Suppose we have two lowercase strings s and t, they are of the same length. We can select one character from s and another from t and swap them. We can do this operation any number of times we want. Finally, we have to check whether it's possible to make the two strings same or not.So, if the input is like s = "abcd" t = "cdab", then the output will be TrueTo solve this, we will follow these steps −fre := a list containing frequencies of each elements present in concatenated string of s and tfor each cnt in ... Read More

Suppose we have a list of numbers called nums and the elements in nums are sorted in ascending order. We also have another value k, we have to check whether any two elements taken from the list add up to k or not. The numbers can also be negative or 0. We have to solve this problem in constant amount of space usage.So, if the input is like nums = [-8, -3, 2, 7, 9] k = 4, then the output will be True, because if we take 7 and -3, then the sum is 7 + (-3) = 4, ... Read More

Suppose we have a list of numbers called nums and also have a target value, we have to find the sum of the largest pair of numbers in nums whose sum is at most (target-1).So, if the input is like nums = [8, 3, 4, 9, 2] target = 8, then the output will be 7, because the sum of the largest pair of numbers less than 8 is 4 + 3 = 7.To solve this, we will follow these steps −sort the list numsp1 := 0p2 := size of nums - 1m := -infwhile p1 < p2, doif nums[p1] ... Read More

Suppose we want to make a data structure that has two methods −add(val) this adds the value val to the data structurefind(val) this checks whether there are two elements whose sum is val or notWe have to design this so that we can get the result on the fly. We will not search for numbers every time when a query comes.So, if the input is like create an object obj and add few numbers 6, 14, 3, 8, 11, 15, then check like obj.find(9), obj.find(11), obj.find(15), then the output will be True, True, False as 9 can be formed with ... Read More

Suppose we have two strings s and t. We want to form t, but there are some problems in the keyboard where some of characters stuck so they may be written 1 or more times. We have to check whether it's possible that typed s was meant to write t or not.So, if the input is like s = "appppleee" t = "apple", then the output will be True.To solve this, we will follow these steps −i := 0, j := 0s_len := size of st_len := size of tt_last := blank stringwhile j < t_len, doif i is same ... Read More

Suppose we have a list of numbers called nums. We have to check whether the list alternates starting from strictly increasing then strictly decreasing and then strictly increasing and so on. And also if the list is only strictly increasing, it will be valid.So, if the input is like nums = [2, 4, 8, 7, 5, 1, 5, 7, 2, 1], then the output will be True, because [2, 4, 8] are increasing, then [7, 5, 1] is decreasing, then again [5, 7] is increasing and [2, 1] is decreasing.To solve this, we will follow these steps −if nums[1] = ... Read More

Suppose we have a list of numbers called nums, where elements are sorted in ascending order, we have to square the elements and return the result in sorted order.So, if the input is like nums = [-8, -3, 0, 5, 6], then the output will be [0, 9, 25, 36, 64]To solve this, we will follow these steps −n := size of numsl := 0r := n - 1index := n - 1res := a list of size same as nums and fill it with 0while index >= 0, doif |nums[l]| > |nums[r]|, thenres[index] := nums[l] * nums[l]l := l ... Read More

Suppose we have a lists of numbers in nums. We have to sort the list in ascending order by the number of 1s present in the binary representation for each number. If two numbers have same number of 1s, then arrange them based on their values.So, if the input is like nums = [4, 1, 12, 7, 6], then the output will be [1, 4, 6, 12, 7], because −Binary form of 4 is 0100Binary form of 1 is 0001Binary form of 6 is 0110Binary form of 12 is 1100Binary form of 7 is 0111So the arrangement is [1, 4, ... Read More

Suppose we have a list of numbers called. Now let us consider any pair of indices (i, j) where i < j and j - i > 1. Then find the smallest pair sum.So, if the input is like nums = [3, 4, 2, 2, 4], then the output will be 5, we can select values 3 and 2 so the total sum is 5. We cannot select 2 and 2 because they are adjacent, and violating the j - i > 1 constraint.To solve this, we will follow these steps −n := size of numsmin_seen := nums[0]ans := inffor ... Read More

Suppose we have a string s with four possible characters "1", "2", "3" and "?". We can place any one of "1", "2" and "3", in the place of "?". We have to find the smallest possible number that we can make such that no two adjacent digits are same.So, if the input is like s = "2??3?", then the output will be 21231To solve this, we will follow these steps −i := 0s := a list of elements from sif size of s < 2, thenif s[i] is same as "?", thenreturn "1"while i < size of s, doif ... Read More