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Suppose we have a linked list, we have to reverse it. So if the list is like 1 → 3 → 5 → 7, then the new reversed list will be 7 → 5 → 3 → 1To solve this, we will follow this approach −Define one procedure to perform list reversal in a recursive way as to solve(head, back)if the head is not present, then return headtemp := head.nexthead.next := backback = headif temp is empty, then return headhead = tempreturn solve(head, back)ExampleLet us see the following implementation to get a better understanding − Live Democlass ListNode: def __init__(self, ... Read More
Suppose we have a limit n. We have to count the number of primes present in the range 2 to n. So if n = 10, the result will be 4. As there are four primes before 10, they are 2, 3, 5, 7.To solve this, we will follow this approach −count = 0take one array prime = of size n + 1, and fill it with Falsefor i = 0 to n, doif prime[i] = false, thenincrease count by 1set j = 2while j * i
Suppose there is a city, and each house in the city has a certain amount. One robber wants to rob the money in one single night. The city has one security system, that is as if two consecutive houses are broken on the same night, then it will automatically call the police. So we have to find how the maximum amount the robber can rob?One array is provided, at index i, the A[i] is the amount that is present in i-th house. Suppose the array is like: A = [2, 7, 10, 3, 1], then the result will be 13. ... Read More
Suppose we have an unsigned number n. We have to find the number of 1s in a binary representation of this number. This is also known as Hamming Weight. So if the number is like 000000101101, then the result will be 4.To solve this, we will use these steps −Take the number and convert it into a binary stringset count = 0for each character e in a binary stringif the character is ‘1’, then increase count by 1return countExampleLet us see the following implementation to get a better understanding − Live Democlass Solution(object): def hammingWeight(self, n): """ ... Read More
Suppose we have an array A. We have to rotate right it k steps. So if the array is A = [5, 7, 3, 6, 8, 1, 5, 4], and k = 3, then the output will be [1, 5, 4, 5, 7, 3, 6, 8]. The steps are like[4, 5, 7, 3, 6, 8, 1, 5][5, 4, 5, 7, 3, 6, 8, 1][1, 5, 4, 5, 7, 3, 6, 8]To solve this, we will follow these steps.let n is the size of the arrayk = k mod nA = subarray of A from n – k to end + ... Read More
Here we will see how to calculate the number of trailing 0s for the result of factorial of any number. So if the n = 5, then 5! = 120. There is only one trailing 0. For 20! it will be 4 zeros as 20! = 2432902008176640000.The easiest approach is just calculating the factorial and count the 0s. But this approach fails for a large value of n. So we will follow another approach. The trailing zeros will be there if the prime factors are 2 and 5. If we count the 2s and 5s we can get the result. ... Read More
We know that the excel column numbers are alphabetic. It starts from A, and after Z, it will AA, AB, to ZZ, then again AAA, AAB, to ZZZ and so on. So column 1 is A, column 27 is Z. Here we will see how to get the column letter if a number of columns is given. So if the column number is 80, then it will be CB.Suppose we have a number n, and its value is 28, then we need to take a reminder with 26. If the remainder is 0, then the number is 26, 52 and ... Read More
Suppose we have an array; we have to check whether given number x is the majority element of that array or not. The array is sorted. One element is said to be the majority element when it appears n/2 times in the array. Suppose an array is like {1, 2, 3, 3, 3, 3, 6}, x = 3, here the answer is true as 3 is the majority element of the array. There are four 3s. The size of the array is 7, so we can see 4 > 7/2.We can count the occurrences of x in the array, and ... Read More
Suppose we have an array A. In this array there are many numbers that occur twice. Only one element can be found a single time. We have to find that element from that array. Suppose A = [1, 1, 5, 3, 2, 5, 2], then the output will be 3. As there is each number twice, we can perform XOR to cancel out that element. because we know y XOR y = 0To solve this, we will follow these steps.Take one variable res = 0for each element e in array A, preform res = res XOR ereturn resExampleLet us see ... Read More
Suppose we have a string with alphanumeric values and symbols. There are lower case and uppercase letters as well. We have to check whether the string is forming a palindrome or not by considering only the lowercase letters (uppercases will be converted into lower case), other symbols like a comma, space will be ignored.Suppose the string is like “A Man, a Plan, a Canal: Panama”, then by considering these rules, it will be “amanaplanacanalpanama”. This is a palindrome.To solve this, follow these steps −define x = “”read each character c in str −if c is lowercase letter or number, then ... Read More