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Convert all lowercase characters to uppercase whose ASCII value is co-prime with k in C++

Ayush Gupta
Ayush Gupta
Updated on 11-Mar-2026 184 Views

In this tutorial, we will be discussing a program to convert all lowercase characters to uppercase whose ASCII value is co-prime with k.For this we will be provided with a string and an integer value k. Our task is to traverse through the given string and change to uppercase all those characters whose ASCII value is co-prime with the given integer k.Example#include using namespace std; //modifying the given string void convert_string(string s, int k){    int l = s.length();    for (int i = 0; i < l; i++) {       int ascii = (int)s[i];     ...

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Gas Station in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 720 Views

Suppose there is a circle, and there are n gas stations on the circle. We have two sets of data like −The amount of gas that every gas stations hasDistance from one gas stations to another.Calculate the first point, from where a car will be able to complete the circle. Assume for 1 unit of gas, the car can go 1 unit of distance. Suppose there are four gas stations, and the amount of gas, and distance from the next gas stations is as like [(4, 6), (6, 5), (7, 3), (4, 5)], the first point from where car can ...

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Evaluate Reverse Polish Notation in C++ Program

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 6K+ Views

Suppose we have Reverse polish notation and we have to evaluate the value. The reverse polish notation is also known as postfix expression. Here we have to use the stack data structure to solve the postfix expressions.From the postfix expression, when some operands are found, pushed them in the stack. When some operator is found, two items are popped from stack and then the operation is performed in the correct sequence. After that, the result is also pushed in the stack for future use. After completing the whole expression, the final result is also stored in the stack top. So ...

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Letter Case Permutation in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 348 Views

Suppose we have a string with letters and numbers. We have to generate all possible combinations of that string by taking uppercase and lowercase versions of letters that are present in the string. So if one string has only numbers, only that will be returned. Suppose the string is like “1ab2”, then the strings will be [“1ab2”, “1Ab2”, “1aB2”, “1AB2”]To solve this problem, we will use recursive approach. It takes the index parameter to start work from that index. It also takes a temp string up to which the result is created. When the index is same as the string ...

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Print All Distinct Elements of a given integer array in C++

sudhir sharma
sudhir sharma
Updated on 11-Mar-2026 932 Views

In this problem, we are given an array of integer values. Our task is to print all distinct elements of the array. The output should contain only distinct values.Let’s take an example to understand the problemInput: array = {1, 5, 7, 12, 1, 6, 10, 7, 5} Output: 1 5 7 12 6 10To solve this problem, we will have to check elements of the array for uniqueness. For this, we will use two nested loops, the outer one will take values and the inner one will check the rest of the values with it. If more than one values ...

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jQuery Traversing Siblings

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 243 Views

With jQuery, you can easily find siblings of an element using the following methods: next(), nextAll(), prev(), prevAll(), siblings(), etc. Let us see some of them traverse siblings−next() methodThe next() method is used to return the next sibling element of the selected element. Let us see an example−Example div {    width:600px; } .demo * {    display: block;    border: 2px solid orange;    color: blue;    padding: 10px;    margin: 10px; }    $(document).ready(function(){       $("h3").next().css({"color": "gray", "border": "3px dashed blue"});    }); parent sibling sibling ...

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Rotate String in Python

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 611 Views

Suppose we have two strings, A and B. We will rotate the string A and check whether it matches with B at any position of rotating, if so then return true, otherwise false. For example, if A = 'abcde', and B = 'bcdea' So answer will be true, as A can be converted to B after rotating it.To solve this, we will follow these steps −When both A and B are empty, then return true, when both are of different length then return falseA := concatenate A after Ai := 0, and j := 0while i < length of Aif ...

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Find N integers with given difference between product and sum in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 222 Views

Suppose we have two integers N and D. We have to find a set of N integers, where the difference between their sum and product is the same as D. Suppose the N = 3, and D = 5, then the output will be 1, 2, 8. Here the sum is 1 + 2 + 8 = 11, and product is 1 * 2 * 8 = 16, the difference between 16 and 11 is 5.We have to solve this problem; we will use one tricky method. Here we will try to find N–2 number of 1s, one 2, and ...

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Print all distinct circular strings of length M in lexicographical order in C++

sudhir sharma
sudhir sharma
Updated on 11-Mar-2026 283 Views

In this problem, we are given a string and an integer M. our task is to print all distinct circular strings of length M in lexicographical order (alphabetical order).Let’s take an example to understand the problem, Input: str= “ssssn” M=3 Output: nss sns ssn sssExplanation − all possible circular strings of length 3 are: sss sss ssn sns nss. Distinct elements in lexicographical order are sss ssn sns nss.To solve this problem, we will iterate over the elements of the string and generate all possible substrings of length M. we will store this generated string in a set that stores ...

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Find n-th element from Stern's Diatomic Series in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 199 Views

Here we will see how to find the nth term in Stern’s Diatomic series. The series is like 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, … This is also known as fusc function. This series can be defined as −𝑝(𝑛)=$p\lgroup\frac{n}{2}\rgroup$ 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛𝑝(𝑛)=$p\lgroup\frac{n-1}{2}\rgroup+p\lgroup\frac{n+1}{2}\rgroup$ 𝑤ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑜𝑑𝑑𝑝(0)=0 𝑎𝑛𝑑 𝑝(1)=1Here we will use the Dynamic programming approach to reduce the number of computations. After saving the base case for p(0) and p(1), we will iterate from index i = 2 to n, and compute p(i)Example#include using namespace std; int findTerm(int n) ...

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