Suppose we have a 2D list of numbers called intervals where each row represents [start, end] (inclusive) interval. For an interval [a, b] (a < b), its size is (b - a). We must add one interval to the given list such that, after merging all the intervals, we get exactly one range left. We have to find the minimum possible size of the added interval.So, if the input is like intervals = [[15, 20], [30, 50]], then the output will be 10, as we can add the interval [20, 30] which is the smallest possible interval.To solve this, we ... Read More

Suppose we have a list of numbers called nums. We can reduce the length of nums by taking any two numbers, removing them, and appending their sum at the end. The cost of doing this operation is the sum of the two integers we removed. We have to find the minimum total cost of reducing nums to one integer.So, if the input is like nums = [2, 3, 4, 5, 6], then the output will be 45, as we take 2 and 3 then remove to get [4, 5, 6, 5], then we take 4 and 5 then remove to ... Read More

Suppose we have two strings S and T we have to check whether they are one or zero edit distance away or not. An edit operation can be defined as deleting a character, adding a character, or replacing a character with another character.So, if the input is like S = "hello", T = "hallo", then the output will be True, as these two strings have edit distance of 1.To solve this, we will follow these steps −m := size of S, n := size of Ti := 0, j := 0count := 0if |m - n| > 1, thenreturn Falsewhile ... Read More

Suppose we have a list of numbers called nums, we have to find a new list such that each element at index i of the newly generated list is the product of all the numbers in the original list except the one at index i. Here we have to solve it without using division.So, if the input is like nums = [2, 3, 4, 5, 6], then the output will be [360, 240, 180, 144, 120]To solve this, we will follow these steps −if size of nums < 1, thenreturn numsl := size of numsleft := a list of size ... Read More

Suppose we have a string s, we have to check whether all its palindromic substrings have odd lengths or not.So, if the input is like s = "level", then the output will be TrueTo solve this, we will follow these steps −for i in range 1 to size of s, doif s[i] is same as s[i - 1], thenreturn Falsereturn TrueLet us see the following implementation to get better understanding −Example Live Democlass Solution:    def solve(self, s):       for i in range(1, len(s)):          if s[i] == s[i - 1]:           ... Read More

In Python, finding palindromic substrings (sequences that read the same forwards and backwards) can be done using various methods, such as expanding from the centre and the brute force method etc. Some of the common approaches to finding palindromic substrings in Python are as follows: Expand around the centre: It involves taking each character in the word as the centre of the potential palindrome. Brute force: This method ... Read More

Suppose we have one array nums, where all elements are positive. We are at index 0. Here, each element in the array represents our maximum jump length at that position. Our goal is to reach to the final index (n-1, where n is size of nums) with less number of jumps. So if the array is like [2, 3, 1, 1, 4], and then the output will be 2, as we can jump to index 1 from 0, then jump to index 4, that is the last index.To solve this, we will follow these steps −end := 0, jumps := ... Read More

Suppose we have n people represented as a number from 0 to n - 1, we also have a list of friend’s tuples, where friends[i][0] and friends[i][1] are friends. We have to check whether everyone has at least one friend or not.So, if the input is like n = 3 friends = [ [0, 1], [1, 2] ], then the output will be True, as Person 0 is friends of Person 1, Person 1 is friends of Person 0 and 2, and Person 2 is friends of Person 1.To solve this, we will follow these steps −people := a list ... Read More

Suppose we have a binary tree, we have to find the most frequent subtree sum. The subtree sum of a node is actually the sum of all values under a node, including the node itself.So, if the input is likethen the output will be 3 as it occurs twice − once as the left leaf, and once as the sum of 3 - 6 + 6.To solve this, we will follow these steps −count := an empty mapDefine a function getSum() . This will take nodeif node is null, thenreturn 0mySum := getSum(left of node) + getSum(right of node) + ... Read More

Suppose we have a list of numbers called nums and another value k, we have to find the number of non-empty subsets S such that min of S + max of S K, doj := j - 1if i