C++ Articles

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Product of first N factorials in C++

Sunidhi Bansal
Sunidhi Bansal
Updated on 11-Mar-2026 442 Views

Given a number N, the task is to find the product of first N factorials modulo by 1000000007. . Factorial implies when we find the product of all the numbers below that number including that number and is denoted by ! (Exclamation sign), For example − 4! = 4x3x2x1 = 24.So, we have to find a product of n factorials and modulo by 1000000007..Constraint 1 ≤ N ≤ 1e6.Input n = 9Output 27Explanation 1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! Mod (1e9 + 7) = 27Input n = 3Output 12Explanation 1! * 2! * 3! mod (1e9 ...

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Write a C program that won't compile in C++

sudhir sharma
sudhir sharma
Updated on 11-Mar-2026 330 Views

Here, we will write some c programs that won’t compile in c++. Though c++ is considered as a successor of c that has all its features and has compatibility with c code there are some programs that won’t compiler or given a compilation error when compiled with c++ compiler.A list of some of the C programs that won’t compile in c++ are −Call to a function before declaration − In c++, function call before declaration gives compilation error. But this works fine in c.Example#include int main(){    printHello();    return 0; } void printHello(){    printf("TutorialsPoint"); }OutputTutorialsPointUsing typecasted pointers ...

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Shortest Path to Get All Keys in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 640 Views

Suppose we have a grid. There are few symbols. "." is indicating empty cell, "#" is for wall, "@" is for starting point, ("a", "b", ...) all are keys, and ("A", "B", ...) all are locks. We will start from the starting point, and one move consists of walking one space in one of the 4 directions (left, right, top, bottom). We will not go outside the grid, and there are walls to block our way. If we walk over a key, we pick it up. We can't walk over a lock unless we have the corresponding key.For each lock ...

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Verbal Arithmetic Puzzle in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 561 Views

Suppose we have an equation, expressions are represented by words on left side and the result on right side. We have to check whether the equation is solvable under the following rules or not −Each character is decoded as one digit (0 to 9).Every pair of different characters must map to different digits.Each words[i] and result are decoded as a number where no leading zeros are present.Sum of numbers on left side will equal to the number on right side.We will check whether the equation is solvable or not.So, if the input is like words = ["SEND", "MORE"], result = ...

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Find All Good Strings in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 515 Views

Suppose we have two strings s1 and s2. The size of these strings is n, and we also have another string called evil. We have to find the number of good strings.A string is called good when its size is n, it is alphabetically greater than or equal to s1, it is alphabetically smaller than or equal to s2, and it has no evil as a substring. The answer may be very large, so return the answer modulo 10^9 + 7.So, if the input is like n = 2, s1 = "bb", s2 = "db", evil = "a", then the ...

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Minimum Time Difference in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 517 Views

Suppose we have a list of 24-hour clock time points in "Hour:Minutes" format, we have to find the minimum minutes difference between any two time points in the list. So if the input is like [“12:30”, ”15:17”], so this will return 167.To solve this, we will follow these steps −Define an array called ok of size 24*60 + 1, and initially all are false.n := size of tpfor i in range 0 to n – 1hr := the hour part from the timemin := minute part of the stringtime := hr * 60 + minif ok[time] is true, then return ...

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Optimal Division in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 372 Views

Suppose we have a list of positive integers; the adjacent integers will perform the float division. So for example, [2, 3, 4] -> 2 / 3 / 4. Now, we can add any number of parenthesis at any position to change the priority of these operations. We should find out how to add parenthesis to get the maximum result, we have to find the corresponding expression in string format. Our expression should NOT contain redundant parenthesis. So if the input is like [1000, 100, 10, 2], then the result will be “1000/(100/10/2)”.To solve this, we will follow these steps −n ...

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Next Greater Element III in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 324 Views

Suppose we have a positive 32-bit integer n, we need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If we have no such positive 32-bit integer number, then return -1.So if the number is 213, then the result will be 231.To solve this, we will follow these steps −s := n as string, sz := size of s, ok := falsefor i in range sz – 2 to 0if s[i] < s[i + 1], then ok := true and break the loopif of is ...

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Array Nesting in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 644 Views

Suppose we have a zero-indexed array A of length N that contains all integers from 0 to N-1. We have to find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below. Now consider the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S. So if the array is like A = [5, 4, 0, 3, 1, 6, ...

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Valid Triangle Number in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 684 Views

Suppose we have an array consists of non-negative integers, our task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. So if the input is like [2, 2, 3, 4], then the result will be 3 as [2, 3, 4] using first 2, [2, 3, 4] using second 2, and [2, 2, 3].To solve this, we will follow these steps −ret := 0, n := size of nums, sort numsfor i in range n – 1 down to 0right := i – 1, left ...

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