Programming Articles - Page 2417 of 3366

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statement −Given an integer, we need to calculate if the difference between the sum of odd digits and sum of even digits is 0 or not.The brute-force approach will be calculating the sum of all even and odd digits in the numbers and subtracting them to compute the answer.To reduce the computational time we use the concept of mental mathematics .The above constraints holds true only if the numbers are divisible by 11. So here in the implementation given below we check the ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statement −Given an input n, we need to print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.Here we will discuss two approach to reach the solution of the problem statement −Brute-force approach using loops.Mathematical solution of sum of n numbers.Approach 1 −Computing sum of each term by adding by iterating over the numbersExample Live Demodef sumOfSeries(n):    sum = 0    for i in range(1, n+1):       sum +=i*i*i    return sum # ... Read More

In this section, we will see how to find the 1’s complete of an integer. We can use the complement operator to do this task very fast, but it will make 32bit complemented value (4-bype integer). Here we want complement of n bit numbers.Suppose we have a number say 22. The binary equivalent is 10110. The complemented value is 01001 which is same as 9. Now the question comes, how to find this value? At first we have to find number of bits of the given number. Suppose the count is c (here c = 5 for 22). We have ... Read More

In this article, we will learn about the solution and approach to solve the given problem statement.Problem statement −We are given with three input values i.e principle, rate & time and we need to compute compound interest.The code given below shows the process of computation of compound interest.The formula used here isCompound Interest = P(1 + R/100)rWhere, P is the principal amountR is the rate andT is the time spanThe implementation is given belowExample Live Demodef compound_interest(principle, rate, time):    CI = principle * (pow((1 + rate / 100), time))    print("Compound interest : ", CI) # main compound_interest(10000, 7.78, 2)OutputCompound ... Read More

Here we will see another problem, where one string and another integer value say k is given. We have to find the number of substrings of length k, whose sum of ASCII values of characters is divisible by k.Suppose a string is “BCGABC”. And the value of k is 3. Here string BCG has ASCII sum 300, ABC has ASCII sum 294, both are divisible by k = 3.The approach is simple. At first we have to find the ASCII value of characters of first substring, whose length is k. We have to use the sliding window technique and subtract ... Read More

A number is given. We have to find two pairs, that can represent the number as sum of two cubes. So we have to find two pairs (a, b) and (c, d) such that the given number n can be expressed as n = a3 + b3 = c3 + d3The idea is simple. Here every number a, b, c and d are all less than n1/3. For every distinct pair (x, y) formed by number less than n1/3, if their sum (x3 + y3) is equal to the given number, we store them into hash table with the sum ... Read More

In this article, we will learn about the implementation of bubble sort sorting technique.The figure shown below illustrates the working of this algorithm −ApproachStarting with the first element(index = 0), compare the current element with the next element of the array.If the current element is greater than the next element of the array, swap them.If the current element is less than the next element, move to the next element.Repeat Step 1.Now let’s see the implementation below −Exampledef bubbleSort(ar):    n = len(arr)    # Traverse through all array elements    for i in range(n):    # Last i elements are ... Read More

In this section we will see how to get number of pairs using the given GCD and LCM values. Suppose the GCD and LCM values are 2 and 12. Now the possible pairs of numbers are (2, 12), (4, 6), (6, 4) and (12, 2). So our program will find the count of pairs. That is 4.Let us see the algorithm to understand what will be the technique to solve this problem.AlgorithmcountPairs(gcd, lcm): Begin    if lcm is nit divisible by gcd, then       return 0    temp := lcm/gcd    c := primeFactorCount(temp)    res := shift ... Read More

In this section we will see another interesting problem. Suppose we have an array of N elements. We have to find minimum number of intersection points to make this array as co-prime array. In the co-prime array gcd of every two consecutive elements is 1. We have to print the array also.Suppose we have elements like {5, 10, 20}. This is not co-prime array. Now by inserting 1 between 5, 10 and 10, 20, it will be co-prime array. So the array will be like {5, 1, 10, 1, 20}AlgorithmmakeCoPrime(arr, n): begin    count := 0    for i in ... Read More

Suppose we have an array where positive and negative numbers are stored. The array is representing the checkpoint from one end to another end of the streets. The positive and negative values are representing the energy at the checkpoints. The positive values can increase energy, and negative number decreases energy. We have to find the initial energy level to cross the street, such that energy level never becomes 0 or less than 0.Suppose we have an array A = {4, -6, 2, 3}. Let the initial energy is 0. So after reaching at first check point, the energy is 4. ... Read More