Suppose we have a list of intervals, where each interval is like [start, end] this is representing start and end times of an intervals (inclusive), We have to find their intersection, i.e. the interval that lies within all of the given intervals.So, if the input is like [[10, 110], [20, 60], [25, 75]], then the output will be [25, 60]To solve this, we will follow these steps −start, end := interval after deleting last element from intervals listwhile intervals is not empty, dostart_temp, end_temp := interval after deleting last element from intervals liststart := maximum of start, start_tempend := minimum ... Read More

Suppose we have two strings s and t, we have to find two strings interleaved, starting with first string s. If there are leftover characters in a string they will be added to the end.So, if the input is like s = "abcd", t = "pqrstu", then the output will be "apbqcrdstu"To solve this, we will follow these steps −res:= blank stringi:= 0m:= minimum of size of s, size of twhile i < m, dores := res concatenate s[i] concatenate t[i]i := i + 1return res concatenate s[from index i to end] concatenate t [from index i to end]Example Live Democlass ... Read More

Suppose we have a number n, we have to find the base 3 equivalent of this number as string.So, if the input is like 17, then the output will be 122.To solve this, we will follow these steps −if n

Suppose we have a list of numbers nums, we have to put all the zeros to the end of the list by updating the list in-place. And the relative ordering of other elements should not be changed. We have to try to solve this in O(1) additional space.So, if the input is like [2, 0, 1, 4, 0, 5, 6, 4, 0, 1, 7], then the output will be [2, 1, 4, 5, 6, 4, 1, 7, 0, 0, 0]To solve this, we will follow these steps −if size of L is same as 0, thenreturn a blank listk := ... Read More

Suppose we have a string s and two integers i and j (i < j). Now suppose p is an infinite string of s repeating forever. We have to find the substring of p from indexes [i, j).So, if the input is like s = "programmer", i = 4, j = 8, then the output will be "ramm".To solve this, we will follow these steps −p:= blank stringfor t in range i to j, dop := p concatenate a character from s at index (t mod size of s)return pLet us see the following implementation to get better understanding −Example Live ... Read More

Suppose we have a list of numbers called nums, we have to find the most frequently present element and get the number of occurrences of that element.So, if the input is like [1, 5, 8, 5, 6, 3, 2, 45, 7, 5, 8, 7, 1, 4, 6, 8, 9, 10], then the output will be 3 as the number 5 occurs three times.To solve this, we will follow these steps −max:= 0length:= size of numsfor i in range 0 to length-2, docount:= 1for j in range i+1 to length-1, doif nums[i] is same as nums[j], thencount := count + 1if ... Read More

Suppose we have a number n, we have to find the maximum number we can make by inserting 5 anywhere in the number.So, if the input is like n = 826, then the output will be 8526.To solve this, we will follow these steps −temp := n as a stringans := -inffor i in range 0 to size of temp, docand := substring of temp from index 0 to i concatenate '5' concatenate substring of temp from index i to endif i is same as 0 and temp[0] is same as '-', thengo for the next iterationans := maximum of ... Read More

Suppose we have a non-negative number n, we have to find a number r such that r * r = n and we have to round down to the nearest integer. We have to solve this problem without using the builtin square-root function.So, if the input is like 1025, then the output will be 32.To solve this, we will follow these steps −if n

Suppose we have a list of numbers called nums, we have to check whether we can split the list into 1 or more groups such that: 1. Size of each group is greater than or equal to 2. 2. Sizes of all groups are same. 3. All the numbers present in each group are the same.So, if the input is like [3, 4, 6, 9, 4, 3, 6, 9], then the output will be True.To solve this, we will follow these steps −counts := a map where each key are distinct element and values are their frequenciestemp := 0for each ... Read More

Suppose we have a list of positive numbers called nums, we have to find the largest positive number that divides each of the number.So, if the input is like [14,28,70,56], then the output will be 14.To solve this, we will follow these steps −ans := first element of numsfor each x in nums, doans := gcd of ans and xreturn ansLet us see the following implementation to get better understanding −Exampleimport math class Solution:    def solve(self, nums):       ans = nums[0]       for x in nums:          ans = math.gcd(ans, x)       return ans ob = Solution() print(ob.solve([14,28,70,56]))Input[14,28,70,56]Output14