Programming Articles - Page 1689 of 3366

Suppose we have a strings older and another string newer. These two are representing software package versions in the format "major.minor.patch", we have to check whether the newer version is actually newer than the older one.So, if the input is like older = "7.2.2", newer = "7.3.1", then the output will be TrueTo solve this, we will follow these steps −older := a list of major, minor, patch code of oldernewer:= a list of major, minor, patch code of newerfor i in range the size of list older, do:= older[i], n := newer[i]if n > o, thenreturn Trueotherwise when n ... Read More

Suppose we have a list of positive numbers called nums, we have to find the number of elements that have odd number of digits.So, if the input is like [1, 300, 12, 10, 3, 51236, 1245], then the output will be 4To solve this, we will follow these steps −c:= 0for i in range 0 to size of nums, dos:= digit count of nums[i]if s is odd, thenc:= c+1return cLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, nums):       c=0       for i in range(len(nums)):       ... Read More

Suppose we have one number n, we have to find the sum of the first n positive odd numbers.So, if the input is like 7, then the output will be 49 as [1+3+5+7+9+11+13] = 49To solve this, we will follow these steps −if n is same as 0, thenreturn 0sum := 1, count := 0, temp := 1while count < n-1, dotemp := temp + 2sum := sum + tempcount := count + 1return sumLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, n):       if n == 0:          return 0          sum = 1          count = 0          temp = 1          while(count

Suppose we have a string s of lowercase letters, we have to find the total number of substrings that contain one unique character.So, if the input is like "xxyy", then the output will be 6 as substrings are [x, x, xx, y, y, yy]To solve this, we will follow these steps −total := 0previous := blank stringfor each character c in s, doif c is not same as previous, thenprevious := ctemp := 1otherwise, temp := temp + 1total := total + tempreturn totalLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, s): ... Read More

Suppose we have a binary tree which contains unique values and we also have another value k, we have to find the number of k-length unique paths in the tree. The paths can go either from parent to child or from child to parent. We will consider two paths are different when some node appears in one path but not the other.So, if the input is likek = 3, then the output will be 4, as the paths are [12, 8, 3], [12, 8, 10], [8, 12, 15], [3, 8, 10].To solve this, we will follow these steps−Define a function ... Read More

Suppose we have a number n, we have to find the number of bit 1 present in the binary representation of that number.So, if the input is like 12, then the output will be 2To solve this, we will follow these steps −count := 0while n is non-zero, docount := count + (n AND 1)n := floor of (n / 2)return countLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, n):       count = 0       while (n):          count += n & 1          n >>= 1       return count ob = Solution() print(ob.solve(12))Input12Output2

Suppose we have a number n, we have to find the nth Fibonacci term. As we know the ith Fibonacci term f(i) = f(i-1) + f(i-2), the first two terms are 0, 1.So, if the input is like 15, then the output will be 610To solve this, we will follow these steps −first := 0, second := 1for i in range 2 to n, dotemp := first + secondfirst := secondsecond := tempreturn secondLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, n):       first = 0       second ... Read More

Suppose we have a number n representing a chessboard of size n x n. We have to find the number of ways we can place n rooks, such that they cannot attack each other. Here two ways will be considered different if in one of the ways, some cell of the chessboard is occupied, and another way, the cell is not occupied. (We know that rooks can attack each other if they are either on the same row or on the same column).So, if the input is like 3, then the output will be 6To solve this, we will follow ... Read More

Suppose we have a BST, and we also have left and right bounds l and r, we have to find the count of all nodes in root whose values are present in between l and r (inclusive).So, if the input is likel = 7, r = 13, then the output will be 3, as there are three nodes: 8, 10, 12.To solve this, we will follow these steps−stack := a stack and insert root at first, count := 0while stack is not empty, donode := top element of stack, and pop elementif node is not null, thenif l

Suppose we have a number n; we have to check whether it is equal to the sum of the digits of n to the power of the number of digits.So, if the input is like 9474, then the output will be True as 9^4 + 4^4 + 7^4 + 4^4 = 6561 + 256 + 2401 + 256 = 9474.To solve this, we will follow these steps −s := a list of digits in of nreturn true if n is same as sum of x*(size of s) for all x in s, otherwise falseLet us see the following implementation to ... Read More