Programming Articles - Page 1740 of 3366

Suppose we have a positive number n, we have to find that we can make n by summing up some non-negative multiple of 3 and some non-negative multiple of 7 or not.So, if the input is like 13, then the output will be True, as 13 can be written as 1*7+2*3 = 13To solve this, we will follow these steps −for i in range 0 to n+1, increase by 7, doif n-i is divisible by 3, thenreturn Truereturn FalseLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, n):       for i ... Read More

Suppose we have a number n, we have to construct a list with each number from 1 to n, except when it is multiple of 3 or has a 3, 6, or 9 in the number, it should be the string "no-fill".So, if the input is like 20, then the output will be ['1', '2', 'clap', '4', '5', 'clap', '7', '8', 'clap', '10', '11', 'clap', 'clap', '14', 'clap', 'clap', '17', 'clap', 'clap', '20']To solve this, we will follow these steps −string := "no-fill"ls:= make a list of numbers as string from 1 to nfor i in range 0 to size ... Read More

Suppose we have a string s. Here s is representing a 12-hour clock time with suffixes am or pm, we have to find its 24-hour equivalent.So, if the input is like "08:40pm", then the output will be "20:40"To solve this, we will follow these steps −hour := (convert the substring of s [from index 0 to 2] as integer) mod 12minutes := convert the substring of s [from index 3 to 5] as integerif s[5] is same as 'p', thenhour := hour + 12return the result as hour:minutesLet us see the following implementation to get better understanding −Example Live Democlass Solution: ... Read More

Suppose we have an integer n, where only 1, 2, and 3 these digits are present. We can flip one digit to a 3. Then find the maximum number we can make.So, if the input is like 11332, then the output will be 31332To solve this, we will follow these steps −li := a list by digits of nfor x in range 0 to size of li - 1, doif li[x] is not '3', thenli[x] := '3'return the number by merging digits from lireturn nLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, ... Read More

Suppose we have a lowercase alphabet string s, we have to find the length of the shortest substring (minimum length is 2) such that some letter appears more than the other letters combined. If we cannot find any solution, then return -1.So, if the input is like "abbbcde", then the output will be 2, the substring "bb" has minimum length and this appears more than other letters.To solve this, we will follow these steps −Define a function ok(), this will take an array cnt, total := 0, maxVal := 0for each element it in cnt, dototal := total + itmaxVal ... Read More

Suppose we have a list of numbers. We have to define a method that can rotate a list of numbers to the left by k elements.So, if the input is like [5, 4, 7, 8, 5, 6, 8, 7, 9, 2], k = 2, then the output will be [8, 5, 6, 8, 7, 9, 2, 5, 4, 7]To solve this, we will follow these steps −Define an array retn := size of numsk := k mod nfor initialize i := k, when i < n, update (increase i by 1), do −insert nums[i] at the end of retfor initialize ... Read More

Suppose we have a list of intervals where each interval contains the [start, end] times. We have to find the minimum total size of any two non-overlapping intervals, where the size of an interval is (end - start + 1). If we cannot find such two intervals, return 0.So, if the input is like [[2, 5], [9, 10], [4, 6]], then the output will be 5 as we can pick interval [4, 6] of size 3 and [9, 10] of size 2.To solve this, we will follow these steps −ret := infn := size of vsort the array v based ... Read More

Suppose we have a string s containing only '(' and ')', we have to find the minimum number of brackets that can be inserted to make the string balanced.So, if the input is like "(()))(", then the output will be 2 as "(()))(", this can be made balanced like "((()))()".To solve this, we will follow these steps −:= 0, cnt := 0for initialize i := 0, when i < size of s, update (increase i by 1), do −if s[i] is same as '(', then −(increase o by 1)Otherwiseif o is non-zero, then −(decrease o by 1)Otherwise(increase cnt by 1)return ... Read More

Suppose we have a singly linked list node containing positive numbers. We have to find the same linked list where every node's next points to the node val nodes ahead. If we cannot find such node, next will be null.So, if the input is like [2, 3, 10, 5, 9], then the output will be [2, 3, 15, ]To solve this, we will follow these steps −Define an array vwhile node is not null, do −insert value of node into vnode := next of noderet = new list node with value 0temp = reti := 0while i < size of ... Read More

Suppose we have two binary trees called source and target; we have to check whether there is some inversion T of source such that it is a subtree of the target. So, it means there is a node in target that is identically same in values and structure as T including all of its descendants.As we know a tree is said to be an inversion of another tree if either −Both trees are emptyIts left and right children are optionally swapped and its left and right subtrees are inversions.So, if the input is like sourceTargetthen the output will be TrueTo ... Read More