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Suppose we have a number n. We have to check whether n is Euclid number or not. As we know Euclid numbers are integer which can be represented as n= Pn+1where is product of first n prime numbers.So, if the input is like n = 211, then the output will be True n can be represented as 211=(2×3×5×7)+1To solve this, we will follow these steps −MAX := 10000primes := a new listDefine a function generate_all_primes() . This will takeprime := a list of size MAX and fill with Truex := 2while x * x < MAX, doif prime[x] is True, thenfor i ... Read More
Suppose we have a string s. We have to check whether the occurrences of each character in s is prime or notSo, if the input is like s = "apuuppa", then the output will be True as there are two 'a's, three 'p's and two 'u's.To solve this, we will follow these steps −freq := a map containing all characters and their frequenciesfor each char in freq, doif freq[char] > 0 and freq[char] is not prime, thenreturn Falsereturn TrueLet us see the following implementation to get better understanding −Example CodeLive Demofrom collections import defaultdict def isPrime(num): if num ... Read More
Suppose we have a string s that contains alphanumeric characters, we have to check whether the average character of the string is present or not, if yes then return that character. Here the average character can be found by taking floor of average of each character ASCII values in s.So, if the input is like s = “pqrst”, then the output will be 'r' because the average of character ASCII values are (112 + 113 + 114 + 115 + 116)/5 = 570/5 = 114 (r).To solve this, we will follow these steps −total := 0for each ch in s, ... Read More
Suppose we have a number n and another number k, we have to check whether the sum of digits of n at it's odd places (from right side to left side) is divisible by k or not.So, if the input is like n = 2416 k = 5, then the output will be True as sum of odd placed numbers from right to left is 4 + 6 = 10. Which is divisible by 5.To solve this, we will follow these steps −total := 0, pos := 1while n > 0 , doif pos is odd, thentotal := total + ... Read More
Suppose we have two strings s and t. We have to check whether t can be formed using characters of s or not.So, if the input is like s = "owleh" t = "hello", then the output will be True.To solve this, we will follow these steps −freq := a map containing all characters and their frequenciesfor i in range 0 to size of t - 1, doif freq[t[i]] is 0, thenreturn Falsefreq[t[i]] := freq[t[i]] - 1return TrueLet us see the following implementation to get better understanding −Example CodeLive Demofrom collections import defaultdict def solve(s, t): freq = ... Read More
Suppose we have a number n. We have to check whether we can make express it like a^b or not.So, if the input is like 125, then the output will be True as 125 = 5^3, so a = 5 and b = 3To solve this, we will follow these steps −if num is same as 1, then:return truefor initialize i := 2, when i * i
Suppose we have three sides in a list. We have to check whether these three sides are forming a right angled triangle or not.So, if the input is like sides = [8, 10, 6], then the output will be True as (8^2 + 6^2) = 10^2.To solve this, we will follow these steps −sort the list sidesif (sides[0]^2 + sides[1]^2) is same as sides[2]^2, thenreturn Truereturn FalseLet us see the following implementation to get better understanding −Example CodeLive Demodef solve(sides): sides.sort() if (sides[0]*sides[0]) + (sides[1]*sides[1]) == (sides[2]*sides[2]): return True return False sides = [8, 10, 6] print(solve(sides))Input[8, 10, 6] OutputTrue
Suppose we have lower limit and upper limit of a range [l, u]. We have to check whether the product of the numbers in that range is positive or negative or zero.So, if the input is like l = -8 u = -2, then the output will be Negative, as the values in that range is [-8, -7, -6, -5, -4, -3, -2], then the product is -40320 so this is negative.To solve this, we will follow these steps −if l and u both are positive, thenreturn "Positive"otherwise when l is negative and u is positive, thenreturn "Zero"otherwise, n := ... Read More
Suppose we have a number n, and another number k, we have to check whether the product of digits at even places of n is divisible by k or not. Places are started counting from right to left. Right most is at place 1.So, if the input is like n = 59361, then the output will be True as (1*3*5) is divisible by 3.To solve this, we will follow these steps −digit_count := digit count of given number nprod := 1while n > 0, doif digit_count is even, thenprod := prod * last digit of nn := quotient of (n ... Read More
Suppose we have a number n, we have to check whether the product of digits at even places of n is divisible by sum of digits at odd place of n or not. Places are started counting from right to left. Right most is at place 1.So, if the input is like n = 59361, then the output will be True as (1*3*5) = (6+9).To solve this, we will follow these steps −digit_count := digit count of given number ntotal := 0, prod := 1while n > 0, doif digit_count is even, thenprod := prod * last digit of notherwise, ... Read More