Server Side Programming Articles - Page 1538 of 2650

Suppose we have a string of words delimited by spaces; we have to reverse the order of words.So, if the input is like "Hello world, I love python programming", then the output will be "programming python love I world, Hello"To solve this, we will follow these steps −temp := make a list of words by splitting s using blank spacetemp := reverse the list tempreturn a string by joining the elements from temp using space delimiter.Let us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, s):       temp = s.split(' ')   ... Read More

Suppose we have two lists of numbers called A, and B. We have to take some sublist in A and reverse it. Then check whether it is possible to turn A into B or not. We can take sublist and reverse it any number of times.So, if the input is like A = [2, 3, 4, 9, 10], B = [4, 3, 2, 10, 9], then the output will be True as we can reverse [2, 3, 4] and [9, 10].To solve this, we will follow these steps −res := a map, initially emptyfor each n in nums, dores[n] := ... Read More

Suppose we have a string, we have to check whether it's a repeating string or not.So, if the input is like string = "helloworldhelloworld", then the output will be TrueTo solve this, we will follow these steps −n := size of sDefine a function findFactors() . This will take nf := a new seti := 1while i * i

Suppose we have a string s and a number k, we have to find the number of k-length substrings of s, that occur more than once in s.So, if the input is like s = "xxxyyy", k = 2, then the output will be 2To solve this, we will follow these steps −seen := a new listfor i in range 0 to size of s - k, dot := substring of s [from index i to i + k - 1]insert t at the end of seenmp := a map for all distinct element in seen and their occurrencesreturn sum ... Read More

Suppose we have a positive number n, we will add all of its digits to get a new number. Now repeat this operation until it is less than 10.So, if the input is like 9625, then the output will be 4.To solve this, we will follow these steps −Define a method solve(), this will take nif n < 10, thenreturn ns := 0l := the floor of (log(n) base 10 + 1)while l > 0, dos := s + (n mod 10)n := quotient of n / 10l := l - 1return solve(s)Let us see the following implementation to get ... Read More

Suppose we have two strings s and t, we have to check whether we can get t by removing 1 letter from s.So, if the input is like s = "world", t = "wrld", then the output will be True.To solve this, we will follow these steps −i:= 0n:= size of swhile i < n, dotemp:= substring of s[from index 0 to i-1] concatenate substring of s[from index i+1 to end]if temp is same as t, thenreturn Truei := i + 1return FalseLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, s, t): ... Read More

Suppose we have a list of numbers A, we have to find all duplicate numbers and remove their last occurrences.So, if the input is like [10, 30, 40, 10, 30, 50], then the output will be [10, 30, 40, 50]To solve this, we will follow these steps −seen:= a new mapd:= a new mapfor i in range 0 to size of nums, doif nums[i] is not in d, thend[nums[i]]:= 1otherwise, d[nums[i]] := d[nums[i]] + 1i:= 0while i < size of nums, don:= d[nums[i]]if nums[i] is not in seen, thenseen[nums[i]]:= 1otherwise, seen[nums[i]] := seen[nums[i]] + 1if n is same as seen[nums[i]] ... Read More

Suppose we have a list of numbers called A and another number k, we have to make a new set of possible elements {A[k], A[A[k]], A[A[A[k]]], ... } stopping before it's out of index. We have to find the size of this set, otherwise -1 when there is a cycle.So, if the input is like A = [1, 2, 3, 4, 5, 6, 7], k = 1, then the output will be 6 as A[1] = 2, A[2] = 3, A[3] = 4, A[4] = 5, A[5] = 6, A[6] = 7, So the set is {2, 3, 4, 5, ... Read More

Suppose we have a string s, we have to find the index of the first recurring character in it. If we cannot find no recurring characters, then return -1.So, if the input is like "abcade", then the output will be 3, as 'a' is again present at index 3.To solve this, we will follow these steps −define a map charsfor i in range 0 to size of s, doif s[i] in chars, thenreturn iotherwise, chars[s[i]] := chars[s[i]] + 1return -1Let us see the following implementation to get better understanding −Example Live Demofrom collections import defaultdict class Solution:    def solve(self, s): ... Read More

Suppose we have a rectangle that is represented as a list with four elements [x1, y1, x2, y2], where (x1, y1) is the coordinates of its bottom-left corner, and (x2, y2) is the coordinates of its top-right corner. Two rectangles overlap when the area of their intersection is positive. So, two rectangles that only touch at the corner or edges do not overlap.So, if the input is like R1 = [0,0,2,2], R2 = [1,1,3,3], then the output will be True.To solve this, we will follow these steps −if R1[0]>=R2[2] or R1[2]=R2[2]) or (R1[2]