Server Side Programming Articles

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Count number of occurrences (or frequency) in a sorted array in C++

Sunidhi Bansal
Sunidhi Bansal
Updated on 11-Mar-2026 2K+ Views

We are given a sorted array of integer type elements and the number let’s say, num and the task is to calculate the count of the number of times the given element num is appearing in an array.Input − int arr[] = {1, 1, 1, 2, 3, 4}, num = 1Output − Count of number of occurrences (or frequency) in a sorted array are − 3Input − int arr[] = {2, 3, 4, 5, 5, 6, -7}, num = 5Output − Count of number of occurrences (or frequency) in a sorted array are − 2Input − int arr[] = {-1, ...

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Flipped Matrix Prequel in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 340 Views

Suppose we have one binary matrix. We have to find the maximum number of 1s we can get if we flip a row and then flip a column.So, if the input is like101010100then the output will be 8To solve this, we will follow these steps −n := size of rows in matrixm := size of columns in matrixret := 0Define an array row of size nDefine an array col of size ntotal := 0for initialize i := 0, when i < n, update (increase i by 1), do −for initialize j := 0, when j < m, update (increase j ...

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Program to find equation of a plane passing through 3 points in C++

Ayush Gupta
Ayush Gupta
Updated on 11-Mar-2026 493 Views

In this tutorial, we will be discussing a program to find equation of a plane passing through 3 points.For this we will be provided with 3 points. Our task is to find the equation of the plane consisting of or passing through those three given points.Example#include #include #include #include using namespace std; //finding the equation of plane void equation_plane(float x1, float y1, float z1, float x2, float y2, float z2, float x3, float y3, float z3){    float a1 = x2 - x1;    float b1 = y2 - y1;    float c1 = z2 - z1; ...

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Revolving Door in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 393 Views

Suppose we have a list of requests, where requests[i] contains [t, d] indicating at time t, a person arrived at the door and either wanted to go inside (inside is indicating using 1) or go outside (outside is indicating using 0).So if there is only one door and it takes one time unit to use the door, there are few rules that we have to follow −The door starts with 'in' position and then is set to the position used by the last participant.If there's only one participant at the door at given time t, they can use the door.If ...

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Unique Fractions in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 349 Views

Suppose we have a list of fractions where each fraction contains [numerator, denominator] (numerator / denominator). We have ti find a new list of fractions such that the numbers in fractions are −In their most reduced terms. (20 / 14 becomes 10 / 7).Any duplicate fractions (after reducing) will be removed.Sorted in ascending order by their actual value.If the number is negative, the '-' sign will be with the numerator.So, if the input is like {{16, 8}, {4, 2}, {7, 3}, {14, 6}, {20, 4}, {-6, 12}}, then the output will be [[-1, 2], [2, 1], [7, 3], [5, 1]]To ...

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Program to find first N Iccanobif Numbers in C++

Ayush Gupta
Ayush Gupta
Updated on 11-Mar-2026 171 Views

In this tutorial, we will be discussing a program to find N lccanobif numbers.For this we will be provided with an integer. Our task is to find the lccanobif number at that position. They are similar to the fibonacci number except the fact that we add the previous two numbers after reversing their digits.Example#include using namespace std; //reversing the digits of a number int reverse_digits(int num){    int rev_num = 0;    while (num > 0) {       rev_num = rev_num * 10 + num % 10;       num = num / 10;    } ...

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Inverted Subtree in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 248 Views

Suppose we have two binary trees called source and target; we have to check whether there is some inversion T of source such that it is a subtree of the target. So, it means there is a node in target that is identically same in values and structure as T including all of its descendants.As we know a tree is said to be an inversion of another tree if either −Both trees are emptyIts left and right children are optionally swapped and its left and right subtrees are inversions.So, if the input is like sourceTargetthen the output will be TrueTo ...

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Minimum Size of Two Non-Overlapping Intervals in C++

Arnab Chakraborty
Arnab Chakraborty
Updated on 11-Mar-2026 387 Views

Suppose we have a list of intervals where each interval contains the [start, end] times. We have to find the minimum total size of any two non-overlapping intervals, where the size of an interval is (end - start + 1). If we cannot find such two intervals, return 0.So, if the input is like [[2, 5], [9, 10], [4, 6]], then the output will be 5 as we can pick interval [4, 6] of size 3 and [9, 10] of size 2.To solve this, we will follow these steps −ret := infn := size of vsort the array v based ...

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Program to find GCD of floating point numbers in C++

Ayush Gupta
Ayush Gupta
Updated on 11-Mar-2026 576 Views

In this tutorial, we will be discussing a program to find GCD of floating point numbers.For this we will be provided with two integers. Our task is to find the GCD (Greatest common divisor) of those two provided integers.Example#include using namespace std; //returning GCD of given numbers double gcd(double a, double b){    if (a < b)       return gcd(b, a);    if (fabs(b) < 0.001)       return a;    else       return (gcd(b, a - floor(a / b) * b)); } int main(){    double a = 1.20, b = 22.5;    cout

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Program to find greater value between a^n and b^n in C++

Ayush Gupta
Ayush Gupta
Updated on 11-Mar-2026 214 Views

In this tutorial, we will be discussing a program to find greater value between a^n and b^n.For this we will be provided with three numbers. Our task is to calculate a^n and b^n and return back the greater of those values.Example#include using namespace std; //finding the greater value void findGreater(int a, int b, int n){    if (!(n & 1)) {       a = abs(a);       b = abs(b);    }    if (a == b)       cout b)       cout

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