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Suppose there is a string S of lowercase letters, these letters form consecutive groups of the same character. So, when a string like S is like "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy". A group will be a large group when it has 3 or more characters. We would like the starting and ending positions of every large group.So, if the input is like "abcdddeeeeaabbbcd", then the output will be [[3, 5], [6, 9], [12, 14]]To solve this, we will follow these steps −ans := a new listcsum := 0for each a, b in make group of ... Read More
Suppose we have a list of points on a plane. We have to find the area of the largest triangle that can be formed by any 3 of the points.So, if the input is like [[0, 0], [0, 1], [1, 0], [0, 2], [2, 0]], then the output will be 2To solve this, we will follow these steps −res := 0N := size of points listfor i in range 0 to N - 2, dofor j in range i + 1 to N - 1, dofor k in range i + 2 to N, do(x1, y1) := points[i], (x2, y2) ... Read More
Suppose we have a string S and we have to write the letters of that given string, from left to right into lines. Here each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, that will be written on the next line. We also have an array widths, here widths[0] is the width of 'a', widths[1] is the width of 'b'and so on.We have to find the answers of two questions −How many lines have at least one character from SWhat is the width used by the ... Read More
Suppose we have a list of words, here each word can be written as a concatenation of the Morse code of each letter. For example, the word "cba" can be written as "-.-..--...", this is the concatenation "-.-." | "-..." | ".-"). This kind of concatenation is called the transformation of a word.We know that International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on.Here is the list of all 26 letters of the ... Read More
Following is the Java program to find the vertex, focus and directrix of a parabola −Example Live Demopublic class Demo{ public static void find_values(float val_1, float val_2, float val_3){ System.out.println("The value of vertex is (" + (-val_2 / (2 * val_1)) + ", "+ (((4 * val_1 * val_3) - (val_2 * val_2)) / (4 * val_1)) + ")"); System.out.println("The value of focus is (" + (-val_2 / (2 * val_1)) + ", " + (((4 * val_1 * val_3) - (val_2 * val_2) + 1) / (4 * val_1)) + ")"); ... Read More
Suppose we have two integers L and R, we have to find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary form.So, if the input is like L = 6 and R = 10, then the output will be 4, as there are 4 numbers 6(110), 7(111), 9(1001), 10(1010), all have prime number of set bits.To solve this, we will follow these steps −count := 0for j in range L to R, doif set bit count of j is in [2, 3, 5, 7, 11, 13, 17, 19], thencount ... Read More
Following is the Java code to find the perimeter of a cylinder −Example Live Demoimport java.io.*; public class Demo{ static int find_peri(int dia, int ht){ return 2*(dia + ht); } public static void main(String[] args){ int dia = 7; int ht = 15; System.out.println("The perimeter of the cylinder is " + find_peri(dia, ht) + " units"); } }OutputThe perimeter of the cylinder is 44 unitsA class named Demo defines a static function that takes in two values, diameter and height. This function computes the sum ... Read More
Suppose we have a dictionary words, and we have to find the minimum length word from a given dictionary words, it has all the letters from the string licensePlate. Now such a word is said to complete the given string licensePlate. Here, we will ignore case for letters. And it is guaranteed an answer exists. If there are more than one answers, then return answer that occurs first in the array.The license plate there may be same letter occurring multiple times. So when a licensePlate of "PP", the word "pile" does not complete the licensePlate, but the word "topper" does.So, ... Read More
Following is the Java code to find the largest prime factor of a number −Example Live Demoimport java.io.*; import java.util.*; public class Demo{ static long maxPrimeFactors( long val){ long max_prime = -1; while (val % 2 == 0) { max_prime = 2; val >>= 1; } for (int i = 3; i 2) max_prime = val; return max_prime; } public static void main(String[] args){ int val = 148592; ... Read More
Suppose we have an integer array called nums, now there is always exactly one largest element. We have to check whether the largest element in the array is at least twice as much as every other number in the array. If it is so, then we have to find the index of the largest element, otherwise return -1.So, if the input is like [3, 6, 1, 0], then the output will be 1, as 6 is the largest number, and for every other number in the array x, 6 is more than twice as big as x. As the index ... Read More