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

Suppose we have a matrix of tasks where each row has 3 values. We also have another value k. We have to select k rows from tasks, call it S, such that the following sum is minimized and return the sum as: maximum of (S[0, 0], S[1, 0], ...S[k - 1, 0]) + maximum of (S[0, 1], S[1, 1], ...S[k - 1, 1]) + maximum of (S[0, 2], S[1, 2], ...S[k - 1, 2]) We can also say like: Each of the 3 columns contribute to a cost, and is calculated by taking the max value of that column in ... Read More

Suppose we have a list of numbers called nums, now consider an operation where we can update an element to any value. We can perform at most 3 of such operations, we have to find the resulting minimum difference between the max and the min value in nums.So, if the input is like nums = [2, 3, 4, 5, 6, 7], then the output will be 2, as we can change the list to [4, 3, 4, 5, 4, 4] and then 5 - 3 = 2.To solve this, we will follow these steps −if size of nums

Suppose we have two lists L1 and L2, we have to find the smallest difference between a number from L1 and a number from L2.So, if the input is like L1 = [2, 7, 4], L2 = [16, 10, 11], then the output will be 3, as the smallest difference is 10 - 7 = 3.To solve this, we will follow these steps −sort the list L1 and sort the list L2ans := infinityi := 0, j := 0while i < size of L1 and j < size of L2, doans := minimum of ans and |L1[i] - L2[j]|if L1[i] ... Read More

Suppose we have a list containing 0s and 1s, we have to remove values from the front or from the back of the list. Finally, we have to find the minimum number of deletions required such that the remaining list has an equal number of 0s and 1s.So, if the input is like nums = [1, 1, 1, 0, 0, 1], then the output will be 2, as we can delete the first one 1 and last one 1 so that there's two 1s and two 0s.To solve this, we will follow these steps −longest := 0d := a map ... Read More