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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
Suppose we have a number n. We have to check whether n is dihedral prime or not. A number is said to be dihedral prime when that number is itself prime and also shown same number or any other prime number using 7-segment display regardless the orientation of the display (normal or up-side down).So, if the input is like n = 1181, then the output will be Truesecond one is up-side down format of the first one and both are prime.To solve this, we will follow these steps −Define a function up_side_down() . This will take ntemp := n, total ... Read More
Suppose we have a number n and another value k. We have to check whether the kth bit in n is set (1) or not. The value of k is considered from right hand side.So, if the input is like n = 23, k = 3, then the output will be True as binary form of 23 is 10111 so the third last bit is 1 (set).To solve this, we will follow these steps −temp := n after shifting bits (k - 1) times to the rightif temp AND 1 is 1, thenreturn Truereturn FalseLet us see the following implementation ... Read More
Suppose we have two arrays nums1 and nums2 and another value k. We have to check whether both arrays can be made equal by modifying any one element from the nums1 in the following way (only once): We can add any value from the range [-k, k] to any element of nums1.So, if the input is like nums1 = [5, 7, 11] nums2 = [5, 5, 11] k = 8, then the output will be True as we can add -2 (in range [-8, 8]) with nums1[1] to make it 5 then it will be same as nums2.To solve this, ... Read More