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Suppose we have an array that represents elements of arithmetic progression in order. One element is missing. We have to find the missing element. So if arr = [2, 4, 8, 10, 12, 14], output is 6, as 6 is missing.Using binary search, we can solve this problem. We will go to the middle element, then check whether the difference between the middle and next to the middle is the same as diff or not. If not, then the missing element is present between indices mid and mid + 1. If the middle element is the n/2th element in the ... Read More
Suppose we have a value n, we have to generate n-th Tribonacci number. The Tribonacci numbers are similar to the Fibonacci numbers, but here we are generating a term by adding three previous terms. Suppose we want to generate T(n), then the formula will be like below −T(n) = T(n - 1) + T(n - 2) + T(n - 3)The first few numbers to start, are {0, 1, 1}We can solve them by following this algorithm −Algorithm• first := 0, second := 1, third := 1 • for i in range n – 3, do o next := first ... Read More
Consider we have two integers. We have to find the Hamming distance of them. The hamming distance is the number of bit different bit count between two numbers. So if the numbers are 7 and 15, they are 0111 and 1111 in binary, here the MSb is different, so the Hamming distance is 1.To solve this, we will follow these steps −For i = 31 down to 0b1 = right shift of x (i AND 1 time)b2 = right shift of y (i AND 1 time)if b1 = b2, then answer := answer + 0, otherwise answer := answer + ... Read More
Suppose we have a number n. We have to display a string representation of all numbers from 1 to n, but there are some constraints.If the number is divisible by 3, write Fizz instead of the numberIf the number is divisible by 5, write Buzz instead of the numberIf the number is divisible by 3 and 5 both, write FizzBuzz instead of the numberTo solve this, we will follow these steps −For all number from 1 to n, if a number is divisible by 3 and 5 both, print “FizzBuzz”otherwise when the number is divisible by 3, print “Fizz”otherwise when ... Read More
Suppose we have a string and we have to find the first unique character in the string. So if the string is like “people”, the first letter whose occurrence is one is ‘o’. So the index will be returned, that is 2 here. If there is no such character, then return -1.To solve this, we will follow these steps −create one frequency mapfor each character c in the string, doif c is not in frequency, then insert it into frequency, and put value 1otherwise, increase the count in frequencyScan the frequency map, if the value of a specific key is ... Read More
Suppose we have two integers a and b. Our task is to find the sum of these two integers. One constraint is that, we cannot use any operator like + or -. So if a = 5 and b = 7, the result will be 12.To solve this, we will follow these steps −For solving we will use the bitwise logical operatorsIf b = 0, then return aotherwise, recursively use the sum function by providing an XOR b, and a AND b after left shifting the result one timeExample (Python)Let us see the following implementation to get a better understanding ... Read More
Suppose we have an array of characters. We have to reverse the string without using any additional space. So if the string is like [‘H’, ‘E’, ‘L’, ‘L’, ‘O’], the output will be [‘O’, ‘L’, ‘L’, ‘E’, ‘H’]To solve this, we will follow these steps −Take two pointers to start = 0 and end = length of the string – 1swap first and last charactersincrease start by 1 and decrease-end by 1ExampleLet us see the following implementation to get a better understanding − Live Democlass Solution(object): def reverseString(self, s): """ :type s: List[str] ... Read More
Suppose we have a number n. We have to check whether the number is the power of 3 or not. So if the number is like n = 27, that is the power of 3, the result will be true, if n = 15, it will be false.To solve this, we will follow these steps −We will use the Logarithm to solve thisif [log10(n) / log10(3)] mod 1 == 0, then it will be power of three, otherwise notExampleLet us see the following implementation to get a better understanding − Live Democlass Solution(object): def isPowerOfThree(self, n): """ :type n: int :rtype: bool """ if not n or n
Suppose we have an array of integers. We have to find the sum of the elements present from index i to j. Two things we have to keep in mind that the array will be immutable, so elements will not be altered, and there will be multiple such queries. So we have to care about the execution time for a large number of queries. Suppose the array is like A = [5, 8, 3, 6, 1, 2, 5], then if query is (A, 0, 3), then it will be 5 + 8 + 3 + 6 = 22.To solve this, ... Read More
Suppose we have an array to hold some numbers. There are non-zero values as well as zero values. So we have to send all zeros to the right without changing the relative order of other numbers. So if the array is like [0, 1, 5, 0, 3, 8, 0, 0, 9], then the final array will be [1, 5, 3, 8, 9, 0, 0, 0, 0]To solve this, we will follow these steps −Suppose index = 0for i = 0 to the length of Aif A[i] != 0, thenA[index] := A[i]index := index + 1for i = index to the ... Read More