C++ Articles

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Different Ways to Add Parentheses in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 1K+ Views

Suppose we have a string of numbers and operators, we have to find all possible results from computing all the different possible ways to group the numbers and operators. Here the valid operators are +, - and *. So if the input is like “2*3-4*5”, then the output will be [-34, -14, -10, -10, 10]. This is because −(2*(3-(4*5))) = -34((2*3)-(4*5)) = -14((2*(3-4))*5) = -10(2*((3-4)*5)) = -10(((2*3)-4)*5) = 10To solve this, we will follow these steps −Define a map called a memo.Define a method called solve(). This will take the input string as input.create an array called retif the memo ...

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Super Pow in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 628 Views

Suppose we have to calculate a^b mod 1337 where a is one positive integer and b is an extremely large positive integer given in the form of an array. So if a = 2 and b = [1, 0] then the output will be 1024To solve this, we will follow these steps −Define powerMod() method this takes base and powerm := 1337, ret := 1while power is not 0if power is odd, then ret := ret * base mod mbase := base^2 mod mpower := power / 2return retDefine superPower(), this takes a and bif size of b = 0, ...

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Maximum Product of Word Lengths in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 550 Views

Suppose we have a string array called words, find the maximum value of length(word[i]) * length(word[j]) where the two words will not share the common letters. We can assume that each word will contain only lower case letters. If no such two words exist, then return 0. So if the input is like [“abcw”, “baz”, “foo”, “bar”, “xtfn”, “abcdef”], then the output will be 16, as two words can be “abcw”, “xtfn”.To solve this, we will follow these steps −Define a method called getRev(), this will take x as inputret := 0for i in range 0 to 25if x / ...

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Wiggle Subsequence in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 361 Views

Suppose we have a sequence of numbers that is called a wiggle sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference may be either positive or negative. A sequence with less than two elements is trivially a wiggle sequence. So for example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because if you see, the differences (6, -3, 5, -7, 3) are alternately positive and negative. But, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences, the first one because its first two differences are ...

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Largest Divisible Subset in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 180 Views

Suppose we have a set of distinct positive integers, we have to find the largest subset such that every pair like (Si, Sj) of elements in this subset satisfies: Si mod Sj = 0 or Sj mod Si = 0.So if the input is like [1, 2, 3], the possible result may come like [1, 2] or [1, 3]To solve this, we will follow these steps −Create an array ret, set endpoint := 0, retLen := 1, n := size of numsif n is 0, then return empty setsort nums arraycreate two arrays len and par of size n, initialize ...

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Linked List Random Node in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 462 Views

Suppose we have a singly linked list, we have to find a random node's value from the linked list. Here each node must have the same probability of being chosen. So for example, if the list is [1, 2, 3], then it can return random node in range 1, 2, and 3.To solve this, we will follow these steps −In the getRandom() method, do the following −ret := -1, len := 1, v := xwhile v is not nullif rand() is divisible by len, then ret := val of vincrease len by 1v := next of vreturn retExample(C++)Let us see ...

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Elimination Game in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 960 Views

Suppose we have a list of sorted integers from 1 to n. That is starting from left and ending at right, we have to remove the first number and every other number afterward until we reach the end of the list. We will repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers. We will repeat the steps again, alternating left to right and right to left, until one single number remains. We have to find the last number that remains starting with a list ...

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UTF-8 Validation in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 3K+ Views

Suppose we have a list integers representing the data. We have to check whether it is valid UTF-8 encoding or not. One UTF-8 character can be 1 to 4-byte long. There are some properties −For 1-byte character, the first bit is a 0, followed by its unicode code.For n-bytes character, the first n-bits are all 1s, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.So the encoding technique is as follows −Character Number RangeUTF-8 octet sequence0000 0000 0000 007F0xxxxxxx0000 0080 0000 07FF110xxxxx 10xxxxxx0000 0800 0000 FFFF1110xxxx 10xxxxxx 10xxxxxx0001 0000 0010 FFFF11110xxx 10xxxxxx 10xxxxxx ...

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Rotate Function in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 420 Views

Suppose we have Given an array of integers A and let n is the length of array A. Now assume Bk to be an array obtained by rotating the array A, k positions clock-wise. Here the rotation can be defined as −F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].Now find the maximum value of F(0), F(1), ..., F(n-1).So if the input is like A = [4, 3, 2, 6], then −F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 ...

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Integer Replacement in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 798 Views

Suppose we have a positive integer n and we can do these operations as follow −If n is even, replace n with n/2.If n is odd, you can replace n with either n + 1 or n - 1.We have to find the minimum number of replacements needed for n to become 1?So if the number is 7, then the answer will be 4, as 7 → 8 → 4 → 2 → 1 or 7 → 6 → 3 → 2 → 1To solve this, we will follow these steps −ret := 0, n := xwhile n > 1if ...

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