Suppose we have an array nums and one number. We can add elements in the array, such that any number in range [1, n] (both are inclusive) can be formed by the sum of some elements in the array. We have to find the minimum number of required patches. So when the array is like [1, 4] and given number is n = 7, then output will be 1, as initially the nums are [1], [4] and [1, 4] = 5, now if we add 2 into array, then the nums will be [1], [2], [4], [1, 2], [1, 4], ... Read More

Suppose we have a string consisting of only lowercase letters. We have to remove all duplicate letters such that all letters only occur once. And we have to display the result in the smallest lexicographic sequence. So if the input is like “abccb”, then the result will be “abc”To solve this, we will follow these steps −ans := one empty stringDefine one stack stDefine an array onStack of size 26Define one map mn := size of sfor initializing i := 0, when i < n, increase i by 1 do −increase m[s[i]] by 1for initializing i := 0, when i ... Read More

Suppose we have an array nums, we have to find another array called count, in this count array, the count[i] stores the number of smaller elements to the right of nums[i].So if the input is like: [5, 2, 7, 1], then the result will be [2, 1, 1, 0].To solve this, we will follow these steps −Define one method called update(), this will take index, array bit and nwhile index 0, do −ans = ans + bit[index]index = index – (index AND - index)return ansFrom the main method, do the following −n := size of numsDefine an array res ... Read More

Suppose we have n balloons, these are indexed from 0 to n-1. Here each balloon is painted with a number on it represented by one array called nums. we have to burst all the balloons. If we burst balloon i we will get nums[i – 1] * nums[i] * nums[i + 1] number of coins. After the burst, the i – 1 and i + 1 then becomes adjacent. We have to find the maximum coins to collect by bursting the balloons wisely.So if the input is like [3, 1, 5, 7], then the result will be 148. Initially the ... Read More

Suppose we have a data stream, in that stream some data element may come and join, we have to make one system, that will help to find the median from the data. As we know that the median is the middle data of a sorted list, if it list length is odd, we can get the median directly, otherwise take middle two elements, then find the average. So there will be two methods, addNum() and findMedian(), these two methods will be used to add numbers into the stream, and find the median of all added numbersTo solve this, we will ... Read More

Suppose we have a string that holds only digits from 0 to 9. And one target value is given. We have to return all possibilities to add binary operators +, - and * into the digits to get the target values. So if the input is like “232” and the target is 8, then the answer will be [“2*3+2”, “2+3*2”]To solve this, we will follow these steps −Define a method called solve(), this will take index, s, curr, target, temp, mult −if idx >= size of s, then, if target is same as curr, then, insert temp at the end ... Read More

Suppose we have an array called nums, there is a sliding window of size k which is moving from the left of the array to the right. We can only see the k numbers in the window. Each time the sliding window moves to the right side by one position. We have to find the max sliding window. So if the input is like −[1, 3, -1, -3, 5, 3, 6, 8] and k is 3, then the window will be like −Window PositionMax13-1-35368313-1-35368313-1-35368313-1-35368513-1-35368613-1-353688To solve this, we will follow these steps −Define an array ansDefine one double ended queue dqif ... Read More

Suppose we have a number n, We have to count the total number of digit 1 appearing in all non-negative numbers less than or equal to n. So if the input is 15, then the output will be 8, because the numbers containing 1 is [1,10,11,12,13,14,15], there are 8 1s.To solve this, we will follow these steps −ret := 0for initializing i := 1, when i

Suppose we have a string s. We can convert it to palindrome by adding characters in front of it. We have to find the shortest palindrome, that we can find performing this information. So if the string is like “abcc”, then the result will be − "ccbabcc".To solve this, we will follow these steps −n := size of s, s1 := s, s2 := sReverse the string s2s2 := s concatenate "#" concatenate s2Define an array lps of size same as s2j := 0, i := 1while i < size of s2, do −if s2[i] is same as s2[j], then, ... Read More

Suppose we have a 2D board and a list of words. So from the dictionary, we have to find all words in the board. Here each word must be constructed from letters of sequentially adjacent cell, where the adjacent cells are those horizontally or vertically neighboring. We have to keep in mind that the same letter cell may not be used more than once in a word.So if the input is like −To solve this, we will follow these steps −make an array resultDefine a method called solve(), this will take board, d, i, j swhen either i or j ... Read More