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Suppose we have two strings that are representing complex numbers, we have to parse them and perform complex number multiplication, then return result as a string.So if the input is like “1+-1i” and “1+-1i”, then the result will be “0+-2i”.To solve this, we will follow these steps −aa := a pair of real and imaginary of first complex numberbb := a pair of real and imaginary of second complex numberx := aa.real * bb.real – aa.img*bb.imgy := aa.real * bb.img + aa.img*bb.realreturn the string as “x+yi”Let us see the following implementation to get better understanding −Example Live Demo#include using namespace ... Read More
In this tutorial, we will be discussing a program to find N-th term of series 1, 6, 17, 34, 56, 86, 121, 162, …For this, we will be provided with a number. Our task is to find the term for the given series at that particular position.Example Live Demo#include #include using namespace std; //calculating nth term of given series int nthTerm(int n) { return 3 * pow(n, 2) - 4 * n + 2; } int main() { int N = 4; cout
In this tutorial, we will be discussing a program to find N-th term of series 0, 10, 30, 60, 99, 150, 210, 280...For this, we will be provided with a number. Our task is to find the term for the given series at that particular position.Example Live Demo#include #include using namespace std; //calculating nth term of series int nthTerm(int n) { return 5 * pow(n, 2) - 5 * n; } int main() { int N = 4; cout
In this tutorial, we will be discussing a program to find N-th term of series 0, 7, 8, 33, 51, 75, 102, 133...For this, we will be provided with a number. Our task is to find the term for the given series at that particular position.Example Live Demo#include #include using namespace std; //calculating nth term of series int nthTerm(int n) { return 2 * pow(n, 2) + n - 3; } int main() { int N = 4; cout
Suppose we have a non-empty array of integer numbers. we have to return the kth most frequent elements. So if the elements are [1, 1, 1, 1, 2, 2, 3, 3, 3] and k = 2, then the result will beFormally the function should −Return true if there exists i, j, ksuch that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.To solve this, we will follow these steps −num_freq = an empty map, freq_list := an empty mapfor each element i in numsif i is not in num_freq, then num_freq[i] ... Read More
Suppose there is an unsorted array. We have to check whether an increasing subsequence of length 3 exists or not in that array.Formally the function should −Return true if there exists i, j, ksuch that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.To solve this, we will follow these steps −small := infinity, big := infinityfor each element i in arrayif i
Suppose we have a singly linked list, we have to group all odd nodes together followed by the even nodes. Here we are talking about the node position not the value in the nodes. We should try to do it in place. So if the nodes are [1, 22, 13, 14, 25], the result will be [1, 13, 25, 22, 14]To solve this, we will follow these steps −if head is null or the next of head is null, then return headhead1 := head, head2 := next of head, head_beg := next of headwhile next of head2 is nor null ... Read More
Suppose we have an array for which the ith element is the price of a given stock on the day i. We have to design an algorithm to find the maximum profit. We may complete as many transactions as we want (So, buy one and sell one share of the stock multiple times). But we have to follow these rules −We may not engage in multiple transactions at the same time (So, we must sell the stock before you buy again).After we sell our stock, we cannot buy stock on next day. (So cool down 1 day)If the input is ... Read More
Suppose we have an unsorted list of integers. We have to find the longest increasing subsequence. So if the input is [10, 9, 2, 5, 3, 7, 101, 18], then the output will be 4, as the increasing subsequence is [2, 3, 7, 101]To solve this, we will follow these steps −trail := an array of length 0 to length of nums – 1, and fill this with 0size := 0for x in numsi := 0, j := sizewhile i is not jmid := i + (j - i) / 2if trails[mid] < x, then i := mid + 1, ... Read More
Suppose we have an array nums containing n + 1 integers. The members are in range 1 to n. prove that at least one duplicate number must be there. Assume that there is only one duplicate number, we have to find that duplicate element. So if the array is like [1, 3, 4, 2, 2], then the duplicate element will be 2.To solve this, we will follow these steps −a := nums[0] and b := nums[0]while Truea := nums[nums[a]]b := nums[b]if a = b, then breakptr := nums[0]while ptr is not bptr := nums[ptr]b := nums[b]return ptrLet us see the ... Read More