = max(profit[i][j - 1], max_so_far);
        }
    }

    return profit[k][n - 1];
}

// Driver code
int main()
{
    int k = 3;
    int price[] ={ 100, 30, 15, 10, 8, 25, 80};
    int n = sizeof(price) / sizeof(price[0]);

    cout << "Maximum profit is: "
        << maxProfit(price, n, k);

    return 0;
}

C++ PROGRAM

def maxProfit(prices, n, k):
    
    # Bottom-up DP approach
    profit = [[0 for i in range(k + 1)]
                for j in range(n)]
    
    # Profit is zero for the first
    # day and for zero transactions
    for i in range(1, n):
        
        for j in range(1, k + 1):
            max_so_far = 0
            
            for l in range(i):
                max_so_far = max(max_so_far, prices[i] -
                            prices[l] + profit[l][j - 1])
                            
            profit[i][j] = max(profit[i - 1][j], max_so_far)
    
    return profit[n - 1][k]

# Driver code
k = 3
prices = [100, 30, 15, 10, 8, 25, 80]
n = len(prices)

print("Maximum profit is:",
    maxProfit(prices, n, k))

Output

Maximum profit is: 72

QUESTIONS)   Number of subarrays having product less than K

Input

100

[1, 9, 2, 8, 6, 4, 3] 

Solution:

PYTHON PROGRAM

#include <iostream>
using namespace std;

int countsubarray(int array[], int n, int k)
{
    int count = 0;
    int i, j, mul;

    for (i = 0; i < n; i++) {
        // Counter for single element
        if (array[i] < k)
            count++;

        mul = array[i];

        for (j = i + 1; j < n; j++) {
            // Multiple subarray
            mul = mul * array[j];
        
            if (mul < k)
                count++;
            else
                break;
        }
    }

    return count;
}

// Driver Code
int main()
{
    int array[] = {1, 9, 2, 8, 6, 4, 3};
    int k = 100;
    int size = sizeof(array) / sizeof(array[0]);
    int count = countsubarray(array, size, k);
    cout << count << "\n";
}

C++ PROGRAM

def countsubarray(array, n, k):
    count = 0
    for i in range(0, n):

        # Counter for single element
        if array[i] < k:
            count += 1

        mul = array[i]

        for j in range(i + 1, n):

            # Multiple subarray
            mul = mul * array[j]

    
            if mul < k:
                count += 1
            else:
                break
    return count


# Driver Code
array = [1, 9, 2, 8, 6, 4, 3]
k = 100
size = len(array)
count = countsubarray(array, size, k)
print(count, end=" ")

Output

16

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