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Suppose we have a m x n matrix filled with non-negative integers, find a path from top left corner to bottom right corner which minimizes the sum of all numbers along its path. Movements can only be either down or right at any point in time. So for example, if the matrix is like below131151421The output will be 7, the path will be 1, 3, 1, 1, 1, this will minimize the sumLet us see the steps −a := number of rows, b := number of columnsi := a – 1, j := b - 1while j >= 0matrix[a, j] ... Read More
Suppose we have one sorted array A. We have to generate one height-balanced binary search. In this problem, a height-balanced binary tree is actually a binary tree in which the depth of the two subtrees of every node never differs by more than 1. Suppose the array is like [-10, -3, 0, 5, 9]. So one possible output will be like: [0, -3, 9, -10, null, 5]To solve this, we will follow these steps.If A is empty, then return Nullfind the mid element, and make it rootDivide the array into two sub-arrays, left part of the mid element, and right ... Read More
Suppose there are three integer arrays arr1, arr2 and arr3 and they are sorted in strictly increasing order, we have to return a sorted array of only the integers that appeared in all of these three arrays. So if arrays are [1, 2, 3, 4, 5], [1, 2, 5, 7, 9], and [1, 3, 4, 5, 8], so the output will be [1, 5]To solve this, we will follow these steps −define an array called rescreate three maps f1, f2 and f3for i in range 0 to length of arr1f1[arr1[i]] increase by 1for i in range 0 to length of ... Read More
Suppose a dieter consumes calories[i], this indicates the calories on the i-th day. If we have an integer k, for every consecutive sequence of k days that is (calories[i], calories[i+1], ..., calories[i+k-1] for all 0 upper, so increase one point, lower = length of C, then come out from the looptemp := temp + C[right]return pointsExampleLet us see the following implementation to get better understanding − Live Democlass Solution(object): def dietPlanPerformance(self, c, k, l, u): temp = 0 for i in range(k): temp += c[i] right ... Read More
Suppose we have one binary tree. We have to find the maximum depth of that tree. The maximum depth of a tree is the maximum number of nodes that are traversed to reach the leaf from the root using the longest path. Suppose the tree is like below. The depth will be 3 here.To solve this, we will follow these steps.Here we will use the recursive approach. The method is solve(root, depth = 0)if the root is empty, then return depthotherwise return max of solve(left, depth + 1) and solve(left, depth + 1)Let us see the following implementation to get ... Read More
Suppose, there is a special keyboard with all keys in a single row. So if we have a string of length 26 indicating the layout of the keyboard (indexed from 0 to 25), initially our finger is at index 0. To type a character, we have to move your finger to the index of the next character. The time taken to move your finger from index i to index j is denoted as |i - j|. So if we want to type a string. we have to define a function to calculate how much time it takes to type it ... Read More
Suppose we have an array called, nums and that is sorted in non-decreasing order, and a number target. We have to find if the target is a majority element. In an array a majority element is an element that appears more than N/2 times in an array of length N. So if the array is like − [2, 4, 5, 5, 5, 5, 5, 6, 6] and target is 5, then output is true.To solve this, we will follow these steps −There will be two helping modules, lower() and upper(). These are as follows.The lower() takes two arguments array arr ... Read More
Suppose we have a list of numbers, we have to return the number whose occurrence is 1, if no such element is present, then return -1. So if the list is like [5, 2, 3, 6, 5, 2, 9, 6, 3], then the output will be 9.To solve this, we will follow these steps −We will check each element, and put the elements inside the map, so if the element is not in map, then put a new entry, otherwise increase the valuethen go through the map, when the value is 1, return the key.Example(Python)Let us see the following implementation ... Read More
Suppose we have a string, we have to remove all vowels from that string. So if the string is like “iloveprogramming”, then after removing vowels, the result will be − "lvprgrmmng"To solve this, we will follow these steps −create one array vowel, that is holding [a, e, i, o, u]for v in a vowelreplace v using blank stringExampleLet us see the following implementation to get a better understanding − Live Democlass Solution(object): def removeVowels(self, s): s = s.replace("a", "") s = s.replace("e", "") s = s.replace("i", "") s ... Read More
Suppose we have an expression. The expression has some parentheses; we have to check the parentheses are balanced or not. The order of the parentheses are (), {} and []. Suppose there are two strings. “()[(){()}]” this is valid, but “{[}]” is invalid.The task is simple; we will use the stack to do this. We should follow these steps to get the solution −Traverse through the expression until it has exhaustedif the current character is opening bracket like (, { or [, then push into stackif the current character is closing bracket like ), } or ], then pop from ... Read More